{"id":35430,"date":"2022-09-14T09:48:48","date_gmt":"2022-09-14T04:18:48","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35430"},"modified":"2022-09-14T09:48:48","modified_gmt":"2022-09-14T04:18:48","slug":"ap-inter-1st-year-chemistry-study-material-chapter-7","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-inter-1st-year-chemistry-study-material-chapter-7\/","title":{"rendered":"AP Inter 1st Year Chemistry Study Material Chapter 7 Chemical Equilibrium and Acids-Bases"},"content":{"rendered":"

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material<\/a> 7th Lesson Chemical Equilibrium and Acids-Bases Textbook Questions and Answers.<\/p>\n

AP Inter 1st Year Chemistry Study Material 7th Lesson Chemical Equilibrium and Acids-Bases<\/h2>\n

Very Short Answer Questions<\/span><\/p>\n

Question 1.
\nState law of chemical equilibrium.
\nAnswer:
\nThe ratio of product of molar concentrations of the products to the product of molar concentrations of the reactants has a constant value. This is known as the equilibrium law (or) law of chemical equilibrium.<\/p>\n

Question 2.
\nCan equilibrium be achieved between water and its vapours in an open vessel. Explain.
\nAnswer:
\nEquilibrium does not acheived between water and its vapour in an open vessel.
\nBut the equilibrium achieved between water and its vapour in a closed vessel.
\nH2<\/sub>Ol)<\/sub> \u21cc H2<\/sub>O(g)<\/sub><\/p>\n

\"AP<\/p>\n

Question 3.
\nWhy the concentrations of pure liquids and pure solids are ignored from equilibrium constant expressions ?
\nAnswer:
\nThe concentrations of pure solids and pure liquids are ignored from equilibrium constant expres-sions because the concentrations of pure solids and pure liquids are taken as unity.<\/p>\n

Question 4.
\nWhat is homogenous equilibrium ? Write two homogenous reactions.
\nAnswer:
\nIt the physical states of the participating substances are same they the equilibrium is homogeneous equilibrium
\ne.g.: H2(g)<\/sub> + I2(g)<\/sub> \u21cc 2HI(g)<\/sub>)
\nN2(g)<\/sub> + 3H2(g)<\/sub> \u21cc 2NH3(g)<\/sub><\/p>\n

Question 5.
\nWhat is heterogeneous equilibrium? Write two heterogeneous reactions.
\nAnswer:
\nIf the physical states of all (or) some of the participating substances are same then the equilibrium is Heterogeneous equilibrium.
\n\"AP<\/p>\n

Question 6.
\nWrite reaction quotient, Q, for each of the following reactions.
\na) 3O2<\/sub>(g) \u21cc 2O3<\/sub>(g)
\nb) 4NH3<\/sub>(g) + 7O2<\/sub>(g) \u21cc 4NO2<\/sub>(g) + 6H2<\/sub>O(g)
\nAnswer:
\na) 3O2(g)<\/sub> \u21cc 2O3(g)<\/sub>
\nQ = \\(\\frac{\\left(\\mathrm{O}_3\\right)_{\\mathrm{t}}^2}{\\left(\\mathrm{O}_2\\right)_{\\mathrm{t}}^2}\\)<\/p>\n

b) 4NH3(g)<\/sub> + 7O2(g)<\/sub> \u21cc 4NO2(g)<\/sub> + 6H2<\/sub>O(s)<\/sub>
\nQ = \\(\\frac{\\left[\\mathrm{NO}_2\\right]_t^4\\left[\\mathrm{H}_2 \\mathrm{O}\\right]_t^6}{\\left[\\mathrm{NH}_3\\right]_t^4\\left[\\mathrm{O}_2\\right]_t^7}\\)<\/p>\n

\"AP<\/p>\n

Question 7.
\nDefine equilibrium constant.
\nAnswer:
\nThe ratio of product of molar concentrations of products to the product of molar concentrations of reactants is called equilibrium constant (Kc<\/sub>)<\/p>\n

Question 8.
\nThe equilibrium constant expression for a gas reaction is Kc<\/sub> = \\(\\frac{\\left[\\mathrm{NH}_3\\right]^4\\left[\\mathrm{O}_2\\right]^5}{[\\mathrm{NO}]^4\\left[\\mathrm{H}_2 \\mathrm{O}\\right]^6}\\) Write the balanced chemical equation corresponding to this expression.
\nAnswer:
\nGiven Kc<\/sub> = \\(\\frac{\\left[\\mathrm{NH}_3\\right]^4\\left[\\mathrm{O}_2\\right]^5}{[\\mathrm{NO}]^4\\left[\\mathrm{H}_2 \\mathrm{O}\\right]^6}\\)
\nBalanced chemical equation is
\n4NO + 6H2<\/sub>O \u21cc 4NH3<\/sub> + 5O2<\/sub><\/p>\n

Question 9.
\nWrite the relation between Kp<\/sub> and Kc<\/sub>.
\nAnswer:
\nKp<\/sub> = Kc<\/sub>(RT)\u2206n<\/sup>
\n\u2206n = No. of gaseous products – No. of gaseous reactants
\nR = Universal gas constant<\/p>\n

Question 10.
\nUnder what conditions for a reaction Kp<\/sub> and Kc<\/sub> are numerically equal?
\nAnswer:
\nIf \u2206n = 0 i.e. No. of gaseous products = No. of gaseous reactants
\nthen Kp<\/sub> = Kc<\/sub> (RT)\u2206n<\/sup>
\nKp<\/sub> = Kc<\/sub>(RT)0<\/sup>
\nthen Kp<\/sub> = Kc<\/sub><\/p>\n

Question 11.
\nGive two chemical equilibrium reactions for which Kp<\/sub> = Kc<\/sub>
\nAnswer:<\/p>\n

    \n
  1. H2(g)<\/sub> + I2(g)<\/sub> \u21cc 2 HI(g)<\/sub>
    \nHere \u2206n = 0
    \n\u2234 Kp<\/sub> = Kc<\/sub><\/li>\n
  2. N2(g)<\/sub> + O2(g)<\/sub> \u21cc 2NO(g)<\/sub>
    \nHere \u2206n = 0
    \n\u2234 Kp<\/sub> = Kc<\/sub><\/li>\n<\/ol>\n

    \"AP<\/p>\n

    Question 12.
    \nGive two chemical equilibrium reactions for which Kp<\/sub> > Kc<\/sub>.
    \nAnswer:
    \n1) 2NH3(g)<\/sub> \u21cc N2(g)<\/sub> + 3H2(g)<\/sub>
    \n\u2206n = np<\/sub> – nR<\/sub> = (3 + 1) – 2 = 2
    \n\u2206n = 2 (+ve)
    \n\u2234 Kp<\/sub> > Kc<\/sub><\/p>\n

    2) 2SO3(g)<\/sub> \u21cc 2SO2(g)<\/sub> + O2(g)<\/sub>
    \n\u2206n = np<\/sub> – nR<\/sub> = (2 + 1) – 2 = 1
    \n\u2206n = 1 (+ve)
    \n\u2234 Kp<\/sub> > Kc<\/sub><\/p>\n

    Question 13.
    \nGive two chemical equilibrium reactions for which Kp<\/sub> < Kc<\/sub>
    \nAnswer:
    \n1) N2(g)<\/sub> + 3H2(g)<\/sub> \u21cc 2NH3(g)<\/sub>
    \n\u2206n = np<\/sub> – nR<\/sub> = 2 – (1 + 3) = -2
    \n\u2206n = -2 (- ve)
    \n\u2234 Kp<\/sub> < Kc<\/sub><\/p>\n

    2) 2SO2(g)<\/sub> + O2(g)<\/sub> \u21cc 2SO3(g)<\/sub>
    \n\u2206n = np<\/sub> – nR<\/sub>
    \n= 2 – (1 + 2) = -1
    \n\u2206n = -1 (-ve)
    \n\u2234 Kp<\/sub> < Kc<\/sub><\/p>\n

    Question 14.
    \nWrite the equations for the conversion of Kc<\/sub> to Kp<\/sub> for each of the following reactions.
    \na) CO(g) + H2<\/sub>O (g) \u21cc CO2<\/sub> (g) + H2<\/sub> (g)
    \nb) C3<\/sub>H8<\/sub>(g) + 5O2<\/sub>(g) \u21cc 3CO2<\/sub> (g) + 4H2<\/sub>O (g)
    \nAnswer:
    \na) CO(g)<\/sub> + H2<\/sub>O(g)<\/sub> \u21cc CO2(g)<\/sub> + H2(g)<\/sub>
    \nKp<\/sub> = \\(\\frac{\\mathrm{P}_{\\mathrm{CO}_2} \\times \\mathrm{P}_{\\mathrm{H}_2}}{\\mathrm{P}_{\\mathrm{CO}} \\times \\mathrm{P}_{\\mathrm{H}_2 \\mathrm{O}}}\\)
    \nKc<\/sub> = \\(\\frac{\\left[\\mathrm{CO}_2\\right]\\left[\\mathrm{H}_2\\right]}{[\\mathrm{CO}]\\left[\\mathrm{H}_2 \\mathrm{O}\\right]}\\)
    \nKc<\/sub> to Kp<\/sub> conversion<\/p>\n

    b) C3<\/sub>H8<\/sub>(g) + 5O2<\/sub>(g) \u21cc 3CO2<\/sub> (g) + 4H2<\/sub>O (g)
    \nKp<\/sub> = \\(\\frac{\\mathrm{P}_{\\mathrm{CO}_2}^3 \\times \\mathrm{P}_{\\mathrm{H}_2 \\mathrm{O}^4}}{\\mathrm{P}_{\\mathrm{C}_3 \\mathrm{H}_8} \\times \\mathrm{P}_{\\mathrm{O}_2}^5}\\)
    \nKc<\/sub> = \\(\\frac{\\left[\\mathrm{CO}_2\\right]^3\\left[\\mathrm{H}_2 \\mathrm{O}\\right]^4}{\\left[\\mathrm{C}_3 \\mathrm{H}_8\\right]\\left[\\mathrm{O}_2\\right]^5}\\)
    \nKc<\/sub> to Kp<\/sub> conversion<\/p>\n

    \"AP<\/p>\n

    Question 15.
    \nWhat are the factors which influence the chemical equilibrium?
    \nAnswer:
    \nThe factors influencing chemical equilibrium are<\/p>\n

      \n
    1. Concentration of reactants and the products<\/li>\n
    2. Temperature of reaction<\/li>\n
    3. Pressure of reaction<\/li>\n
    4. Inert gas addition etc.<\/li>\n<\/ol>\n

      Question 16.
      \nWhat is the effect of pressure on a gaseous chemical equilibrium?
      \nAnswer:
      \nIncrease of external pressure of the reaction at equlibrium favours the reaction in the direction in which the volume (or) the No. of molecules decreases.
      \nDecrease of external pressure of reaction at equilibrium favour the reaction in the direction in which the volume (or) no. of molecules increase.<\/p>\n

      Question 17.
      \nWhat is the effect of increase in concentration of reactants of a chemical reaction at equilibrium?
      \nAnswer:
      \nIncrease of concentration of the reactants in the reaction mixture at equilibrium favours the forward reaction (Reactants \u2192 products)<\/p>\n

      Question 18.
      \nCan catalyst disturb the state of equilibrium?
      \nAnswer:
      \nAddition of catalyst to the reaction does not effect the equilibrium position. It only speedy up the attainment of the equilibrium.<\/p>\n

      Question 19.
      \nOn which factor, the equilibrium constant value changes?
      \nAnswer:
      \nThe numerical value of equilibrium constant depends on the standard state of products and reactants chosen.<\/p>\n

      Question 20.
      \nThe equilibrium constants of a reaction at 27\u00b0 C and at 127\u00b0C are 1.6 \u00d7 10-3<\/sup> and 7.6 \u00d7 10-2<\/sup> respectively. Is the reaction exothermic or endothermic.
      \nAnswer:
      \nGiven equilibrium constants
      \nAt 27\u00b0C – 1.6 \u00d7 10-3<\/sup>
      \nAt 127\u00b0C -7.6 \u00d7 10-2<\/sup>
      \nFrom the above values as the temperature raised euilibrium constant value increased. Here \u2206H = +ve
      \n\u2234 The reaction is endothermic.<\/p>\n

      \"AP<\/p>\n

      Question 21.
      \nWhat is the effect of temperature on a system at equilibrium?
      \nAnswer:<\/p>\n

        \n
      • Increase in temperature of the reaction mixture at equilibrium favours the endothermic reaction.<\/li>\n
      • Decrease in temperature of the reaction mixture at equilibrium favours the exothermic reaction.<\/li>\n<\/ul>\n

        Question 22.
        \nFor an exothermic reaction, what happens to the equilibrium constant if temperature is raised?
        \nAnswer:
        \nFor an exothermic reaction when the temperature is raised equilibrium constant value decreased.<\/p>\n

        Question 23.
        \nWhat kind of equilibrium constant can be calculated form \u2206G\u00b0 value for a reaction involving only gases?
        \nAnswer:
        \nMathematical expression in the thermodynamic view of equilibrium is
        \n\u2206G = \u2206G\u00b0 + RTlnQ
        \nAt equilibrium \u2206G = 0 and Q = K
        \n\u2234 \u2206G = \u2206G\u00b0 + RTlnK = 0
        \n\u2206G\u00b0 = -RTlnK
        \nknK = \\(\\frac{-\\Delta G^{\\circ}}{R}\\)
        \nK = e-\u2206G\u00b0\/RT<\/sup>
        \nK = Equilibrium constant using the above equation the sponteinity of reaction can be predicted.<\/p>\n

        Question 24.
        \nWhat is a Bronsted base ? Give one example.
        \nAnswer:
        \nThe substance which accepts a proton from the other substance is called Bronsted base
        \ne.g.: NH3<\/sub>, H2<\/sub>O etc.<\/p>\n

        Qeustion 25.
        \nWhat is Lewis acid ? Give one example.
        \nAnswer:
        \nA substance which can accept an electron pair to from a co-ordinate covalent bond with donor is called Lewis acid.
        \ne.g. : H+<\/sup>, BF3<\/sub>, SnCl2<\/sub> etc.<\/p>\n

        Question 26.
        \nWhat is meant by ionic product of water ?
        \nAnswer:
        \nAt a given temp, the product of the concentrations of H+<\/sup> and OH–<\/sup> ions in water is called ionic product
        \nIonic product Kw<\/sub> = [H+<\/sup>] [OH–<\/sup>]
        \nAt 25\u00b0C Kw<\/sub> = 1.008 \u00d7 10-14<\/sup> mole2<\/sup>\/lit2<\/sup><\/p>\n

        \"AP<\/p>\n

        Question 27.
        \nWhat is the value of Kw<\/sub>? What are its units ?
        \nAnswer:
        \nAt 25\u00b0C Ionic product of water Kw<\/sub> = 1.008 \u00d7 10-14<\/sup> mole2<\/sup>\/lit2<\/sup>
        \nUnits : mole2<\/sup>\/lit2<\/sup><\/p>\n

        Question 28.
        \nWhat is the effect of temperature on ionic product of water ?
        \nAnswer:
        \nAs the temperature increases ionisation of water increases and hence Ionic product alue (Kw<\/sub>) also increases.<\/p>\n

        Question 29.
        \nH2<\/sub>O + H2<\/sub>O \u21cc H2<\/sub>O+<\/sup> + OH–<\/sup>
        \nThe ionic product of water is 1 \u00d7 10-14<\/sup> at 25\u00b0C and 3.0 \u00d7 10-14<\/sup> at 40\u00b0C.
        \nIs the above process endothermic or exothermic ?
        \nAnswer:
        \nGiven reaction
        \nH2<\/sub>O + H2<\/sub>O \u21cc H2<\/sub>O+<\/sup> + OH–<\/sup>
        \nKwat25\u00b0C – 1 \u00d7 10-14<\/sup> mole2<\/sup>\/lit2<\/sup>
        \nKwat40\u00b0C – 3 \u00d7 10-14<\/sup> mole2<\/sup>\/lit2<\/sup>
        \nThe above reaction is endo thermic reaction because Kw<\/sub> value increases with increase of temperature.<\/p>\n

        Question 30.
        \nAll Bronsted bases are Lewis bases. Explain.
        \nAnswer:
        \nBronsted base means proton acceptor and lewis base means electron pair donor.
        \nBut according to two theories base provides a lone pair.
        \n\u2234 All bronsted bases are lewisbases<\/p>\n

        Question 31.
        \nAll Lewis acids are not Bronsted acids. Why ?
        \nAnswer:<\/p>\n

          \n
        • Lewis acid means electron pair acceptor and bronsted acid means proton donar.<\/li>\n
        • According to Lewis concept many acids donot have proton.
          \nSo all Lewis acids are not Bronsted bases.<\/li>\n<\/ul>\n

          \"AP<\/p>\n

          Question 32.
          \nWhat is degree of ionization?
          \nAnswer:<\/p>\n

            \n
          • It is the expression used to find the extent of ionisation of electrolytes.<\/li>\n
          • It is devoted by ‘\u03b1’.<\/li>\n<\/ul>\n

            Question 33.
            \nWhat is the measure of strength of an acid and base ?
            \nAnswer:<\/p>\n

              \n
            • Strength of an acid can be measured by the acid disociation constant (Ka<\/sub>)<\/li>\n
            • Strength of a base can be measured by the base disaciating constant (kb<\/sub>)<\/li>\n<\/ul>\n

              Question 34.
              \nGive two examples of salts whose aqueous solutions are basic.
              \nAnswer:
              \nSodium acetate (CH3<\/sub>COONa), Sodium carbonate (Na2<\/sub>CO3<\/sub>) salts aqueous solutions are basic in nature due to anionic hydrolysis.<\/p>\n

              Question 35.
              \nGive two examples of salts whose aqueous solutions are acidic.
              \nAnswer:
              \nAmmonium chloride (NH4<\/sub>Cl), Ammonium sulphate [(NH4<\/sub>)2<\/sub> SO4<\/sub>] salts aqueous solutions are acidic in nature.<\/p>\n

              Question 36.
              \nWhat equation is used for calculating the pH of an acid buffer ?
              \nAnswer:
              \nHendersen’s equation is used to calculate the PH<\/sup> of an acid buffer
              \nPH<\/sup> = PKa + log \\(\\frac{\\text { [Salt] }}{[\\text { Acid] }}\\)<\/p>\n

              \"AP<\/p>\n

              Question 37.
              \nPhosphoric acid (H3<\/sub>PO4<\/sub>) have three ionization constants Ka1<\/sub>, Ka2<\/sub> and Ka3<\/sub>. Among these ionization constants which has a lower value ? Give reason for it ? Answer:
              \nPhosphoric acid (H3<\/sub>PO4<\/sub>) has
              \nKa1<\/sub> – 7.5 \u00d7 10-3<\/sup>
              \nKa2<\/sub> – 6.2 \u00d7 10-8<\/sup>
              \nKa3<\/sub> – 4.2 \u00d7 10-13<\/sup>
              \nKa3<\/sub> has lower value.
              \nReason :<\/p>\n

                \n
              • It is more difficult to remove a proton from the negativeion due to electro static forces.<\/li>\n
              • It is difficult to remove a proton from \\(\\mathrm{HPO}_4^{-2}\\) anion.<\/li>\n<\/ul>\n

                Question 38.
                \nIce melts slowly at high altitudes. Explain. Why ?
                \nAnswer:
                \nIce melts slowly at high altitudes
                \nReason :<\/p>\n

                  \n
                • Ice and water are equilibrium only at a particular temperature and pressure conditions.<\/li>\n
                • At high attitude temp and pressure conditions are changed.<\/li>\n<\/ul>\n

                  Short Answer Questions<\/span><\/p>\n

                  Qeustion 1.
                  \nWrite expression for the equilibrium constant, Kc<\/sub> for each of the following reactions.
                  \n(i) 2NOCl(g)<\/sub> \u21cc 2NO(g)<\/sub> + Cl2(g)<\/sub>
                  \n(ii) 2Cu(NO3<\/sub>)2(s)<\/sub> \u21cc 2CUO(s)<\/sub> + 4NO2(g)<\/sub> + O2(g)<\/sub>
                  \n(iii) CH3<\/sub>COOC2<\/sub>H5(aq)<\/sub> + H2<\/sub>O(l) \u21cc CH3<\/sub>COOH(aq)<\/sub> + C2<\/sub>H5<\/sub>OH(aq)<\/sub>
                  \n(iv) Fe+3<\/sup>(aq)<\/sub> + 3OH–<\/sup>(aq)<\/sub> \u21cc Fe(OH)3(s)<\/sub>
                  \nAnswer:
                  \ni) Given equation
                  \n2NOCl(g)<\/sub> \u21cc 2NO(g)<\/sub> + Cl2(g)<\/sub>
                  \nEquilibrium constant Kc<\/sub> = \\(\\frac{\\left[\\mathrm{NO}^2\\left[\\mathrm{Cl}_2\\right]\\right.}{[\\mathrm{NoCl}]}\\)
                  \nii) Given equation
                  \n2Cu(NO3<\/sub>)2(s)<\/sub> \u21cc 2CUO(s)<\/sub> + 4NO2(g)<\/sub> + O2<\/sub>
                  \nKc<\/sub> = (NO2<\/sub>)4<\/sup> (O2<\/sub>)<\/p>\n

                  iii) CH3<\/sub>COOC2<\/sub>H5(aq)<\/sub> + H2<\/sub>O(l) \u21cc CH3<\/sub>COOH(aq)<\/sub> + C2<\/sub>H5<\/sub>OH(aq)<\/sub>
                  \nKc<\/sub> = \\(\\frac{\\left[\\mathrm{CH}_3 \\mathrm{COOH}\\right]\\left[\\mathrm{C}_2 \\mathrm{H}_5 \\mathrm{OH}\\right]}{\\left[\\mathrm{CH}_3 \\mathrm{COO} \\mathrm{C}_2 \\mathrm{H}_5\\right]}\\)<\/p>\n

                  iv) Fe+3<\/sup>(aq)<\/sub> + 3OH–<\/sup>(aq)<\/sub> \u21cc Fe(OH)3(s)<\/sub>
                  \nKc<\/sub> = \\(\\frac{1}{\\left[\\mathrm{Fe}^{+3}\\right]\\left[\\mathrm{OH}^{-}\\right]^3}\\)<\/p>\n

                  \"AP<\/p>\n

                  Question 2.
                  \nDerive the relation between Kp<\/sub> and Kc<\/sub> for the equilibrium reaction
                  \nN2<\/sub>(g) + 3H2<\/sub>(g) \u21cc 2NH3<\/sub>(g) [Mar. 13]
                  \nAnswer:
                  \nGiven equation (Equilibrium reaction)
                  \nN2(g)<\/sub> + 3H2(g)<\/sub> \u21cc 2NH3(g)<\/sub>
                  \nWe can make the following generalisation concerning the composition of equilibrium mixtures : If Kc<\/sub> > 103<\/sup> products predominate over reactants, i.e., if Kc<\/sub> is very large, the reaction proceeds nearly to completion. Consider the following examples :
                  \na) The reaction of H2<\/sub> with O2<\/sub> at 500 K has a very large equilibrium constant Kc<\/sub> = 2.4 \u00d7 1047<\/sup>
                  \nb) H2(g)<\/sub> + Cl2(g)<\/sub> \u21cc 2Hl(g) at 300 K has Kc<\/sub> = 4.0 \u00d7 1031<\/sup>
                  \nc) H2(g)<\/sub> + Br2(g)<\/sub> \u21cc 2HBr(g)<\/sub> at 300 K. Kc<\/sub> = 5.4 \u00d7 1018<\/sup>
                  \nIf Kc<\/sub> < 10-3<\/sup> reactants predominate over products i.e, if Kc<\/sub> is very small the reaction proceeds rarely.<\/p>\n

                  Consider the following examples :
                  \na) The decomposition of H2<\/sub>O into H2<\/sub> and O2<\/sub> at 500 K has a very small equilibrium constant.
                  \nKc<\/sub> = 4.1 \u00d7 10-48<\/sup><\/p>\n

                  b) N2(g)<\/sub> + O2(g)<\/sub> \u21cc 2NO(g)<\/sub>
                  \nat 298 K has Kc<\/sub> = 4.8 \u00d7 10-31<\/sup>
                  \nIf Kc<\/sub> is in the range of 10-3<\/sup> to 103<\/sup> appreciable concentrations of both reactants and products are present. Consider the following examples :
                  \na) For reaction of H2<\/sub> with I2<\/sub> to give HI.
                  \n\"AP
                  \nb) Also, gas phase decomposition of N2<\/sub>O4<\/sub> to NO2<\/sub> is another reaction with a value of Kc<\/sub> = 4.64 \u00d7 10-3<\/sup> at 25\u00b0C which is neither too small nor too large. Hence equilibrium mixtures contain appreciable concentrations of both N2<\/sub>O4<\/sub> and NO2<\/sub>.
                  \nThese generarlisations are illustrated.<\/p>\n

                  Question 3.
                  \nState law of chemical equilibrium ? What is Kc for the following equilibrium when the equilibrium concentration of each substance is [SO2<\/sub>] = 0.60 M,
                  \n[O2<\/sub>] = 0.82 M and [SO3<\/sub>] = 1.90 M
                  \n2SO2<\/sub>(g) + O2<\/sub>(g) \u21cc 2SO3<\/sub>(g)
                  \nAnswer:
                  \nThe ratio of product of molar concentrations of the products to the product of molar concentrations of the reactants has a constrant value. This is known as the equilibrium law (or) law of chemical equilibrium.
                  \nGiven equation is
                  \n2SO2(g)<\/sub> + O2(g)<\/sub> \u21cc 2SO3(g)<\/sub>
                  \nGiven [SO2<\/sub>] = 0.60M, [O2<\/sub>] = 0.82M, [SO3<\/sub>] – 1.90 M
                  \nKc<\/sub> = \\(\\frac{\\left[\\mathrm{SO}_3\\right]^2}{\\left[\\mathrm{SO}_2\\right]^2\\left[\\mathrm{O}_2\\right]}=\\frac{(1.9)^2}{(0.6)^2(0.82)}\\)
                  \n= \\(\\frac{3.61}{0.36 \\times 0.86}=\\frac{3.61}{0.2952}\\)
                  \n= 12.22<\/p>\n

                  \"AP<\/p>\n

                  Question 4.
                  \nWhy sealed soda water bottle on opening shows the evolution of gas with effervescence ?
                  \nAnswer:<\/p>\n

                    \n
                  • Sealed soda bottle on opening shows evolution of gas with effervescence<\/li>\n
                  • This is due to difference in solubility of CO2<\/sub> at different pressures there is equilibrium between molecules of gaseous state and molecules of dissolved in the liquid state.
                    \nCO2(g)<\/sub> \u21cc O2<\/sub> (Jusolutcoiny)<\/li>\n<\/ul>\n

                    Question 5.
                    \nExplain the significance of
                    \na) a very large value of K,
                    \nb) a very small value of K and
                    \nc) a value of K of about 1.0
                    \nAnswer:
                    \na) A very large value of K indicates reaction proceeds almost to complete.
                    \nb) A very small value of K indicates reaction hardly proceeds.
                    \nc) A value of K about 1.0 indicates that reactants and products are at equilibrium.<\/p>\n

                    Qeustion 6.
                    \nWhy is it useful to compare Q with K ? What is the situation when
                    \n(a) Q = K
                    \n(b) Q < K (c) Q > K
                    \nAnswer:
                    \nQ and K are compared to predict the direction of reaction
                    \na) Q = K indicates that the reaction mixture is already at equilibrium
                    \nb) Q < K indicates that the reaction proceeds in the direction. c) Q > K indicates that the reaction proceeds in the direction of reactants (Reverse reaction)<\/p>\n

                    Qeustion 7.
                    \nFor the reaction Cl2<\/sub>(g) + F2<\/sub>(g) 2Cl F (g), Kc<\/sub> = 19.9
                    \nWhat will happen in a mixture originally containing [Cl2<\/sub>] = 0.4 mol L–<\/sup>; [F2<\/sub>] = 0.2 mol L-1<\/sup> and [Cl F] = 7.3 mol L–<\/sup>?
                    \nAnswer:
                    \nGiven reaction
                    \nCl2(g)<\/sub> + F2(g)<\/sub> \u21cc 2ClF(g)<\/sub>, Kc<\/sub> = 19.9
                    \nGiven [Cl2<\/sub>] = 0.4 mole\/lit, [F2<\/sub>] = 0.2 mole\/lit
                    \n[ClF] = 7.3 moles\/lit
                    \n\u2234 \\(\\frac{[\\mathrm{ClF}]^2}{\\left[\\mathrm{Cl}_2\\right]\\left[\\mathrm{F}_2\\right]}=\\frac{[7.3]^2}{0.4 \\times 0.2}=\\frac{53.29}{0.8}\\) = 66.6125
                    \nQc<\/sub> > Kc<\/sub>
                    \n\u2234 The reaction proceeds in the backward direction (Reactants side, Reverse Reaction)<\/p>\n

                    \"AP<\/p>\n

                    Question 8.
                    \nPredict which of the following reaction will have appreciable concentration of reactants and products :
                    \na) Cl2<\/sub>(g) 2Cl (g), Kc<\/sub> = 5 \u00d7 10-39<\/sup>
                    \nb) Cl2<\/sub>(g) + 2NO(g) \u21cc 2NOG (g), Kc<\/sub> = 3.7 \u00d7 108<\/sup>
                    \nc) Cl2<\/sub>(g) + 2NO2<\/sub> (g) \u21cc 2N02a(g) Kc<\/sub> = 1.8
                    \nAnswer:
                    \na) Cl2(g)<\/sub> \u21cc 2ClF(g)<\/sub> Kc<\/sub> = 5 \u00d7 10-39<\/sup>
                    \nb) Cl2(g)<\/sub> +2NO(g)<\/sub> \u21cc 2NOCl(g)<\/sub> Kc<\/sub> = 3.7 \u00d7 108<\/sup>
                    \nc) Cl2(g)<\/sub> + 2 NO2(g)<\/sub> \u21cc 2NO2<\/sub>Cl(g) Kc<\/sub> = 1.8<\/p>\n

                      \n
                    • This equation (c) have appreciable concentrations of reactants and products.<\/li>\n
                    • ‘Kc<\/sub>‘ value is neither too small nor too large. Hence equilibrium mixtures contain appreciable concentrations of reactants and products.<\/li>\n<\/ul>\n

                      Question 9.
                      \nHow to recognise the conditions under which changes in pressure would effect system in equilibrium.
                      \nAnswer:
                      \nEffect of pressure charge : A pressure change obtained by changing the volume can affect the yield of products in case of a gaseous reaction where the total number of moles of gaseous reactants and total number of moles of ga seous products are different. In applying Le Chatelier’s principle to a heterogeneous equilibrium the effect of pressure changes on solids and liquids can be ignored because the volume (and concentration) of a solution\/liquid is nearly independent of pressure.
                      \nConsider the reaction.
                      \nCO(g)<\/sub> + 3H2(g)<\/sub> \u21cc CH4(g)<\/sub> + H2<\/sub>O(g)<\/sub>
                      \nHere, 4 mol of gaseous reactants (CO + 3H2<\/sub>) become 2 mol of gaseous products (CH4<\/sub> + H2<\/sub>O). Suppose equilibrium mixture (for above reaction) kept in a cylinder fitted with a piston at constant temperature is compressed to one half of its original volume. Then, total pressure will be doubled (according to pV = constant). The partial pressure and therefore, concentration of reactants and products have changed and the mixture is no longer at equilibrium. The direction in which the reaction goes to re-establish equilibrium can be predicted by applying the Le Chatelier’s principle.<\/p>\n

                      Since pressure is doubled, the equilibrium now shifts in the forward direction, a direction in which the number of moles of the gas or pressure decreases (we know pressure is proportional to moles of the gas). The can also be understood by using reaction quotient, Qc<\/sub>. Let [CO], [H2<\/sub>], [CH4<\/sub>] and [H2<\/sub>O] be the molar concentrations at equilibrium for methanation reaction. When volume of the reaction mixture is halved, the partial pressure and the concentration are doubled. We obtain the reaction quotient by replacing each equilibrium concentration by double its value.
                      \nQC<\/sub> = \\(\\frac{\\left[\\mathrm{CH}_{4(\\mathrm{~g})}\\right]\\left[\\mathrm{H}_2 \\mathrm{O}_{(\\mathrm{g})}\\right]}{\\left[\\mathrm{CO}_{(\\mathrm{g})}\\right]\\left[\\mathrm{H}_{2_{(\\mathrm{g})}}\\right]^3}\\)
                      \nAs QC<\/sub> < KC<\/sub>, the reaction proceeds in the forward direction.
                      \nIn reaction C(s)<\/sub> + CO2(g)<\/sub> \u21cc 2CO(g)<\/sub>, when pressure is increased, the reaction goes in the reverse direction because the number of moles of gas increases in the forward direction.<\/p>\n

                      \"AP<\/p>\n

                      Question 10.
                      \nWhat property of a reaction can be used to predict the effect of a change in temperature on the magnitude of an equilibrium constant ?
                      \nAnswer:<\/p>\n

                        \n
                      • Change in temperature leads to the charge in the value of equilibrium constant (Kc<\/sub>)<\/li>\n
                      • The temperature dependence of the equilibrium constant depends on the sign of \u2206H for the reaction.<\/li>\n
                      • The equilibrium constant for an exothermic reactions (\u2206H = -ve) decreases as the temperature increases.<\/li>\n
                      • The equilibrium constant for an endothermic reaction (\u2206H = +ve) increases as the temperature increases.<\/li>\n
                      • Temp changes affect the equilibrium constant and rates of reactions.<\/li>\n<\/ul>\n

                        Question 11.
                        \nDoes the number of moles of reaction products increase, decrease or remains same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume ?
                        \ni) PCl5<\/sub>(g) \u21cc PCl3<\/sub> (g) + Cl2<\/sub> (g)
                        \nii) CaO(s) + CO3<\/sub>(g) \u21cc CaCO3<\/sub> (s)
                        \nAnswer:
                        \ni) PCl5(g)<\/sub> \u21cc PCl3(g)<\/sub> + Cl2(g)<\/sub>
                        \nDecrease in pressure by increasing volume leads to the reaction goes in forward direction be cause no. of moles of gas decrease in the forward direction.
                        \n\u2234 No. of moles of reaction products decreases.<\/p>\n

                        ii) CaO(s)<\/sub> + CO2(g)<\/sub> \u21cc CaCO3(s)<\/sub>
                        \nDecrease in pressure by increasing volume the reaction does not move in any direction because only one gaseous product present i.e. CO2(g)<\/sub>. So nochange in no. of moles of reaction products.<\/p>\n

                        Question 12.
                        \nWhich of the following reactions will get affected by increasing the pressure ? Also mention whether change will cause the reaction to go into forward or back ward direction.
                        \ni) COCl2<\/sub>(g) \u21cc CO(g) + Cl2<\/sub> (g)
                        \nii) CH4<\/sub>(g) + 2S2<\/sub>(g) \u21cc CS2<\/sub> (g) + 2H2<\/sub>S(g)
                        \niii) CO2<\/sub>(g) + C(s) \u21cc 2CO(g)
                        \niv) 4NH3<\/sub>(g) + 5O2<\/sub>(g) \u21cc 4NO(g) + 6H2<\/sub>O(g)
                        \nAnswer:
                        \ni) COCl2(g)<\/sub> \u21cc CO(g)<\/sub> + Cl2(g)<\/sub> (affected by pressaure)
                        \nIncrease in pressure the reaction goes in reverse direction and No. of moles of gas increase in the forward direction. .<\/p>\n

                        ii) CH4(g)<\/sub> + 2S2(g)<\/sub> \u21cc CS2(g)<\/sub> + 2H2(s)<\/sub>
                        \nNo. of reactants and No. of products are same, pressure has no affect.<\/p>\n

                        iii) CO2(g)<\/sub> + C(s)<\/sub> \u21cc 2 CO(g)<\/sub> (affected by pressure)<\/p>\n

                        iv) 4NH3(g)<\/sub> + 5O2(g)<\/sub> \u21cc 4 NO(g)<\/sub> + 6H2<\/sub>O(g)<\/sub> (affected by pressure)
                        \nIncrease in pressure, the reaction goes in reverse direction and No. of moles of gas increase in the forward direction.<\/p>\n

                        \"AP<\/p>\n

                        Question 13.
                        \nHow will an increase in pressure affect each of the following equilibria ? An increase in temperature ?
                        \ni) 2NH3<\/sub>(g) \u21cc N2<\/sub>(g) + 3H2<\/sub>(g) \u2206 H = 92 kJ
                        \nii) N2<\/sub>(g) + O2<\/sub>(g) \u21cc 2NO(g) \u2206 H = 181 kJ
                        \niii) 2O3<\/sub>(g) \u21cc 3O2<\/sub>(g) \u2206 H = – 285 kJ
                        \niv) CaO(S) + CO2<\/sub>(g) \u21cc CaCO2<\/sub>(s) \u2206 H = – 176 kJ
                        \nAnswer:
                        \ni) 2NH3(g)<\/sub> \u21cc N2<\/sub> + 3H2(g)<\/sub> \u2206H = +92KJ<\/p>\n

                          \n
                        • Increase in pressure favours backword reaction<\/li>\n
                        • Increase in temperatures favours to shift the equilibrium right.<\/li>\n<\/ul>\n

                          ii) N2(g)<\/sub> + O2(s)<\/sub> \u21cc 2NO(s)<\/sub> \u2206H = 181 KJ<\/p>\n

                            \n
                          • Pressure has no effect. NP<\/sub> = NR<\/sub><\/li>\n
                          • Increase in temperature favour to shift the equilibrium right<\/li>\n<\/ul>\n

                            jjj) 2O3(g)<\/sub> 3O2(g)<\/sub> \u21cc \u2206H = -285 KJ .<\/p>\n

                              \n
                            • Increase in pressure favours backward reaction<\/li>\n
                            • Increase in temperative favours to shift the equilibrium left<\/li>\n<\/ul>\n

                              iv) CaO(s)<\/sub> + CO2(g)<\/sub> \u21cc CaCO3(s)<\/sub> \u2206H = -176 KJ<\/p>\n

                                \n
                              • Pressure has no effect.<\/li>\n
                              • Increase in temperature favour to shieft the equilibrium left.<\/li>\n<\/ul>\n

                                Question 14.
                                \nThe dissociation of HI is independent of pressure, while the dissociation of PCl5<\/sub> depends upon the pressure applied explain.
                                \nAnswer:
                                \nDissociation of HI
                                \n2HI(g)<\/sub> \u21cc H2(g)<\/sub> + I2(g)<\/sub>
                                \nHere np<\/sub> = nR<\/sub>
                                \n\u2234 Pressure has no effect on the above reaction.<\/p>\n

                                Dissociation of PCl5<\/sub>
                                \nPCl5(g)<\/sub> \u21cc PCl3(g)<\/sub> + Cl2(g)<\/sub>
                                \nHere np<\/sub> \u2260 nR<\/sub>
                                \n\u2234 This reaction has affected by pressure<\/p>\n

                                Question 15.
                                \nExplain the terms :<\/p>\n

                                  \n
                                1. electrolyte<\/li>\n
                                2. Non-electrolyte<\/li>\n
                                3. strong and weak electrolytes<\/li>\n
                                4. ionic equilibrium<\/li>\n<\/ol>\n

                                  Answer:<\/p>\n

                                    \n
                                  1. Electrolyte: The substances either in molten state (or) in solution conducts electricity and at the same time undergoes chemical decomposition, e.g.: HCl, HNO3<\/sub> etc.<\/li>\n
                                  2. Non-electrolytes: The substances which donot ionise (or) chemical decomposed when dis-solved in a solvent.
                                    \ne.g.: Sugar, urea etc.<\/li>\n
                                  3. Strong Electrolytes: The electrolytes which readily under go chemical decomposition are called strong electrolytes.
                                    \neg.: NaOH, HG, H2<\/sub>SO4<\/sub> etc.,
                                    \nWeak electrolystes : The electrolytes which donot readtions undergo chemical decomposition are called weak electrolytes.
                                    \neg. : CH3<\/sub>COOH, NH4<\/sub>OH etc.,<\/li>\n
                                  4. Ionic equilibrium is a state of equilibrium set up between ionised and unionised molecule. In weak electrolytes, equilibrium is established between ions and the unionized molecules. This type of equilibrium involving ions in aq. solution is called ionic equilibrium.
                                    \n\"AP<\/li>\n<\/ol>\n

                                    \"AP<\/p>\n

                                    Question 16.
                                    \nExplain the terms :
                                    \ni) extent of ionization and on what factors it depends
                                    \nii) dissociation
                                    \niii) ionization
                                    \nAnswer:
                                    \nAcids are generally represented as ‘HX’ and bases are represented as ‘BOH’.
                                    \nIonization of acids and bases : HX \"AP H+<\/sup> + X–<\/sup>
                                    \nBOH \"AP B+<\/sup> + OH–<\/sup>
                                    \nThis process of splitting of acids and bases when dissolved in water is called ‘ionization’ or ‘dissociation1. The extent of ionization of acids and bases depends on the polarity of the molecule. For strong acids and strong bases extent of ionization is large. The extent of ionization is less for weak acids or bases.
                                    \nThis extent of ionization depends on the concentration of the solution of the substance. Process of ionization is an equilibrium process.
                                    \n\"AP
                                    \nHence, the equilibrium constants are expressed as Ka<\/sub> = \\(\\frac{\\left[\\mathrm{H}^{+}\\right]\\left[\\mathrm{X}^{-}\\right]}{[\\mathrm{HX}]}\\)
                                    \nKa<\/sub> = acid ionization constant (or) acid dissociation constant.
                                    \nKb<\/sub> = \\(\\frac{\\left[\\mathrm{B}^{+}\\right]\\left[\\mathrm{OH}^{-}\\right]}{[\\mathrm{BOH}]}\\)
                                    \nKb<\/sub> = base ionization constant (or) base dissociation constant.<\/p>\n

                                    Question 17.
                                    \nExplain the Arrhenins concept of acids and bases.
                                    \nAnswer:
                                    \nAccording to Arrhenius
                                    \nAcid : Acids are the substances that dissociates in water to gave H+<\/sup> ions.
                                    \ne.g. : HCl, H2<\/sub>SO4<\/sub> etc..
                                    \nH x(aq)<\/sub> \u2192 H+<\/sup>(aq)<\/sub> + x–<\/sup>(aq)<\/sub>
                                    \nBase : Bases are the substances that produce hydrorylions (OH–<\/sup>) ion dissociation in water
                                    \ne.g. : NaOH, KOH MOH,
                                    \nMOH(aq)<\/sub> \u2192 M+<\/sup>(aq)<\/sub> + OH–<\/sup>(aq)<\/sub><\/p>\n

                                    \"AP<\/p>\n

                                    Question 18.
                                    \nWhat is a, conjugate acid – base pair ? Illustrate with examples. [A.P. Mar. 16]
                                    \nAnswer:
                                    \nConjugate acid – base pair:
                                    \nA pair of a Bronsted acid and a base that differs by one proton (H+<\/sup>) is known as conjugate acid – base pair.
                                    \nConsider the following reaction :
                                    \nCH3<\/sub>COOH + H2<\/sub>O \u21cc H3<\/sub>O\u2295<\/sup> + CH3<\/sub>COO\u2296<\/sup>
                                    \nIt involves two conjugate pairs. The acid – base pairs are :
                                    \n\"AP<\/p>\n

                                    Question 19.
                                    \nAcetic acid is a weak acid. List, in order of descending concentration, all of the ionic and molecular species present in 1 M aqueous solution of acetic acid.
                                    \nAnswer:
                                    \n[H2<\/sub>O] > [CH3<\/sub>COOH] > [H3<\/sub>O+] [CH3<\/sub>COO–<\/sup>] > [OH–<\/sup>]<\/p>\n

                                    Question 20.
                                    \nShow by suitable equations that each of the following species can act as a Brostead acid.
                                    \na) H3<\/sub>O+<\/sup>
                                    \nb) HCl
                                    \nc) NH3<\/sub>
                                    \nd) HSO4<\/sub>–<\/sup>
                                    \nAnswer:
                                    \na) H3<\/sub>O+<\/sup> \u2192 H2<\/sub>O + H+<\/sup>
                                    \nProton donar Hence a Bronsted acid<\/p>\n

                                    b) HCl \u2192 H+<\/sup> + Cl–<\/sup>
                                    \nProton donar Hence a Bronsted acid<\/p>\n

                                    c) NH3<\/sub> is bronsted base it’s not an acid
                                    \n(Sometimes NH3<\/sub> \u2192 NH–<\/sup>2<\/sub> + H+<\/sup> as acid]<\/p>\n

                                    d) HSO4<\/sub>–<\/sup> \u2192 H+<\/sup> + SO4<\/sub>-2<\/sup>
                                    \nproton donar Hence Bronsted acid.<\/p>\n

                                    Question 21.
                                    \nShow by suitable equations that each of the following spices can act as a Bronsted base.
                                    \na) H3<\/sub>O
                                    \nb) OH–<\/sup>
                                    \nc) C2<\/sub>H5<\/sub>OH
                                    \nd) HPO4<\/sub>-2<\/sup>
                                    \nAnswer:
                                    \na) H2<\/sub>O + H+<\/sup> \u2192 H3<\/sub>O+<\/sup>
                                    \nProton acceptor Hence a Bronsted base
                                    \nb) OH–<\/sup> + H+<\/sup> \u2192 H2<\/sub>O
                                    \nProton acceptor Hence a Bronsted base
                                    \nc) C2<\/sub>H5<\/sub>OH is a proton donar
                                    \nSo it is not a bronsted base it is acid
                                    \nd) HPO4<\/sub>-2<\/sup> + H+<\/sup> \u2192 H2<\/sub>PO4<\/sub>–<\/sup>
                                    \nproton acceptor Hence it is Bronsted base.<\/p>\n

                                    \"AP<\/p>\n

                                    Question 22.
                                    \nThe species H2<\/sub>O, HCO4<\/sub>–<\/sup>, HSO4<\/sub>–<\/sup> and NH3<\/sub> can act both as Bronsted acids and bases. Give the corresponding conjugate acid and base for each of them.
                                    \nAnswer:
                                    \nThe answer is given in the following Table :
                                    \n\"AP<\/p>\n

                                    Question 23.
                                    \nWrite equation that shows H2<\/sub>PO4<\/sub>–<\/sup> acting both as an acid and as a base.
                                    \nAnswer:
                                    \nAs an acid :
                                    \nH2<\/sub>Po4<\/sub>–<\/sup> \u2192 H+<\/sup> + HP2<\/sub>PO4<\/sub>-2<\/sup>
                                    \nProton donar
                                    \nAs a base:
                                    \nH2<\/sub>Po4<\/sub>–<\/sup> + H+<\/sup> \u2192 H3<\/sub>PO4<\/sub>
                                    \nProton acceptor<\/p>\n

                                    Question 24.
                                    \nWrite the conjugate acid and conjugate base of each of the following :
                                    \na) OH–<\/sup>
                                    \nb) H2<\/sub>O
                                    \nc) HCO3<\/sub>–<\/sup>
                                    \nd) H2<\/sub>O2<\/sub>
                                    \nAnswer:
                                    \n\"AP<\/p>\n

                                    \"AP<\/p>\n

                                    Question 25.
                                    \nIdentity and label the Bronstead acid and its conjugate base, the Bronsted base and its conjugate acid in each of the following equations.
                                    \na) H2<\/sub>SO4<\/sub> + Cl–<\/sup> \u2192 HCl + HSO4<\/sub>–<\/sup>
                                    \nb) H2<\/sub>S + NH2<\/sub> \u2192 HS–<\/sup> + NH3<\/sub>
                                    \nc) CN–<\/sup> + H2<\/sub>O \u2192 HCN + OH–<\/sup>
                                    \nd) O-2<\/sup> + H2<\/sub>O \u2192 2OH–<\/sup>
                                    \nAnswer:
                                    \n\"AP<\/p>\n

                                    Question 26.
                                    \nClassify the species AlCl3<\/sub>, NH3<\/sub>, Mg+2<\/sup> and H2<\/sub>O into Lewis acids and Lewis bases and justify your answer ?
                                    \nAnswer:<\/p>\n

                                      \n
                                    • AlCl3<\/sub>, Mg+2<\/sup> are lewis acids because These are electron pair acceptory<\/li>\n
                                    • NH3<\/sub>, H2<\/sub>O are lewis bases becuse. These are electron pair donors.<\/li>\n<\/ul>\n

                                      Question 27.
                                      \nWhat are the strengths of conjugate bases of a strong acid and a weak acid ?
                                      \nAnswer:
                                      \nA strong acid has a weak conjugate base
                                      \nEg : HCl, Cl–<\/sup>
                                      \nA weak acid has a strong conjugate base
                                      \nEg : CH3<\/sub>COOH, CH3<\/sub>COO–<\/sup><\/p>\n

                                      Question 28.
                                      \nWhat are the strength of conjugates acids of a strong base and weak base ?
                                      \nAnswer:
                                      \nStrong base has a weak conjugate acid
                                      \nEg : HS–<\/sup>, H2<\/sub>S
                                      \nA weak base has a stong conjugatge acid
                                      \nEg : ClO–<\/sup>4<\/sub>, HClO4<\/sub><\/p>\n

                                      Question 29.
                                      \nDefine ionic product of water. What is its value at room temperature ?
                                      \nAnswer:
                                      \nAt a given temp, the product of the concentrations of H+<\/sup> and OH–<\/sup> ions in water is called ionic product
                                      \nIonic product Kw<\/sub> = [H–<\/sup>] [OH–<\/sup>]
                                      \nAt 25\u00b0C Kw<\/sub> = 1.008 \u00d7 10-14<\/sup> mole2<\/sup>\/lit2<\/sup>
                                      \nAt 25\u00b0C Ionic product of water Kw<\/sub> = 1.008 \u00d7 10-14<\/sup> mole2<\/sup>\/lit2<\/sup>
                                      \nUnits: mole2<\/sup>\/lit2<\/sup><\/p>\n

                                      \"AP<\/p>\n

                                      Question 30.
                                      \nDefine pH. pH cannot be calculated directly from the molar concentration of a weak acid or weak base. Why ? Derive an equation for the pH of a weak acid.
                                      \nAnswer:
                                      \nThe negative logarithm on value to the 10 of the hydrogen ion concentration expressed in moles\/lit, in a solution is know as pH. of the solution.
                                      \npH : – log (H+<\/sup>) = – log10<\/sub> \\(\\frac{1}{\\left(\\mathrm{H}^{+}\\right)}\\)
                                      \nIn case of a weak acid (or) weak base extent of ionisation is low so pH cannot be calculated directly from their molar concentrations.<\/p>\n

                                      Derivation of pH of a weak acid :
                                      \nA general step wise approch can be adopted to evaluate the pH of the weak acid as follows.<\/p>\n

                                      Step 1. The species present before dissociation are identified as Bronsted – Lowry acids\/ bases.
                                      \nStep 2. Balanced equations for all possibel reactions i.e., with a species acting both as acid as well as base are written.
                                      \nStep 3. The reaction with the higher Ka<\/sub> is identified as the primary reaction whilst the other is a subsidiary reaction.
                                      \nStep 4. Enlist in a tabular form the following values for each of the species in the primary reaction.
                                      \na) Initial concentration. C.
                                      \nb) Change in concentration on proceeding to equilibrium in terms of a, degree of ionization.
                                      \nStep 5. Substitute equilibrium concentrations into equilibrium constrant equation for principal reaction and solve for \u03b1.
                                      \nStep 6. Calculate the concentration of species in principal raction.
                                      \nStep 7. Calculate pH = – log [H3<\/sub>O+<\/sup>]
                                      \nThe above mentioned methodology has been elucidated in the following examples.<\/p>\n

                                      Question 31.
                                      \nWrite equations to show the step wise ionization of the polyprotic acids H2<\/sub>SO4<\/sub> and H2<\/sub>PO4<\/sub>.
                                      \nAnswer:
                                      \nStep wise Ionisation H2<\/sub>SO4<\/sub>
                                      \n\"AP<\/p>\n

                                      Question 32.
                                      \nExplain how acid strength changes among<\/p>\n

                                        \n
                                      1. the hydrides of the group elements and<\/li>\n
                                      2. the hydrides in the same row of the periodic table.<\/li>\n<\/ol>\n

                                        Answer:<\/p>\n

                                          \n
                                        1. The hydrides of the group elements acid strength decreases from top to bottom.<\/li>\n
                                        2. The hydrides in the same row of Periodic table i.e., in a period the acid strength increases from left to right.<\/li>\n<\/ol>\n

                                          \"AP<\/p>\n

                                          Question 33.
                                          \nJustify the statement that water behaves like an acid and also like base on the basis of protonic concept.
                                          \nAnswer:<\/p>\n

                                            \n
                                          • Water has the ability to acts as an acid and as well as a base i.e., it behaves as an amphoteric substance.<\/li>\n
                                          • The following are the supporting reactions for the fact that water behaves as an amphoteric substance.<\/li>\n
                                          • As an acid :
                                            \nH2<\/sub>O(l)<\/sub> + NH3(aq)<\/sub> \u21cc OH–<\/sup>(aq)<\/sub> + NH4<\/sub>+<\/sup>(aq)<\/sub>
                                            \nH2<\/sub>O is proton do nor i.e., an acid according to Bronsted sense<\/li>\n
                                          • As a base:
                                            \nH2<\/sub>O(l)<\/sub> + H2<\/sub>S(aq)<\/sub> \u21cc H3<\/sub>O+<\/sup>(aq)<\/sub> + HS–<\/sup>(aq)<\/sub>
                                            \nH2<\/sub>O is proton acceptor i.e., a base according to Bronsted Sense<\/li>\n<\/ul>\n

                                            Question 34.
                                            \nWhat is common ion effect ? Illustrate.
                                            \nAnswer:
                                            \nCommon ion effect : The decrease in the ionisation (dissociation) of a weak electrolyte, by the addition of a strong electrolyte having an ion common with the weak electrolyte, is known as common ion effect.
                                            \nExample : The dissociation of NH4<\/sub>OH is diminished by the addition of NH4<\/sub>Cl due to the common ion, NH4<\/sub>+<\/sup> ion
                                            \nNH4<\/sub>OH \u21cc NH4<\/sub>\u2295<\/sup> + OH\u2296<\/sup>
                                            \nNH4<\/sub>OH \"AP NH4<\/sub>+<\/sup> + OH\u2296<\/sup> (addition of NH4<\/sub>Cl)
                                            \nImportance of this effect In qualitative analysis :<\/p>\n

                                              \n
                                            1. This principle is used in the systematic qualitative analysis of cations.<\/li>\n
                                            2. The concentration of S2-<\/sup> in II group or OH–<\/sup> ion in III group of chemical analysis is controlled by HCl and NH4<\/sub>OH on the basis of common ion effect.
                                              \nH+<\/sup> ion is common to H2<\/sub>S (II group)
                                              \nNH4<\/sub>+<\/sup> ion is common to NH4<\/sub>OH (IV group) .<\/li>\n
                                            3. The common ion effect principle is also used in controlling the H+<\/sup> ion concentration in buffer solutions.<\/li>\n
                                            4. It is also used in the purification of common salts by passing dry HCl gas into impure salt solution.<\/li>\n<\/ol>\n

                                              \"AP<\/p>\n

                                              Question 35.
                                              \nDefine solubility product ? Write solubility product expressions for the following :
                                              \n(i) Ag2<\/sub>Cr2<\/sub>O7<\/sub>
                                              \nii) Zr3<\/sub>(PO4<\/sub>)4<\/sub>
                                              \nAnswer:
                                              \nSolubility product (KSP<\/sub>) : The product of the concentrations of the cation and the anion in a saturated solution of a salt at room temperature is called solubility product (Ksp<\/sub>).
                                              \nKsp<\/sub> = [Mn+<\/sup>][An-<\/sup>]
                                              \nSignificance of solubility product in chemical analysis : In any solution of a salt, if the product of [Mn+<\/sup>] and [An-<\/sup>] is greater than the solubility product (KSP<\/sub>) of the salt, then only the salt gets precipitated. For example,
                                              \ni) If the product of [Cu 2+<\/sup>] & [S-2<\/sup>] in the II group (in presence of acid) of chemical analysis exceeds the value of KSP<\/sub> of CuS, then only CuS will be precipitated in the group. Similar argument applies to other metal ions in the group namely (Cd2+<\/sup>, Bi3+<\/sup>, Hg2+<\/sup>, Sb2+<\/sup>, etc).<\/p>\n

                                              ii) If the product of [Zn2+<\/sup>] and [S2-<\/sup>] in ammonical medium in IV group of chemical analysis exceeds the value of KSP<\/sub> of ZnS, it will be precipitated. Similar arguments apply to Mn2+<\/sup>, Ni2+<\/sup> etc.
                                              \na) Solubility product expression for Ag2<\/sub>Cr2<\/sub>O7<\/sub>
                                              \n\"AP<\/p>\n

                                              Question 36.
                                              \nGive the classification of salts. What type of salts undergo hydrolysis ?
                                              \nAnswer:
                                              \nThe following are the types of salts : –<\/p>\n

                                                \n
                                              1. Salts of a strong acid and a strong base. Ex : NaCl<\/li>\n
                                              2. Salts of a strong acid and a weak base. Ex : NH4<\/sub>Cl<\/li>\n
                                              3. Salts of a weak acid and a strong base. Ex : CH3<\/sub>COONa<\/li>\n
                                              4. Salts of a weak acid and a weak base. Ex : CH3<\/sub>COONH4<\/sub>
                                                \nTypes 2, 3, 4 under go hydrolysis reactions.<\/li>\n<\/ol>\n

                                                Question 37.
                                                \nWhat must be true of the value of \u2206G\u00b0 for a reaction if
                                                \na) K > 1
                                                \nb) K = 1
                                                \nc) K < 1
                                                \nAnswer:
                                                \na) If K > 1 then \u2206 G\u00b0 < 0
                                                \nb) If K = 1 then \u2206 G\u00b0 = 0
                                                \nc) If K < 1 then \u2206 G\u00b0 > 0<\/p>\n

                                                \"AP<\/p>\n

                                                Question 38.
                                                \nAqueous solution of NH4<\/sub>Cl is acidic. Explain.
                                                \nAnswer:
                                                \nGiven salt NH4<\/sub>Cl is obtained by the reaction of strong acid (HCl) and weak base (NH4<\/sub>OH).
                                                \nThe ionisation of NH4<\/sub>Cl as follows
                                                \n\\(\\mathrm{NH}_4^{+}\\) + Cl–<\/sup> \u21cc NH4<\/sub>Cl
                                                \nThe above salt under goes cationic hydrolysis
                                                \n\\(\\mathrm{NH}_4^{+}\\) + H2<\/sub>O \u21cc NH4<\/sub>OH + H+<\/sup>
                                                \n\u2234 The equation solution of salt is acidic due to greater H+<\/sup> ion concentration than OH–<\/sup> ion concentration.
                                                \n\u2234 PH<\/sub> < 7<\/p>\n

                                                Question 39.
                                                \nAqueous solution of CH3<\/sub>COONa is basic explain.
                                                \nAnswer:
                                                \nGiven salt CH3<\/sub>COONa is obtained by the reaction of weak acid (CH3<\/sub>COOH) and strong base (NaOH).
                                                \nThe ionisation CH3<\/sub>COONa as follows
                                                \nCH3<\/sub>COONa \u21cc CH3<\/sub>COO–<\/sup> + Na+<\/sup>
                                                \nThe above salt under go anionic hydrolysis
                                                \nCH3<\/sub>COO–<\/sup> + H2<\/sub>O \u21cc CH3<\/sub>COOH + OH–<\/sup>
                                                \n\u2234 The equation solution of salt is basic due to greater OH–<\/sup> ion concentration than H+<\/sup> ion concentration.
                                                \n\u2234 PH<\/sup> > 7<\/p>\n

                                                Question 40.
                                                \nGive reason that acetic acid is less acidic in sodium acetate solution than in sodium chloride solution.
                                                \nAnswer:
                                                \nAcetic acid in sodium acetate solution is less acidic.
                                                \nReasons:<\/p>\n

                                                  \n
                                                • Because of common ion effect ionisation process is decreased.<\/li>\n
                                                • Sodium acetate is obtained by the reaction of weak acid and strong base:
                                                  \nAcetic acid in NaG solution is more acidic.
                                                  \nHere no common ion effect observed.
                                                  \nNaCl is the salt obtained by strong acid and strong base.<\/li>\n<\/ul>\n

                                                  Question 41.
                                                  \nAgCl is less soluble in AgNO3<\/sub> solution than in pure water explain.
                                                  \nAnswer:<\/p>\n

                                                    \n
                                                  • AgCl is less soluble in AgNO3<\/sub> because of common ion effect solubility is decreased.<\/li>\n
                                                  • AgCl is more soluble in pure water because common ion effect not observed here.<\/li>\n<\/ul>\n

                                                    \"AP<\/p>\n

                                                    Question 42.
                                                    \nPredict whether the following reaction will proceed from left to right to any measurable extent:
                                                    \nCH3<\/sub>COOH (aq) + Cl–<\/sup>(aq) \u2192
                                                    \nAnswer:
                                                    \nGiven CH3<\/sub>COOH(aq)<\/sub> + \\(\\mathrm{Cl}_{(\\mathrm{aq})}^{-}\\) \u2192
                                                    \nThe reaction does not proceed from left to right because Cl–<\/sup> ion is obtained from strong acid do not under go hydrolysis and CH3<\/sub>COOH has low extent of ionisation.<\/p>\n

                                                    Question 43.
                                                    \nAqueous solution of H2<\/sub>S contains H2<\/sub>S, HS–<\/sup>, S-2<\/sup>, H3<\/sub>O+<\/sup>, OH–<\/sup> and H2<\/sub>O in varying concentrations. Which of these species can act only as a base ? Which can act only as an acid ? Which can act both an acid and as a base ?
                                                    \nAnswer:<\/p>\n

                                                      \n
                                                    • H23<\/sub>S, H3<\/sub>O+<\/sup> acts as only acids<\/li>\n
                                                    • HS–<\/sup>, OH–<\/sub>, H2<\/sub>O acts as acids as well as bases<\/li>\n
                                                    • S-2<\/sup> acts as only base<\/li>\n<\/ul>\n

                                                      Long Answer Questions<\/span><\/p>\n

                                                      Question 1.
                                                      \nWhat are equilibrium processes ? Explain equilibrium in Physical and Chemical pro-cesses with examples.
                                                      \nAnswer:
                                                      \nEquilibrium State (or) Equilibrium Process : The state at which the velocity of forward reaction becomes equal to the velocity of backward reaction or reverse reaction is called ‘equilibrium state’. Consider a reversible reaction.
                                                      \n\"AP
                                                      \nA + B \u21cc C + D taking place in a closed vessel. At the beginning we have only the reactants A and B. Their concentrations are maximum. As the reaction proceeds the reactants A and B change into the products C and D. The concentrations of the products increase gradually.
                                                      \nThe rate of forward reaction diminishes while the reverse reaction sets in and proceeds with increasing speed. A state is soon reached where the speeds of forward and backward reactions become equal. If the rate of forward reaction Vf<\/sub> = The rate of reverse reaction (Vb<\/sub>).<\/p>\n

                                                      The system is said to have attained a state of equilibrium. Once equilibrium is reached there is no further change in the composition of the system. The system appears to be stand still although it is dynamic. The products are formed by forward reaction just as fast as they change back into reactants by the reverse reaction.<\/p>\n

                                                      The state at which the rate of forward reaction is equal to the rate of the reverse reaction in a reversible reaction is known as the equilibrium state or chemical equilibrium.<\/p>\n

                                                      Examples :
                                                      \ni) Equilibrium physical processes :
                                                      \na) Equilibrium between two phases (i.e.,) physical transformation processes.
                                                      \nSolid \u21cc Liquid (melting or fusion)
                                                      \nSolid \u21cc Vapour (sublimation)
                                                      \nb) Equilibrium between two different allotropic forms of same substance
                                                      \n\u03b1 – Sulphur \u21cc \u03b2 – Sulphur<\/p>\n

                                                      ii) Equilibrium chemical processes :
                                                      \na) N2<\/sub> (g) + 3H2<\/sub> (g) \u21cc 2NH3<\/sub> (g)
                                                      \nb) 2SO2<\/sub> (g) + O2<\/sub> (g) \u21cc 2SO3<\/sub> (g)
                                                      \nc) H2<\/sub>(g) + I2<\/sub>(g) \u21cc 2HI(g)<\/p>\n

                                                      \"AP<\/p>\n

                                                      Question 2.
                                                      \nWhat is meant by dynamic equilibrium ? Explain with suitable examples.
                                                      \nAnswer:
                                                      \nDynamic equilibrium : The forward and reverse reactions of a reversible reaction continue to take place with equal rates simultaneously at the equilibrium stage also. Hence the equilibrium is called Dynamic equilibrium.<\/p>\n

                                                      Explanation : In order to understand the dynamic nature of the reaction, synthesis of ammonia is carried with exactly the same starting conditions but using D2<\/sub> (deuterium) in place of H2<\/sub>. The reaction mixtures starting either with H2<\/sub> or D2<\/sub> reach equilibrium with the same composition, except that D2<\/sub> and ND3<\/sub> are present instead of H2<\/sub> and NH3<\/sub>. After equilibrium is attained, these two mixtures (H2<\/sub>, N2<\/sub>, NH3<\/sub> and D2<\/sub>, N2<\/sub>, ND3<\/sub>) are mixed together and left for a while. Later when this mixture analysed, it is found that the concentration of ammonia is just the same as before.<\/p>\n

                                                      However, when this mixture is analysed by a moss spectrometer, it is found that ammonia and all deuterium containing forms of ammonia (NH3<\/sub>, NH2<\/sub>D, NHD2<\/sub> and ND3<\/sub>) and dihydrogen and its deutrated forms (H2<\/sub>, HD and D2<\/sub>) are present. Thus one can conclude that scrambling of H and D atoms in the molecules must result from a continuation of the forward and reverse reactions in the mixture. If the reaction had simply stopped when they reached equilibrium, then there would have been no mixing of isotopes in this way.<\/p>\n

                                                      Use of isotope (deuterium) in the formation of ammonia clearly indicates that chemical reaction react a state of dynamic equilibrium in which the rates of forward and reverse reactions are equal and there is no net change in composition.<\/p>\n

                                                      Question 3.
                                                      \nGive the General characteristics of equilibria involving physical processes ?
                                                      \nAnswer:
                                                      \nCharacteristics of general equilibria :<\/p>\n

                                                        \n
                                                      1. The rate of the forward reaction is equal to the rate of the reverse reaction (uf = ub).<\/li>\n
                                                      2. The concentrations of the reactants and the products remain unchanged with time.<\/li>\n
                                                      3. The observable properties such as pressure, concentration, density, colour also remain unchanged with time.<\/li>\n
                                                      4. The equilibrium is dynamic in nature. Both the forward reaction and the reverse reaction continue to take place simultaneously with equal rates.<\/li>\n
                                                      5. A catalyst does not alter the state of equilibrium or the composition of the chemical substances of the reaction at the equilibrium. It only speeds up the attainment of the equilibrium.<\/li>\n
                                                      6. Chemical equilibrium can be established from either side of the reversible reaction.<\/li>\n<\/ol>\n

                                                        \"AP<\/p>\n

                                                        Question 4.
                                                        \nWhat are the important features of equilibrium constant ? Discuss any two applications of equilibrium constant.
                                                        \nAnswer:
                                                        \nThe ratio of product of molar concentrations of products to the product of molar concentrations of reactants is called equilibrium constant (Kc<\/sub>)
                                                        \nEx : aA + bB \u21cc cC + dD
                                                        \nKC<\/sub> = \\(\\frac{[C]^c[D]^d}{[A]^a[B]^b}\\)
                                                        \nKC<\/sub> for reverse reaction is the inverse of the KC<\/sub> for the reaction in the forward direction.
                                                        \nConcentration equilibrium constant (KC<\/sub>) and pressure equilibrium constant (Kp<\/sub>) are related as follows
                                                        \nKp<\/sub> = KC<\/sub>(RT)\u2206n<\/sup>
                                                        \n\u2206n = np<\/sub> – nR<\/sub>
                                                        \nWhile writing KC<\/sub> expression pure liquids, pure solids concentration are ignored Applications :<\/p>\n

                                                        a) In predicting the extent of reaction :
                                                        \nThe numerical value of the equilibrium constant for a reaction indicates the extent of the reac-tion. But it is important to note that an equilibrium constant does not give any information about the rate at which the equilibrium is reached the Magnitude of ke<\/sub> or Kp<\/sub> is directly proportional to the concentrations of products (as these appear in the numerator of equilibrium constant expression) and inversely proportional to the concentrations of the reactants (these appear in the denominator). This implies that a high value of K is suggestive of a high concentration of products and vice- versa.<\/p>\n

                                                        We can make the following generalisation concerning the composition of equilibrium mixtures : If Kc<\/sub> > 103<\/sup> products predominate over reactants, i.e., if Kc<\/sub> is very large, the reaction proceeds nearly to completion. Consider the following examples :<\/p>\n

                                                        a) The reaction of H2<\/sub> with O2<\/sub> at 500 K has a very large equilibrium constant Kc<\/sub> = 2.4 \u00d7 1047<\/sup>
                                                        \nb) H2(g)<\/sub> + Cl2(g)<\/sub> \u21cc 2HCl(g) at 300 K has Kc<\/sub> = 4.0 \u00d7 1031<\/sup>
                                                        \nc) H2(g)<\/sub> + Br2(g)<\/sub> \u21cc 2HBr(g)<\/sub> at 300 K. Kc<\/sub> = 5.4 \u00d7 1018<\/sup>
                                                        \nIf Kc<\/sub> = < 10-3<\/sup> reactants predominate over products i.e, if Kc<\/sub> is very small the reaction proceeds rarely. Consider the following examples : –
                                                        \na) The decomposition of H2<\/sub>O into H2<\/sub> and O2<\/sub> at 500 K has a very small equilibrium constant.
                                                        \nKc<\/sub> = 4.1 \u00d7 10-48<\/sup>
                                                        \nb) N2(g)<\/sub> + O2(g)<\/sub> \u21cc 2NO(g)<\/sub>
                                                        \nat 298 K has Kc<\/sub> = 4.8 \u00d7 10-31<\/sup>
                                                        \nIf Kc<\/sub> is in the range of 10-3<\/sup> to 103<\/sup> appreciable concentrations of both reactants and products are present. Consider the following examples :
                                                        \na) For reaction of H2<\/sub> with I2<\/sub> to give HI.
                                                        \n\"AP
                                                        \nb) Also, gas phase decomposition of N2<\/sub>O4<\/sub> to NO2<\/sub> is another reaction with a value of Kc<\/sub> = 4.64 \u00d7 10-3<\/sup> at 25\u00b0C which is neither too small nor too large. Hence equilibrium mixtures contain appreciable concentrations of both N2<\/sub>O4<\/sub> and NO2<\/sub>.
                                                        \nThese generarlisations are illustrated.
                                                        \nb) In predicting the direction of reaction : A and K are compared to predict the direction of reaction
                                                        \na) Q = K indicates that the reaction mixture is already at equilibrium
                                                        \nb) Q < K indicates that the reaction proceeds in the direction. c) Q > K indicates that the reaction proceeds in the direction of reactants (Reverse reaction)<\/p>\n

                                                        \"AP<\/p>\n

                                                        Question 5.
                                                        \nWhat is Lechatlier’s principle ? Discuss briefly the factors which can influence the equilibrium.
                                                        \nAnswer:
                                                        \nLe Chatelier’s Principle – Statement: “If a system at equilibrium is subjected to the change of pressure, temperature (or) concentration, the system is shifted in such a way as to nullify the effect of change”.
                                                        \nExplanation:
                                                        \n1) Concentration: Increase of reactant concentrations pushes the equilibrium state to the products side and increase of the reactions concentrations pushes to the equilibrium to the reactants side. For example in the chemical equilibrium
                                                        \nH2<\/sub>(g) + I2<\/sub>(g) \u21cc 2HI(g)
                                                        \nIncrease of H2<\/sub> or I2<\/sub> concentrations pushes the equilibrium in favours of HI and similarly the increase of HI concentrations pushes the equilibrium in favour of H2<\/sub> and I2<\/sub>.<\/p>\n

                                                        2) Pressure : Pressure will have no effect on the equilibrium reactions in which there is no change in the number of moles of the reactants and the products.
                                                        \nFor example pressure will have no effect on
                                                        \nH2<\/sub>(g) + I2<\/sub>(g) \u21cc 2HI(g) equilibrium
                                                        \n(or) N2<\/sub> (g) + O2<\/sub> (g) \u21cc 2NO (g) equilibrium
                                                        \nBut in the case of reactions in which there is a change in the number of moles of the reactants and the products increase of pressure pushes the reaction equilibrium in the direction in which there is a decrease in the number of moles.
                                                        \nFor example in the reaction PCl5<\/sub> (g) \u21cc PCl3<\/sub> (g) + Cl2<\/sub> (g)
                                                        \nIncrease of pressure favours the backward reaction PCl3<\/sub> (g) + Cl2<\/sub>(g) \u2192 PCl5<\/sub> (g) and the decrease of pressure favour the forward reaction PCl5<\/sub> (g) \u2192 PCl3<\/sub> (g) + Cl2<\/sub> (g)<\/p>\n

                                                        3) Temperature : Increase of temperature favour the endothermic reaction and decrease of temperature favours exothermic reactions.
                                                        \nFor example .
                                                        \nN2<\/sub> (g) + O2<\/sub> (g) \u21cc 2N0(g) – 180.8 kJ
                                                        \nIncrease of temperature favours The forward reaction
                                                        \nN2<\/sub> (g) + O2<\/sub> (g) \u2192 2NO (g) (since endothermic)
                                                        \ndecrease of temperature favour the dissociation of NO into N2<\/sub> and O2<\/sub> since it is exothermic.
                                                        \nThe backward reaction
                                                        \n2NO (g) \u2192 N2<\/sub> (g) + O2<\/sub> (g) (since exothermic)<\/p>\n

                                                        Question 6.
                                                        \nDiscuss the application of Lechaterlier’s principle for the industrial synthesis of Ammonia and sulphur trioxide.
                                                        \nAnswer:
                                                        \nApplications of Le Chatelier’s principle to synthesis of Ammonia by Haber’s process :
                                                        \n\"AP
                                                        \nNitrogen and Hydrogen combine to form ammonia. The formation of ammonia is reversible and exothermic reaction. It is accompanied by decrease in volume.<\/p>\n

                                                        Effect of Pressure: 1 volume of N2<\/sub> combines with 3 volumes of H2<\/sub> to form 2 volumes of NH3<\/sub>. There is decrease in volume in the forward reaction (4 volumes to 2 volumes). According to Le Chatelier’s principle increase of pressure favours the reaction where there is decrease in volume. So higher the pressure, greater the yield of ammonia. In practice 200 atmospheres are used in the manufacture of ammonia by Haber’s process.<\/p>\n

                                                        Low pressures favour the reverse reaction i.e., decomposition of NH3<\/sub> already formed.<\/p>\n

                                                        Effect of Temperature : The formation of ammonia (forward reaction) is exothermic reaction
                                                        \nN2<\/sub> + 3H2<\/sub> \u2192 2 NH3<\/sub>; \u2206H = – 92 kJ.
                                                        \nLow temperatures favour the forward reaction. But at low temperatures the reaction is too slow. Therefore an optimum temperature (725 K- 775 K) is chosen in Haber’s process. To speed up the reaction, a catalyst, finely divided iron is used. To increase the activity of the catalyst molybdenum or a mixture of oxides of K after Al is used as promoter.<\/p>\n

                                                        The reverse reaction (i.e.,) decomposition of NH3<\/sub> is an endothermic reaction. High temperatures favour the decomposition of NH3<\/sub>. Therefore high temperatures are avoided in Haber’s process. Thus the optimum conditions are
                                                        \nPressure : 200 atm
                                                        \nTemperature : 725-775 K
                                                        \nCatalyst : Fe (Powered)
                                                        \nPromoter : Mo (or) (K2<\/sub>O + Al2<\/sub>O3<\/sub>)
                                                        \nApplication of Le Chatelier’s principle to the synthesis of SO3<\/sub> :
                                                        \n\"AP
                                                        \nThe formation of SO3<\/sub> is reversible and exothermic reaction. It is accompanied by decrease in volume.
                                                        \n1. Effect of Pressure : 2 volumes of SO2<\/sub> and one volume of O2<\/sub> combine to give 2 volumes of SO3<\/sub>. According to Le Chatelier’s Principle increase of pressure favours the reaction where there is decrease in volume. The formation of SO3<\/sub> is accompanied by decrease in volume (3 volumes to 2 volumes). Higher the pressure greater is the yield of S03. But in contact process high pressures are not used because towers used in the manufacture are corroded by the acid at these high pressures.<\/p>\n

                                                        Low pressures favour the decomposition of SO3<\/sub> as there is increase in the volume of (2 volumes to 3 volumes). Therefore optimum pressures are used (1.5 to 1.7 atmosphere).<\/p>\n

                                                        2. Effect of Temperature : The formation of SO3<\/sub> is exothermic. 189 Id of heat is evolved. High temperatures favour the reverse reaction which is endothermic and do not favour the forward reaction which is exothermic. Low temperatures are favourable for the formation of SO3<\/sub>. At low temperature the reaction is too slow. Therefore an optimum temperature 673 K is used. To speed up the reaction V205 is used as catalyst.
                                                        \nThus the optimum conditions are
                                                        \nPressure : 1.5 to 1.7 atm
                                                        \nTemperature : 673 K
                                                        \nCatalyst : V2<\/sub>O5<\/sub> (or) Platinised asbestos<\/p>\n

                                                        \"AP<\/p>\n

                                                        Question 7.
                                                        \nDihydrogen gas is obtained from natural gas by partial oxidation with steam as per the following endothermic reaction. .
                                                        \nCH4<\/sub>(g) + H2<\/sub>O(g) \u21cc CO(g) + 3H2<\/sub> (g)
                                                        \na) Write an expression for Kp<\/sub> for the above reaction.
                                                        \nb) How will the values of Kp<\/sub> and composition of equilibrium mixture be affected by<\/p>\n

                                                          \n
                                                        1. increasing the pressure<\/li>\n
                                                        2. increasing the temperature<\/li>\n
                                                        3. using a catalyst ?<\/li>\n<\/ol>\n

                                                          Solution:
                                                          \na) Given equation
                                                          \nCH4(g)<\/sub> + H2<\/sub>O(g)<\/sub> \u21cc CO(g)<\/sub> + 3H2(g)<\/sub>
                                                          \nKp<\/sub> = \\(\\frac{\\mathrm{P}_{\\mathrm{CO}} \\times \\mathrm{P}_{\\mathrm{H}_2}^3}{\\mathrm{P}_{\\mathrm{CH}_4} \\times \\mathrm{P}_{\\mathrm{H}_2 \\mathrm{O}}}\\)<\/p>\n

                                                          b)<\/p>\n

                                                            \n
                                                          1. Increase of pressure the reaction proceeds in reverse direction because the no. of moles of gas increases in the forward direction.
                                                            \nIncrease pressure Kp<\/sub> value increases.<\/li>\n
                                                          2. Increase of temperature of the reaction mixture at equilibrium favours the reaction in the direction in which heat is absorbed.
                                                            \nIncrease of temperature Kp<\/sub> value increases.<\/li>\n
                                                          3. A catalyst does not disturbs (or) change the chemical reaction it just speed up the attaining of equilibrium.<\/li>\n<\/ol>\n

                                                            Question 8.
                                                            \nDescribe the effect of :
                                                            \na) addition of H2<\/sub>
                                                            \nb) addition of CH3<\/sub>OH
                                                            \nc) removal of CO
                                                            \nd) removal of CH3<\/sub>OH on the equilibrium of the reaction
                                                            \n2H2<\/sub>(g) + CO(g) \u21cc CH3<\/sub>OH(g)
                                                            \nSolution:
                                                            \na) Addition of H2<\/sub> (Reactant):
                                                            \nIncreasing the concentration of reactants in the reaction mixture at equilibrium favours the forward reaction.<\/p>\n

                                                            b) Addition of CH3<\/sub>OH (Product):
                                                            \nIncreasing the concentration of products in the reaction mixture at equilibrium favours the reverse reaction.<\/p>\n

                                                            c) Removal of CO (Reactant):
                                                            \nDecreasing the concentration of reactants in the reaction mixture at equilibrium favours the reverse reaction.<\/p>\n

                                                            d) Removal of CH3<\/sub>OH (Product):
                                                            \nDecreasing the concentration of products in the reaction mixture at equilibrium favours the forward reaction.<\/p>\n

                                                            \"AP<\/p>\n

                                                            Question 9.
                                                            \nAt 473K, equilibrium constant Kc<\/sub> for the decomposition of phosphorous pentachloride, PCl5<\/sub>, is 8.3 \u00d7 10-3<\/sup>. If the decomposition is depicted as :
                                                            \npCl5<\/sub> (g) pCl3<\/sub> (g) + Cl2<\/sub> (g) \u2206H = 124.0 kJ mol-1<\/sup>.
                                                            \na) Write an expression of Kc<\/sub> for the reaction
                                                            \nb) What is the value of Kc<\/sub> for the reverse reaction at the same temperature ?
                                                            \nc) What would be effect on Kc<\/sub> if
                                                            \n(i) more PCl5<\/sub> is added
                                                            \n(ii) pressure is increased
                                                            \n(iii) the temperature is increased.
                                                            \nSolution:
                                                            \nGiven
                                                            \nPCl5(g)<\/sub> \u21cc PCl3(g)<\/sub> + Cl2(g)<\/sub> \u2206H = 124.0 kJ\/mole
                                                            \nKC<\/sub> = 8.3 \u00d7 10-3<\/sup>
                                                            \na) KC<\/sub> = \\(\\frac{\\left[\\mathrm{PCl}_3\\right]\\left[\\mathrm{Cl}_2\\right]}{\\left[\\mathrm{PCl}_5\\right]}\\)<\/p>\n

                                                            b) Given KC<\/sub> = 8.3 \u00d7 10-3
                                                            \n<\/sup>We have to calculate KC<\/sub> for reverse reaction
                                                            \ni.e., PCl3(g)<\/sub> + Cl2(g)<\/sub> \u21cc PCl5(g)<\/sub>
                                                            \n\u2234 Kc<\/sub> for the reverse reaction = \\(\\frac{1}{8.3 \\times 10^{-3}}\\)
                                                            \n= 0.12048 \u00d7 103<\/sup>
                                                            \n= 120.48<\/p>\n

                                                            c) i) More PCl5<\/sub> is added KC<\/sub> value is decreased.
                                                            \nii) By the increase of pressure KC<\/sub> value is increased.
                                                            \niii) Given reaction is Endothermic so increase of temperature increases KC<\/sub> value.<\/p>\n

                                                            Question 10.
                                                            \nExplain the concept of Bronsted acids and Bronsted bases. Illustrate the answer with suitable examples.
                                                            \nAnswer:
                                                            \nAccording to Bronsted theory o substance which can donate a proton to the other substance is known as acid. A substance which accept a proton from other substance is a base.
                                                            \n\"AP
                                                            \nHere HCl donates a proton to water and behaves as Bronsted Lowry acid. Similarly NH3<\/sub> accepts a proton from H2<\/sub>O and acts as Bronsted Lowry base.<\/p>\n

                                                            Above reaction is a reversible reaction so that H3<\/sub>O+<\/sup> can donate proton to act as acid. Cl–<\/sup> can accept a proton to act as base. Thus each acid base reaction equilibrium involves two acids and two bases. Each pair differs by a proton, such acid base pair is called Conjugate acid base pain.
                                                            \n\"AP
                                                            \nAccording to this theory strengths of acids and bases can be explained.<\/p>\n

                                                            An acid which show great tendency to donate protons is a strong acid and an acid which shows less tendency to donate proton is a weak acid.<\/p>\n

                                                            A base which shows great tendency to accept a proton is a strong base and a base which shows less tendency to accept a proton is a weak base.<\/p>\n

                                                            \"AP<\/p>\n

                                                            Question 11.
                                                            \nExplain Lewis acid-base theory with suitable example. Classify the following species into Lewis acids and Lewis bases and show these act as Lewis acid \/ base.
                                                            \na) OH–<\/sup>
                                                            \nb) F–<\/sup>
                                                            \nc) H+<\/sup>
                                                            \nd) BCl3<\/sub>
                                                            \nAnswer:
                                                            \nLewis theory of acids and bases : G.N. Lewis proposed a more generalised theory of acids and bases. According to this theory.
                                                            \nAcid : A substance that can accept an electron pair to form a co-ordinate covalent bond is called an acid.
                                                            \nTypes of Lewis acids :
                                                            \nLewis acids are of 5 types.
                                                            \n1) All Cations : Simple cations Ag\u2295<\/sup>, CO+3<\/sup>, Cu+2<\/sup>, Fe+3<\/sup>, Al+3<\/sup> can act as Lewis acids.<\/p>\n

                                                            2) Compounds in which the central atom has an incomplete octet and possessing an empty orbital can act as Lewis acids.
                                                            \nEx : BF3<\/sub>, BCl3<\/sub>, AlCl3<\/sub>, FeCl3<\/sub>.<\/p>\n

                                                            3) Compounds in which the central atom has vacant d-orbitals and may expand its octet can act as lewis acids.
                                                            \nEx : SiF4<\/sub>, SF4<\/sub>, TeF4<\/sub>, SnCl4<\/sub>, FeCl3<\/sub>.<\/p>\n

                                                            4) Molecules having multiple bonds between atoms of dissimilar electronegativities can behave like Lewis acids.
                                                            \nEx : CO2<\/sub>, SO2<\/sub>, SO3<\/sub>, NO2<\/sub>, Cl2<\/sub>O7<\/sub>, P4<\/sub>O10<\/sub><\/p>\n

                                                            5) Elements with six electrons in the valence shell or electron sextet can act as Lewis acids.
                                                            \nEx : O, S
                                                            \nBase : According to Lewis theory a base is a substance which can donate an electron pair to form a co-ordinate covalent bond.<\/p>\n

                                                            Types of Lewis bases : Lewis bases are divided into three types.
                                                            \n1) All anions
                                                            \nEx: Cl\u2296<\/sup>, OH\u2296<\/sup>, CN\u2296<\/sup>, \\(\\mathrm{NH}_2^{\\ominus}\\), F\u2296<\/sup>, SCN\u2296<\/sup> All simple anions can act as Lewis bases.
                                                            \n2) Molecules with one or two lone pairs on the central atom can act as Lewis bases.
                                                            \nEx : \"AP
                                                            \na) Hydrenyl ion is a lewis base as it can donate an electron lone pair (:OH–<\/sup>).
                                                            \nb) Flouride ion acts as a Lewis base as it can donate any one of its four electron lone pairs.
                                                            \nc) A Proton is a lewis acid as it can accept alone pair of electrons from bases like hydrenyl ion and flowride ion.
                                                            \nd) BCl3<\/sub> acts as a lewis acid as it can accept a lane pair electrons from species like ammenia or amine molecules.<\/p>\n

                                                            \"AP<\/p>\n

                                                            Question 12.
                                                            \nWhat is degree of ionization in respect of weak acids and weak bases ? Derive the relationship between degree of ionization (\u03b1) and ionization constant (Ka<\/sub>) for the weak acid HX.
                                                            \nAnswer:
                                                            \nDegree of ionization (\u03b1):
                                                            \nThe extent of ionization (or) dissociation of a weak acid or weak base is termed as its Degree of ionization (\u03b1). Let the concentration of weak acid (HX) be ‘C moles\/lit and the degree of ionization be ‘\u03b1’. Hence at equilibrium,
                                                            \n\"AP
                                                            \nIf ‘\u03b1’ is neglected compared to ‘one’ in the denominator.
                                                            \nWe have Ka<\/sub> = C\u03b12<\/sup> \u21d2 \u03b12<\/sup> = \\(\\frac{K_a}{C} \\Rightarrow \\alpha=\\sqrt{\\frac{K_a}{C}}\\)
                                                            \nSimilar equation can be written for the ionization of weak base (BOH) in aq. solution.
                                                            \n\"AP
                                                            \nTherefore strengths of two acids or two bases are compared generally by their Ka<\/sub> or Kb<\/sub> values. Higher Ka<\/sub> or Kb<\/sub> value, stronger is the acid or base.<\/p>\n

                                                            Question 13.
                                                            \nDefine pH. What is buffer solution ? Derive Henderson – Hasselbalch equation for calculating the pH of an acid buffer solution.
                                                            \nAnswer:
                                                            \npH : “The negative value of the logarithm to the base 10, of the hydrogen ion concentration, expressed in moles \/ lit, in a solution is known as the pH of the solution”.
                                                            \nMathematically, pH = – log10<\/sub> [H+<\/sup>]
                                                            \nBuffer Solution:
                                                            \n“A buffer solution is that solution which resists any change in its pH value on dilution (or) on addition of a small amount of a strong acid or a strong base
                                                            \nEx : Acidic buffer: (CH3<\/sub>COOH + CH3<\/sub>COONa)
                                                            \nBasic buffer: (NH4<\/sub>OH + NH4<\/sub>Cl)<\/p>\n

                                                            Preparation of buffer solutions :
                                                            \n1) Acid buffer solutions : An acid buffer consists of weak acid and its salt with strong base.
                                                            \nEx : (CH3<\/sub>COOH + CH3<\/sub>COONa)
                                                            \nAcid buffer solutions are normally prepared by mixing either equal or different volumes of equimolar solutions of a weak acid and its salt.<\/p>\n

                                                            2) Base buffer solutions : A basic buffer solution consists of a mixture of a weak base and its salt with a strong acid.
                                                            \nEx : (NH4<\/sub>OH + NH4<\/sub>Cl)
                                                            \nBase buffer solutions are prepared generally, by mixing either equal or different volumes of equimolar solutions of a weak base and its salt.<\/p>\n

                                                            Derivation of Hendersen’s equation for an acid buffer :
                                                            \ni) Consider the acid buffer HA + NaA
                                                            \nHA \u21cc H+<\/sup> + A-
                                                            \nNaA \u21cc Na+<\/sup> + A–<\/sup>
                                                            \nAcid dissociation constant Ka<\/sub> = \\(\\frac{\\left[\\mathrm{A}^{+}\\right]\\left[\\mathrm{A}^{-}\\right]}{[\\mathrm{HA}]}\\)
                                                            \nHere due to common ion effect
                                                            \n\"AP<\/p>\n

                                                            Question 14.
                                                            \nExplain the term “Hydrolysis of salts” with examples. Discuss the pH of the following types of salt solutions.
                                                            \n(i) Salts of weak acid and strong base.
                                                            \n(ii) Salts of strong acid and weak base. [T.S. Mar. 16]
                                                            \nSolution:
                                                            \nIt may be defined as a phenomenon in which the anion or cation or both of a salt react with water producing excess of OH\u2296<\/sup> ions or H\u2295<\/sup> ions or both in aq. solution”.
                                                            \ni) -4 Given salt CH3<\/sub>COONa is obtained by the reaction of weak acid (CH3<\/sub>COOH) and strong base (NaOH).
                                                            \n-4 The ionisation CH3<\/sub>COONa as follows
                                                            \nCH3<\/sub>COONa \u21cc CH3<\/sub>COO–<\/sup> + Na+<\/sup>
                                                            \nThe above salt undergo anionic hydrolysis
                                                            \nCH3<\/sub>COO–<\/sup> + H2<\/sub>O \u21cc CH3<\/sub>COOH + OH–<\/sup>
                                                            \nThe aq solution of salt is basic due to greater OH–<\/sup> ion concentration than H+<\/sup> ion concentration.
                                                            \n\u2234 PH<\/sup> > 7<\/p>\n

                                                            ii) -4 Given salt NH4<\/sub>;C3<\/sub> is obtained by the reaction of strong acid (HCl) and weak base (NH4<\/sub>OH).
                                                            \n-4 The ionisation of NH4<\/sub>Cl as follows
                                                            \n\\(\\mathrm{NH}_4^{+}\\) + Cl–<\/sup> \u21cc NH4<\/sub>Cl
                                                            \n-4 The above salt under goes catioric hydrolysis
                                                            \n\\(\\mathrm{NH}_4^{+}\\) + H2<\/sub>O \u21cc NH4<\/sub>OH + H+<\/sup>
                                                            \n\u2234 The equation solution of salt is acidic due to greater H+<\/sup> ion concentration than OH–<\/sup> ion concentration.
                                                            \n\u2234 PH < 7<\/p>\n

                                                            \"AP<\/p>\n

                                                            Question 15.
                                                            \nWhat is solubility product ? Explain the common ion effect on solubility of ionic salts.
                                                            \nSolution:
                                                            \nSolubility product (KSP<\/sub>) : The product of the concentrations of the cation and the anion in a saturated solution of a salt at room temperature is called solubility product (KSP<\/sub>).
                                                            \nKSP<\/sub> = [Mn+<\/sup>][An-<\/sup>]
                                                            \nSignificance of solubility product in chemical analysis :
                                                            \nIn any solution of a salt, if the product of [Mn+<\/sup>] and [An-<\/sup>] is greater than the solubility product (KSP<\/sub>) of the salt, then only the salt gets precipitated. For example,<\/p>\n

                                                            i) If the product of [Cu2+<\/sup>] & [S-2<\/sup>] in the II group (in presence of acid) of chemical analysis exceeds the value of KSP<\/sub> of CuS, then only CuS will be precipitated in the group. Similar argument applies to other metal ions in the group namely (Cd2+<\/sup>, Bi3+<\/sup>, Hg2+<\/sup>, Sb2+<\/sup>, etc).<\/p>\n

                                                            ii) If the product of [Zn2+<\/sup>] and [S2-<\/sup>] in ammonical medium in IV group of chemical analysis exceeds the value of Ksp<\/sub> of ZnS, it will be precipitated. Similar arguments apply to Mn2+<\/sup>, Ni2+<\/sup> etc.
                                                            \n\"AP<\/p>\n

                                                             <\/p>\n

                                                              \n
                                                            • Given salt CH3<\/sub>COONa is obtained by the reaction of weak acid (CH3<\/sub>COOH) and strong base (NaOH).<\/li>\n
                                                            • The ionisation CH3<\/sub>COONa as follows
                                                              \nCH3<\/sub>COONa \u21cc CH3COO–<\/sup> + Na+<\/sup><\/li>\n
                                                            • The above salt undergo anionic hydrolysis
                                                              \nCH3<\/sub>COO–<\/sup> + H2<\/sub>O \u21cc CH3<\/sub>COOH + OH–<\/sup><\/li>\n<\/ul>\n

                                                              \u2234 The aq solution of salt is basic due to greater OH–<\/sup> ion concentration than H+<\/sup> ion concentration.
                                                              \n\u2234 PH<\/sup> > 7<\/p>\n

                                                              Given salt NH4<\/sub>Cl is obtained by the reaction of strong acid (HCl) and weak base (NH4<\/sub>OH).
                                                              \nThe ionisation of NH4<\/sub>Cl as follows
                                                              \n\\(\\mathrm{NH}_4^{+}\\) + Cl–<\/sup> \u21cc NH4<\/sub>Cl
                                                              \nThe above salt under goes catioric hydrolysis
                                                              \n\\(\\mathrm{NH}_4^{+}\\) + H2<\/sub>O \u21cc NH4<\/sub>OH + H+<\/sup>
                                                              \n\u2234 The equation solution of salt is acidic due to greater H+<\/sup> ion concentration than OH–<\/sup> ion concentration.
                                                              \n\u2234 PH < 7<\/p>\n

                                                              Common ion effect on solubility of ionic salts
                                                              \nThe common ion effect is used for almost complete precipitation of a particular ion as its sparingly soluble salt, with every low value of solubility product for gravimetric estimation. Thus we can precipitate silver ion as silver chloride, ferric ion as its hydroxide (or hydrated ferric oxide) and barium ion as its sulphate for quantitative estimations.<\/p>\n

                                                              \"AP<\/p>\n

                                                              Question 16.
                                                              \nWrite notes on
                                                              \n(i) Common ion effect.
                                                              \n(ii) The relation between Ksp<\/sub> and solubility (S) of a spaningly soluble salt BaSO4<\/sub>.
                                                              \nSolution:
                                                              \ni) Common ion effect : The decrease in the ionisation (dissociation) of a weak electrolyte, by the addition of a strong electrolyte having an ion common with the weak electrolyte, is known as common ion effect.
                                                              \nExample : The dissociation of NH4<\/sub>OH is diminished by the addition of NH4<\/sub>Cl due to the common ion, NH4<\/sub>+<\/sup> ion
                                                              \nNH4<\/sub>OH \u21cc NH4<\/sub>\u2295<\/sup> + OH\u2296<\/sup>
                                                              \nNH4<\/sub>OH \"AP NH4<\/sub>+<\/sup> + OH\u2296<\/sup> (addition of NH4<\/sub>Cl)
                                                              \nImportance of this effect In qualitative analysis :<\/p>\n

                                                                \n
                                                              1. This principle is used in the systematic qualitative analysis of cations.<\/li>\n
                                                              2. The concentration of S2-<\/sup> in II group or OH–<\/sup> ion in III group of chemical analysis is controlled by HCl and NH4<\/sub>OH on the basis of common ion effect.
                                                                \nH+<\/sup> ion is common to H2<\/sub>S (II group)
                                                                \nNH4<\/sub>+<\/sup> ion is common to NH4<\/sub>OH (IV group) .<\/li>\n
                                                              3. The common ion effect principle is also used in controlling the H+<\/sup> ion concentration in buffer solutions.<\/li>\n
                                                              4. It is also used in the purification of common salts by passing dry HCl gas into impure salt solution.<\/li>\n<\/ol>\n

                                                                ii) Relation between KSP<\/sub> and ‘S’ of salt at BaSO4<\/sub>
                                                                \n\"AP<\/p>\n

                                                                Problems<\/span><\/p>\n

                                                                Question 1.
                                                                \n1 Mole of PCl5<\/sub> is heated in a closed vessel of 1 litre capacity. At equilibrium 0.4 moles of chlorine is found. Calculate the equilibrium Constant.
                                                                \nSolutioni:
                                                                \nPCl5(g)<\/sub> \u21cc PCl3(g)<\/sub> + Cl2(g)<\/sub>
                                                                \nInitial [PCl5<\/sub>] = 1 mole\/lit
                                                                \nAt equilibrium [Cl2<\/sub>] = 0.4 moles\/lit
                                                                \n[PCl3<\/sub>] at equilibrium = 0.4 moles\/lit
                                                                \n[PCl5<\/sub>] at equilibrium = 0.6 moles\/lit
                                                                \nKC<\/sub> = \\(\\frac{\\left[\\mathrm{PCl}_3\\right]\\left[\\mathrm{Cl}_2\\right]}{\\left[\\mathrm{PCl}_5\\right]}\\)
                                                                \n= \\(\\frac{0.4 \\times 0.4}{0.6}=\\frac{0.16}{0.6}\\)
                                                                \n= 0.2606<\/p>\n

                                                                Question 2.
                                                                \nNitrogen dioxide forms dinitrogen tetroxide according to the equation 2 NO2<\/sub> (g) \u21cc N2<\/sub>O4<\/sub> (g) when 0.1 mole of NO2<\/sub> as added to a 1 litre flask at 25\u00b0C, the concentration changes so that at equilibrium [NO2<\/sub>] = 0.016M and [N2<\/sub>O4<\/sub>] = 0.042 M
                                                                \na. What is the value of the reaction Quotient before any reaction occurs.
                                                                \nb. What is the value of the equilibrium constant for the reaction.
                                                                \nSolution:
                                                                \nGiven equation
                                                                \n2NO2(g)<\/sub> \u21cc N2<\/sub>O4(g)<\/sub>
                                                                \na) The value of reaction quotient before any reaction occurs = 0<\/p>\n

                                                                b) KC<\/sub> = \\(\\frac{\\left[\\mathrm{N}_2 \\mathrm{O}_4\\right]}{\\left[\\mathrm{NO}_2\\right]^2}\\)
                                                                \n= \\(\\frac{0.042}{(0.016)^2}=\\frac{0.042}{0.000256}\\)
                                                                \n= 164.06
                                                                \n= 1.64 \u00d7 102<\/sup><\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 3.
                                                                \nThe equilibrium constant for the reaction:
                                                                \nN2<\/sub>(g) + 3H2<\/sub>(g) \u21cc 2 NH3<\/sub>(g) at 725K is 6.0 \u00d7 10-2<\/sup>. At equilibrium, [H2<\/sub>] = 0.25 mol L-1<\/sup> and [NO3<\/sub>] = 0.06 mol L-1<\/sup>. Calculate the equilibrium concentration of N2<\/sub>.
                                                                \nSolution:
                                                                \nGiven
                                                                \nN2(g)<\/sub> + 3H2(g)<\/sub> \u21cc 2NH3(g)<\/sub>
                                                                \nKC<\/sub> = 6 \u00d7 10-2<\/sup>
                                                                \n[H2<\/sub>] = 0.25 moles\/lit
                                                                \n[NH3<\/sub>] = 0.06 moles\/lit
                                                                \nKC<\/sub> = \\(\\frac{\\left[\\mathrm{NH}_3\\right]^2}{\\left[\\mathrm{~N}_2\\right]\\left[\\mathrm{H}_2\\right]^3}\\)
                                                                \n6 \u00d7 10-2<\/sup> = \\(\\frac{(0.06)^2}{\\left(N_2\\right)(0.25)^3}\\)
                                                                \n[N2<\/sub>] = \\(\\frac{(0.06)^2}{6 \\times 10^{-2} \\times(0.25)^3}\\)
                                                                \n= \\(\\frac{0.0036}{0.015625 \\times 6 \\times 10^{-2}}\\)
                                                                \n= \\(\\frac{0.0036}{0.09375 \\times 10^{-2}}\\)
                                                                \n= 3.84 moles\/lit<\/p>\n

                                                                Question 4.
                                                                \nAt certain temperature, Kc for the re action.
                                                                \nSO2<\/sub> (g) + NO2<\/sub> (g) \u21cc SO3<\/sub> (g) + NO (g)
                                                                \nis 16. If intially one mole each of all the four gases are taken in one litre vessel. What are the equilibrium con centrations of NO and NO2<\/sub>.
                                                                \nSolution:
                                                                \nGiven equations is
                                                                \n\"AP
                                                                \nx = 8 – 4x
                                                                \n5x = 8
                                                                \nx = 1.6 moles\/lit \u2234 (NO] = 1.6 moles\/lit
                                                                \n2 – x = 2 – 1.6 = 0.4 moles\/lit
                                                                \n[NO2<\/sub>] = 0.4 moles\/lit<\/p>\n

                                                                Question 5.
                                                                \nUnder certain conditions, the equilibrium constant for the decomposition of PCl5<\/sub> (g) into PCl3<\/sub> (g) and Cl2<\/sub> (g) is 00211 mol L-1<\/sup>. What are the equilibrium concentrations of PCl5<\/sub>, PCl3<\/sub> and Cl2<\/sub>, if the intial concentration of PCl5<\/sub> was 1.00 M?
                                                                \nSolution:
                                                                \nPCl5(g)<\/sub> \u21cc PCl3<\/sub> + Cl2(g)<\/sub>
                                                                \n\"AP
                                                                \n0.0211 (1 – x) = x2<\/sup>
                                                                \n0.0211 – 0.0211 x = x2<\/sup>
                                                                \nx = 0.14 M
                                                                \n\u2234 1 – x = 1 – 0.14
                                                                \n= 0.86 M
                                                                \n\u2234 1 [PCl5<\/sub>] = 0.86 M
                                                                \n\u2234 [PCl3<\/sub>] = 0.86 M
                                                                \n[PCl3<\/sub>] = [Cl2<\/sub>] = 0.14 M<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 6.
                                                                \nFor the reaction A + B \u21cc 3C at 25\u00b0C, a 3 litre vessel contains 1, 2, 4 mole of A, B and C respectively predict the direction of reaction if
                                                                \na. Kc<\/sub> for the reaction is 10
                                                                \nb. Kc<\/sub> for the reaction is 15
                                                                \nc. Kc<\/sub> for the reaction is 10.66
                                                                \nAnswer:
                                                                \nA + B \u21cc 3C at 25\u00b0C
                                                                \n\"AP
                                                                \na) Given Kc<\/sub> = 10
                                                                \nKc<\/sub>< Qc<\/sub>
                                                                \n\u2234 Backward reaction takes place<\/p>\n

                                                                b) Given Kc<\/sub> = 15
                                                                \nKc<\/sub> > Qc<\/sub>
                                                                \n\u2234 Forward reaction takes place<\/p>\n

                                                                c) Given Kc<\/sub> = 10.66 Q = Kc<\/sub>
                                                                \n\u2234 Represents equilibrium state<\/p>\n

                                                                Question 7.
                                                                \nA mixture of H2<\/sub>, N2<\/sub> and NH3<\/sub> with molar concentrations 5.0 \u00d7 10-3<\/sup> mol L-1<\/sup>, 4.0 \u00d7 10-3<\/sup> mol L-1<\/sup> and 2.0 \u00d7 10-3<\/sup> mol L-1<\/sup> respectively was prepared and heated to 500K. The value of Kc<\/sub> for the reaction : 3H2<\/sub> (g) + N2<\/sub> (g) \u21cc 2NH3<\/sub> (g) at this temperature is 60. Predicts whether ammonia tends to form or decompose at this stage of concentration.
                                                                \nAnswer:
                                                                \n\"AP
                                                                \nGiven Kc<\/sub> = 60
                                                                \nKc<\/sub> < Qc<\/sub>
                                                                \n\u2234 Backward reaction i.e decomposition of NH3<\/sub> takes place<\/p>\n

                                                                Question 8.
                                                                \nAt 500 K, Kp<\/sub> value for the reaction
                                                                \n2 SO2<\/sub> (g) + O2<\/sub> (g) \u21cc 2SO3<\/sub> (g) is 2.5 \u00d7 1010<\/sup>.
                                                                \nFind the value of Kp<\/sub> for each of following reactions at the same temperature.
                                                                \na. SO2<\/sub>(g) + 1\/2O2<\/sub>(g) \u21cc SO3<\/sub> (g)
                                                                \nb. SO3<\/sub>(g) \u21cc SO2<\/sub>(g) + 1\/2O2<\/sub>(g)
                                                                \nc. 3SO2<\/sub>(g) + 3\/2 O2<\/sub>(g) \u21cc 3SO3<\/sub>(g)
                                                                \nSolution:
                                                                \n2SO2(g)<\/sub> + O2(g)<\/sub> \u21cc 2SO3(g)<\/sub>
                                                                \n\"AP
                                                                \n\"AP<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 9.
                                                                \nKc<\/sub> for the reaction N2<\/sub>O4<\/sub> (g) \u21cc 2NO2<\/sub> (g) is 4.63 \u00d7 10-3<\/sup> at 25\u00b0C.
                                                                \na. What is the value of K at this termperature
                                                                \nb. At 25\u00b0C, if the partial pressure of N2<\/sub>O4<\/sub> (g) at equilibrium is 0.2 atm1 calculate equilibrium pressure of NO2<\/sub>(g).
                                                                \nSolution:
                                                                \na) Given
                                                                \nNa2<\/sub>O4(g)<\/sub> \u21cc 2NO2(g)<\/sub>
                                                                \nkc<\/sub> = 4.63 \u00d7 10-3<\/sup>
                                                                \nkp<\/sub> = kc<\/sub> (RT)\u2206n
                                                                \n= 4.63 \u00d7 10-3<\/sup> \u00d7 0.0821 \u00d7 298 [\u2206n = 1]
                                                                \n= 113.27 \u00d7 10-3<\/sup>
                                                                \n= 0.1132<\/p>\n

                                                                b) kp<\/sub> = \\(\\frac{\\mathrm{P}_{\\mathrm{NO}_2{ }^2}}{\\mathrm{P}_{\\mathrm{N}_2 \\mathrm{O}_4}}\\)
                                                                \n0.1132 = \\(\\frac{\\mathrm{P}_{\\mathrm{NO}_2{ }^2}}{0.2}\\)
                                                                \nPNO2<\/sub>2<\/sup> = 0.2264
                                                                \nPNO2<\/sub> = 0.15 atm<\/p>\n

                                                                Question 10.
                                                                \nAt 27\u00b0C, Kp<\/sub> value for the reversible reaction PCl5<\/sub>(g) PCl5<\/sub>(g) + Cl2<\/sub>(g) is 0.65, calculate Kc<\/sub>
                                                                \nSolution:
                                                                \nPCl5(g)<\/sub> \u21cc PCl3(g)<\/sub> +Cl2(g)<\/sub>
                                                                \nGiven kp<\/sub> = 0.65 .
                                                                \nkp<\/sub> = k<subC (RT)\u2206n<\/sup>
                                                                \n0.65 = k<subC 0.082 1 \u00d7 300 [T =27\u00b0C = 300K; \u2206n = 1]
                                                                \nk<subC = \\(\\frac{0.65}{24.63}\\) = 0.02639<\/p>\n

                                                                Question 11.
                                                                \nKc<\/sub> for the reaction, N2<\/sub> (g) + 3H2<\/sub> (g) 2NH3<\/sub>(g) is 0.5 at 400K, find Kp<\/sub>.
                                                                \nSolution:
                                                                \nN2(g)<\/sub> + 3 H2(g)<\/sub> \u21cc 2 NH3(g)<\/sub>
                                                                \nGiven kC<\/sub> = 0.5
                                                                \nT = 400 K
                                                                \nkp<\/sub> = kc<\/sub> (RT)\u2206n<\/sup>
                                                                \nkp<\/sub> = 0.5 \u00d7 (0.0821 \u00d7 400)-2<\/sup> \u2206n = -2
                                                                \nkp<\/sub> = 0.5 \u00d7 (8.21 \u00d7 4)-2<\/sup> = 0.5 \u00d7 (32.84)-2<\/sup>
                                                                \n\\(\\frac{0.5}{1078.46}\\) = 4.63 \u00d7 10-4<\/sup><\/p>\n

                                                                Question 12.
                                                                \n1 mole of A and 1 mole of B are taken in a 5 litre flask, 0.5 mole of C is formed in the equilibrium of
                                                                \nA + B \u21cc C + D.
                                                                \nWhat is molar concentration of each species if the reaction is carried with 2 mole of A 1 mole of B in a 5 litre flask at the same temperature.
                                                                \nSolution:
                                                                \n\"AP
                                                                \n= 0.2 – 0.133 = 0.067 M
                                                                \n[C] = [D] = 0.133 M<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 13.
                                                                \nFor the following reaction
                                                                \nPCl5<\/sub>(g) \u21cc PCl3<\/sub>(g) + Cl2<\/sub>(g) 0.4 mole of PCl5<\/sub> 0.2 mole of PCl3<\/sub> and 0.6 mole of Cl2<\/sub> are taken in a 1 litre flask. If Kc<\/sub> = 0.2, Predict the direction in which reaction proceeds.
                                                                \nSolution:
                                                                \nGiven
                                                                \nPCl(g) \u21cc PCl3(g)<\/sub> + Cl2<\/sub> (g)
                                                                \nQC<\/sub> = \\(\\frac{\\left[\\mathrm{PCl}_3\\right]\\left[\\mathrm{Cl}_2\\right]}{\\left[\\mathrm{PCl}_5\\right]}\\)
                                                                \n= \\(\\frac{0.2 \\times 0.6}{0.4}\\)
                                                                \n= 0.03
                                                                \nGiven KC<\/sub> = 0.2
                                                                \nQC<\/sub> < KC<\/sub>
                                                                \n\u2234 Reaction proceds in forward direction.<\/p>\n

                                                                Question 14.
                                                                \nIn an equilibrium A + B C + D; A and B are mixed in a vessel at temperature T. The initial concentration of A was twice the initial concentration of B. After the attainment of equilibrium, concentration of C was thrice concentration of B, calculate Kc<\/sub>.
                                                                \nSolution:
                                                                \nGiven
                                                                \nA + B \u21cc C + D
                                                                \nThe initial concentrations of A was twice the initial concentration of B and after the attaining of equilibrium concentration of ‘C’ was thrice the concentration of ‘B’.
                                                                \n\"AP<\/p>\n

                                                                Question 15.
                                                                \nA mixture of SO2<\/sub>, SO3<\/sub> and O2<\/sub> gases are maintained at equilibrium in 10 litre flast at a temperature at which Kc<\/sub> for the reactin 2SO2<\/sub>(g) + O2<\/sub>(g) \u21cc 2SO3<\/sub>(g) is 100. At equilibrium.
                                                                \na. If no. of moles of SO3<\/sub> and SO2<\/sub> inflask are same, how many moles of O2<\/sub> are present.
                                                                \nb. If no. of moles of SO3<\/sub> in flask is twice the no. of moles SO2<\/sub>, how many moles of O2<\/sub> are present.
                                                                \nSolution:
                                                                \na) Given
                                                                \n2 SO2(g)<\/sub> + O2(g)<\/sub> \u21cc 2 SO3(g)<\/sub>
                                                                \nGiven KC<\/sub> = 100<\/p>\n

                                                                a) Given no.of moles of SO3<\/sub> = No. of moles of SO2<\/sub>
                                                                \n\"AP
                                                                \ny = \\(\\frac{4}{10}\\) = 0.4
                                                                \n\u2234 0.4 moles of O2<\/sub> present.<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 16.
                                                                \nFor A + B \u21cc C, the equilibrium concentrations of A and B at a temperature are 15 mol L-1<\/sup>. When volume is doubled the reaction has euilibrium concentration of A as 10 mol L-1<\/sup>. Calculate.
                                                                \na) Kc<\/sub>
                                                                \nb) Concentration of C in original equilibrium
                                                                \nSolution:
                                                                \na) Given
                                                                \nA + B \u21cc C + D
                                                                \na) KC<\/sub>
                                                                \nb) Concentration of C in original equilibrium<\/p>\n

                                                                a) KC<\/sub> = \\(\\frac{[\\mathrm{C}]}{[\\mathrm{A}][\\mathrm{B}]}\\)
                                                                \n[A] = 10 moles\/lit
                                                                \n[B] = 10 moles\/lit
                                                                \n[CI = 20 moles\/lit
                                                                \n= \\(\\frac{20}{10 \\times 10}\\)
                                                                \n= \\(\\frac{20}{100}\\) = 0.2<\/p>\n

                                                                b) KC<\/sub> = \\(\\frac{[\\mathrm{C}]}{[\\mathrm{A}][\\mathrm{B}]}\\)
                                                                \n[AI = [B] = 15 moles\/lit
                                                                \n0.2 = \\(\\frac{[C]}{15 \\times 15}\\)
                                                                \n[C] = 225 \u00d7 0.2
                                                                \n= 45 M<\/p>\n

                                                                Question 17.
                                                                \nA vessel at 100 K contains CO2<\/sub> with a pressure of 0.5 atm. Some of the CO2<\/sub> is converted into CO on addition of graphite. Calculate the value of K, if total pressure at equilibnum is 0.8 atm.
                                                                \nSolution:
                                                                \n\"AP
                                                                \n0.5 – x + 2x = 0.8
                                                                \nx = 0.3
                                                                \n\u2234 PCO<\/sub> = 0.2 atm
                                                                \nPCO2<\/sub> = 0.6 atm
                                                                \nK = \\(\\frac{\\mathrm{P}_{\\mathrm{CO}_2^2}}{\\mathrm{P}_{\\mathrm{CO}}}\\)
                                                                \n= \\(\\frac{0.36}{0.2}\\) = 1.8<\/p>\n

                                                                Question 18.
                                                                \nThe K1, va)ue for the reaction H2<\/sub>(g)+ I2<\/sub>(g) \u21cc 2 HI(g) at 460\u00b0C is 49. If the initial pressure of H2<\/sub> and I2<\/sub> are 0.5 atm respectively, determine the partial pressure of each gases a equilibrium.
                                                                \nSolution:
                                                                \n\"AP
                                                                \nx = \\(\\frac{3.5}{9}\\)
                                                                \n= 0.3888
                                                                \n\u2234 PHI<\/sub> = 2 \u00d7 0.3888
                                                                \n= 0.7788 atm
                                                                \nPH2<\/sub> = 0.5 – 0.388
                                                                \n= 0.111 atm
                                                                \nPI2<\/sub> = 0.5 – 0.3888
                                                                \n= 0.111 atm<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 19.
                                                                \n0.5 mol of H2<\/sub> and 0.5 mole of I2<\/sub> react in 10 litre flask at 448\u00b0C. The equilibrium constant Kc<\/sub> is 50 for
                                                                \nH2<\/sub>(g) + I2<\/sub>(g) \u21cc 2HI(g).
                                                                \na) What is the value of Kp<\/sub>
                                                                \nb) Calculate mole of I2<\/sub> at equilibrium.
                                                                \nSolution:
                                                                \n\"AP
                                                                \n\u2234 No of moles of I2<\/sub> = 0.5 – 0.3888
                                                                \n= 0.111 moles<\/p>\n

                                                                Question 20.
                                                                \nHow much PCl5<\/sub> must be added to a one little vesel at 250\u00b0C in order to obtain a concentration of 0.1 mole of Cl2<\/sub> at equilibrium. Kc<\/sub> for
                                                                \nPCl5<\/sub>(g) \u21cc PCl3<\/sub>(g) + Cl2<\/sub>(g) is 0.0414 M.
                                                                \nSolution:
                                                                \n\"AP
                                                                \n\u2234 KC<\/sub> = \\(\\frac{\\left[\\mathrm{PCl}_3\\right]\\left[\\mathrm{Cl}_2\\right]}{\\left[\\mathrm{PCl}_5\\right]}\\)
                                                                \n0.0414 = \\(\\frac{0.1 \\times 0.1}{x-0.0414}\\)
                                                                \n0.0414 x – 0.00414 = 0.01
                                                                \n0.0414 x = 0.01 + 0.00414
                                                                \nx = 0.34154 moles<\/p>\n

                                                                Question 21.
                                                                \nK for the reaction
                                                                \nN2<\/sub>(g) + 3H2<\/sub>(g) \u21cc 2NH3<\/sub>(g) at 400\u00b0C is 1.64 \u00d7 10-4<\/sup>.
                                                                \na) Calculate Kc<\/sub>
                                                                \nb) Calculate \u2206G\u00b0 value using Kc<\/sub> value.
                                                                \nSolution:
                                                                \nGiven equation N2(g)<\/sub> + 3H2(g)<\/sub> \u21cc 2NH3(g)<\/sub>
                                                                \na) Kp<\/sub> = Kc<\/sub> (RT)\u2206n<\/sup>
                                                                \n\u2206n = 2 – (4)
                                                                \n= -2
                                                                \n1.64 \u00d7 10-4<\/sup> = Kc<\/sub> (0.0821 \u00d7 673)-2<\/sup>
                                                                \nKc<\/sub> = 1.64 \u00d7 (55.2533)2<\/sup> \u00d7 10-4<\/sup>
                                                                \nKc<\/sub> = 0.5006<\/p>\n

                                                                b) \u2206G\u00b0 = -2.303 RT log K
                                                                \n= -2.303 \u00d7 0.0821 \u00d7 673 \u00d7 log 0.5005
                                                                \n= 3874 J<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 22.
                                                                \nCalculate the pH of
                                                                \na) 10-3<\/sup> M HCl
                                                                \nb) 10-3<\/sup> M H2<\/sub>SO4<\/sub>
                                                                \nc) 10-6<\/sup> M HNO3<\/sub>
                                                                \nd) 0.02 M H2<\/sub>SO4<\/sub>
                                                                \nSolution:
                                                                \na) 10-3<\/sup> M HCl
                                                                \npH = -log (H+)
                                                                \n= – log 10-3<\/sup>
                                                                \n= 3<\/p>\n

                                                                b) 10-3<\/sup> M H2<\/sub>SO4<\/sub>
                                                                \npH = – log10<\/sub> [H+<\/sup>]
                                                                \n= 0.001 \u00d7 2
                                                                \n= 0.002
                                                                \npH = – log 0.002
                                                                \n= – log2<\/sup> \u00d7 10-3<\/sup>
                                                                \n= – log2<\/sup> – log 10-3<\/sup>
                                                                \n= 3 – log2<\/sup>
                                                                \n= 3 – 0.3010
                                                                \n= 2.699<\/p>\n

                                                                c) 10-6<\/sup> M HNO3<\/sub>
                                                                \n0.0008 M Ba(OH)2<\/sub>)
                                                                \npOH = – log [OH–<\/sup>]
                                                                \n[OH–<\/sup>] = 0.0008 \u00d7 2
                                                                \n= 0.0016 N
                                                                \n\u2234 pOH = – log 0.0016
                                                                \n= – log 16 \u00d7 10-4<\/sup>
                                                                \n= – log 16 + 4 log 10
                                                                \n= – log24<\/sup> + 4
                                                                \n= – 4 log 2 + 4 = 2.796
                                                                \npH + pOH = 14
                                                                \npH = 14 – 2.796
                                                                \npH = 11.204<\/p>\n

                                                                d) 0.02 M H2<\/sub>SO4<\/sub>
                                                                \npH = -log10<\/sub>(H+<\/sup>)
                                                                \n[H+<\/sup>] = 0.02 \u00d7 2
                                                                \n= 0.04N
                                                                \npH = – log 0.04
                                                                \n= -log 4 \u00d7 102<\/sup>
                                                                \n= -log4 – log 10-2<\/sup>
                                                                \n= 2 – log4
                                                                \n= 1.3010<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 23.
                                                                \nCalculate of pH for
                                                                \na) 0.001 M NaOH
                                                                \nb) 0.01 M Ca (OH)2<\/sub>
                                                                \nc) 0.0008 M Ba(OH)2<\/sub>
                                                                \nd) 0.004M NaOH
                                                                \nSolution:
                                                                \na) 0.001 M NaOH
                                                                \npOH = -log[OH–<\/sup>]
                                                                \npOH = – log (0.001)
                                                                \n= – log 10-3<\/sup>
                                                                \n= 3
                                                                \npH = 14 – pOH = 14 – 3 = 11<\/p>\n

                                                                b) 0.01 M Ca (OH)2<\/sub>
                                                                \npOH = -log [OH–<\/sup>]
                                                                \n[OH–<\/sup>] = 0.01 \u00d7 2
                                                                \n= 0.02 N
                                                                \npOH = – log 0.02
                                                                \npOH = -log 2 \u00d7 10-2<\/sup>
                                                                \npOH = – log2 + 2 log 10
                                                                \n= 2 – 0.3010
                                                                \n= 1.699
                                                                \n\u2234 pH = 14 – pOH
                                                                \n= 14 – 1.699 = 12.301<\/p>\n

                                                                c) 00008M Ba(OH)2<\/sub>
                                                                \npOH = -log [OW]
                                                                \n[OH–<\/sup>] = 0.0008 \u00d7 2
                                                                \n= 0.0016N
                                                                \n\u2234 pOH = – log 0.0016
                                                                \n= -log 16 \u00d7 10
                                                                \n= -log 16 + 4log 10
                                                                \n= – log 24<\/sup> + 4
                                                                \n= -4 log 2 + 4 = 2.796
                                                                \npH + pOH = 14
                                                                \npH = 14 – 2.796
                                                                \npH = 11.204
                                                                \n10-6<\/sup>M HNO3
                                                                \nPH = -log [H+<\/sup>]
                                                                \n= -log 10-6<\/sup>
                                                                \n= 6 log 10
                                                                \n= 6<\/p>\n

                                                                d) 0.004M NaOH
                                                                \npOH = – log[OH–<\/sup>]
                                                                \n= – log 0.004
                                                                \n= – log 4 \u00d7 10-3<\/sup>
                                                                \n= 3 – log22<\/sup>
                                                                \n= 3 – 0.6020
                                                                \n= 2.398
                                                                \npH = 14 – POH
                                                                \n= 14 – 2.398
                                                                \n= 11.602<\/p>\n

                                                                Question 24.
                                                                \nThe pH of a solution is 3.6 Calculate H3<\/sub>O+<\/sup> ion concentration.
                                                                \nSolution:
                                                                \npH = – log [H+<\/sup>]
                                                                \nlog [H+<\/sup>] = – 3.6 (or) \\(\\overline{4} .4000\\)
                                                                \n[H+<\/sup>] = anti log of \\(\\overline{4} .4\\)
                                                                \n= 2.512 \u00d7 10-4<\/sup>.<\/p>\n

                                                                Question 25.
                                                                \nThe pH of a solution is 8.6 calculate the OH ion concentration.
                                                                \nSolution:
                                                                \npH = 8.6
                                                                \n\u2234 pOH = 14 – pH
                                                                \n= 14 – 8.6
                                                                \n= 5.4
                                                                \n\u2234 pOH = – log [OH–<\/sup>]
                                                                \nlog [OH–<\/sup>] = -5.4 (or) \\(\\overline{6} .6000\\)
                                                                \n[OH–<\/sup>] = anti log of \\(\\overline{6} .6000\\)
                                                                \n= 3.981 \u00d7 10-6<\/sup><\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 26.
                                                                \nWhat is [H+<\/sup>] for a solution in which
                                                                \na) pH = 3
                                                                \nb) pH = 4.75
                                                                \nc) pH = 4.4
                                                                \nSolution:
                                                                \na) Given
                                                                \npH = 3
                                                                \n\u2234 pH = – log [H+<\/sup>]
                                                                \nlog [H+<\/sup>] = -3
                                                                \n\u2234 [H+<\/sup>] = 10-3<\/sup> M<\/p>\n

                                                                b) Given
                                                                \npH = 4.75
                                                                \nPH = – log [H+<\/sup>]
                                                                \nlog [H+<\/sup>+] = – 4.75 (or) \\(\\overline{5} .2500\\)
                                                                \n[H+<\/sup>] = anti log of 5.2500
                                                                \n= 1.77 \u00d7 10-5<\/sup> M<\/p>\n

                                                                c) Given
                                                                \npH = 4.4
                                                                \nPH = – log [H+<\/sup>]
                                                                \nlog [H+<\/sup>] = – 4.4 (or) \\(\\overline{5} .6\\)
                                                                \n[H+<\/sup>] = anti log of \\(\\overline{5} .6\\)
                                                                \n= 2.512 \u00d7 10-6<\/sup> M<\/p>\n

                                                                Question 27.
                                                                \nA solution of 0.005 M H2<\/sub>SO4<\/sub> is diluted 100 times. Calculate the pH of diluted solution.
                                                                \nSolution:
                                                                \n0.005 M H2<\/sub>SO4<\/sub>
                                                                \n[H+<\/sup>] = 0.005 \u00d7 2 = 0.01
                                                                \nIt is diluted to 100 times
                                                                \n[H+<\/sup>] = \\(\\frac{0.01}{100}\\) = 0.0001
                                                                \npH = – log [H+<\/sup>]
                                                                \n= – log 0.0001
                                                                \n= – log 10-4<\/sup> = 4<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 28.
                                                                \nA solution of HCl has a pH = 3. If one ml of it is diluted to 1 litre, what will be the pH of the resulting solution ?
                                                                \nSolution:
                                                                \nGiven
                                                                \npH = 3 for HCZ solution
                                                                \n\u2234 [H+<\/sup>] = 10-3<\/sup> M
                                                                \nIt is diluted to litre
                                                                \n\u2234 [H+<\/sup>] = \\(\\frac{10^{-3}}{10^3}\\) = 10-6<\/sup>
                                                                \n\u2234 pH = – log [H+<\/sup>]
                                                                \n= – log 10-6<\/sup> = 6<\/p>\n

                                                                Question 29.
                                                                \nWhat is the pH of 10-8<\/sup> M HCl ?
                                                                \nSolution:
                                                                \nThe given acid solution is very dilute. Hence the H+<\/sup> obtained from the acid and water must be taken into consideration.
                                                                \nH+<\/sup> ion concentration of acid is 10-8<\/sup> M
                                                                \nH+<\/sup> ion concentration of water is 10-7<\/sup> M
                                                                \nTotal [H+<\/sup>] = 10-8<\/sup> + 10-7<\/sup>
                                                                \n= 1.1 \u00d7 10-7<\/sup> or 1.1 \u00d7 10-7<\/sup> M
                                                                \n\u2234 pH = – log 1.1 \u00d7 10-7<\/sup>
                                                                \n= 7 – log 1.1 = 7 – 0.0414 = 6.995<\/p>\n

                                                                Question 30.
                                                                \nCalculate the pH of the following basic solutions.
                                                                \na. [OH–<\/sup>] = 0.05 M
                                                                \nb. [OH–<\/sup>] = 2 \u00d7 10-4<\/sup> M
                                                                \nSolution:
                                                                \na) [OH–<\/sup>] = 0.05 M
                                                                \npOH = – log (0.05)
                                                                \n= – log 5 \u00d7 10-2<\/sup>
                                                                \n= – log 5 + 2 log 10
                                                                \n= 2 – log 5
                                                                \n= 1.3010
                                                                \npH = 14 – pOH
                                                                \npH = 14 – 1.3010
                                                                \n= 12.699<\/p>\n

                                                                b) [OH–<\/sup>] = 2 \u00d7 10-4<\/sup> M
                                                                \npOH = – log 2 \u00d7 10-4<\/sup>
                                                                \n= – log 2 + log 10
                                                                \n= 4 – log 2
                                                                \n= 4 – 0.3010
                                                                \n= 3.699
                                                                \npH = 14 – 3.699 = 10.301<\/p>\n

                                                                Question 31.
                                                                \n2g of NaOH is dissolved in water to give 1 litre solution. What is the pH of the solution ?
                                                                \nSolution:
                                                                \nN = \\(\\frac{w t}{G \\in w} \\times \\frac{1}{1}\\)
                                                                \n= \\(\\frac{2}{40}\\) = \\(\\frac{1}{20}\\) = 0.05 N = [OH–<\/sup>]
                                                                \n\u2234 pOH = – log 0.05
                                                                \n= – log 5 \u00d7 10-2<\/sup>
                                                                \n= 2 – log 5
                                                                \n= 1.3010
                                                                \npH = 14 – pOH
                                                                \n= 13 – 1.3010
                                                                \n= 12.699<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 32.
                                                                \nCalculate the pH of the following solutions.
                                                                \na. 0.37 g of Ca(OH)2<\/sub> dissolved in water to give 500 ml solution
                                                                \nb. 0.3 g of NaOH dissolved in water to give 200 ml solution
                                                                \nc. 0.1825% HCl aqueous solution
                                                                \nd. 1ml of 13.6 M HCl is diluted with water to give 1 litre solution.
                                                                \nSolution:
                                                                \na) 0.37 gms of Ca(OH)2<\/sub> in 500 ml
                                                                \nN = \\(\\frac{w t}{G \\times w} \\times \\frac{1}{1}\\)
                                                                \nN = \\(\\frac{0.37}{37} \\times \\frac{1000}{500}\\)
                                                                \n= 0.01 \u00d7 2
                                                                \n= 0.02 = pOH
                                                                \npOH = – log 0.02
                                                                \n= 1.699
                                                                \npH = 14 – pOH = 12.301<\/p>\n

                                                                b) 0.3 gms of NaOH in 200 mZ solution
                                                                \nN = \\(\\frac{0.3}{40} \\times \\frac{1000}{200}\\)
                                                                \n= \\(\\frac{0.3 \\times 5}{40}=\\frac{1.5}{40}\\)
                                                                \n= 0.375 = [OH–<\/sup>]
                                                                \npOH = – log [OH–<\/sup>]
                                                                \n= – log 0.375
                                                                \n= 1.426
                                                                \npH = 14 – pOH
                                                                \n= 14 – 1.426
                                                                \n= 12.574<\/p>\n

                                                                c) 0.1825% HCl solution means 100 ml
                                                                \nsolution contain 0.1825 g. of HCl
                                                                \nNormality (N) = \\(\\frac{W}{E W} \\times \\frac{1000}{V}\\)
                                                                \n= \\(\\frac{0.1825}{36.5} \\times \\frac{1000}{100}\\) = 0.05 N
                                                                \n\u2234 [H+<\/sup>] = 0.05 N
                                                                \n\u2234 pH = – log[H+<\/sup>]
                                                                \n= – log [0.051
                                                                \n= – log [5 \u00d7 10-2<\/sup>]
                                                                \n= 2 – log 5
                                                                \n= 2 – 0.692
                                                                \n\u2234 pH = 1.308<\/p>\n

                                                                d) \u2234 [H+<\/sup>] = 13.6
                                                                \nDiluted with water to 1 litre solution
                                                                \n\u2234 [H+<\/sup>] = \\(\\frac{13.6}{1000}\\) = 0.016
                                                                \npH = – log [H+<\/sup>]
                                                                \n= – log 0.0136 = 1.87<\/p>\n

                                                                Question 33.
                                                                \nHow many grams of NaOH are present in 100 ml solution if pH of the solution is 10?
                                                                \nSolution:
                                                                \nGiven
                                                                \npH = 10
                                                                \npOH = 14 – 10 = 4
                                                                \n\u2234 Normality = 10-4<\/sup> N
                                                                \nN = \\(\\frac{w t}{G M w} \\times \\frac{1000}{V(m l)}\\)
                                                                \n10-4<\/sup> = \\(\\frac{w t}{40} \\times \\frac{1000}{100}\\)
                                                                \n\u2234 wt = 4 \u00d7 10-4<\/sup> gms<\/p>\n

                                                                Question 34.
                                                                \nThe value of Kw<\/sub> is 9.55 \u00d7 10-14<\/sup> at certain temperature. Calculate the pH of water at this temperature.
                                                                \nSolution:
                                                                \nSince Kw<\/sub> = [H+<\/sup>] [OH–<\/sup>]
                                                                \nKw<\/sub> = 9.55 \u00d7 10-14<\/sup> mol2<\/sup>\/lit2<\/sup>
                                                                \n[H+<\/sup>] [OH–<\/sup>] = 9.55 \u00d7 10-14<\/sup> mol2<\/sup>\/lit2<\/sup>
                                                                \n\u2234 [H+<\/sup>] = \\(\\sqrt{9.55 \\times 10^{-14}}\\)
                                                                \n= 3.09 \u00d7 10-7<\/sup> mol\/lit
                                                                \npH = – log[H+<\/sup>]
                                                                \n= – log [3.09 \u00d7 10-7<\/sup>]
                                                                \n= – [log 3.09 + log 10-7<\/sup>]
                                                                \n= – [0.49 – 7]
                                                                \npH = 7 – 0.49 = 6.51<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 35.
                                                                \nCalculate the pH of 10-8<\/sup> M NaOH.
                                                                \nSolution:
                                                                \nThe given box solution is very dilute. Hence OH–<\/sup> obtained from the box and water must be taken into consideration.
                                                                \n\u2234 [OH–<\/sup>] = 10-8<\/sup> + 10-7<\/sup> = 1.1 \u00d7 10-7<\/sup>
                                                                \npOH = -log 1.1 \u00d7 10-7<\/sup>
                                                                \npOH = 6.995
                                                                \n\u2234 pOH = 14 – 6.995 = 7.005<\/p>\n

                                                                Question 36.
                                                                \n150 ml of 0.5 M HCl and 100 ml of 0.2 M HCl are mixed. Find the pH of the resulting solution.
                                                                \nSolution:
                                                                \nN = \\(\\frac{N_1 V_1+N_2 V_2}{V_1+V_2}\\)
                                                                \n= \\(\\frac{150 \\times 0.5+100 \\times 0.2}{250}\\)
                                                                \n= \\(\\frac{75+20}{250}=\\frac{95}{250}\\) = 0.38
                                                                \npH = – log [H+<\/sup>] = – log 1 (0.38) = 0.42<\/p>\n

                                                                Question 37.
                                                                \nCalculate the pH of solution obtained by mixing 10ml of 0.1 M HCl and 40ml of 0.2M H2<\/sub>SO4<\/sub>.
                                                                \nSolution:
                                                                \nM = \\(\\frac{10 \\times 0.1+40 \\times 0.4}{50}\\)
                                                                \n= \\(\\frac{1+16}{50}=\\frac{17}{50}\\) = 0.34
                                                                \npH = – log [H+<\/sup>]
                                                                \n= – log (0.34) = 0.4685<\/p>\n

                                                                Question 38.
                                                                \n100 ml of pH = 4 solution is mixed with 100 ml of pH = 6 solution. What is the pH of resulting solution.
                                                                \nSolution:
                                                                \n\"AP<\/p>\n

                                                                Question 39.
                                                                \nEqual volumes of 0.5 M NaOH and 0.3 M KOH are mixed in an experiment. Find the pOH and pH of the resulting solution.
                                                                \nSolution:
                                                                \nN = \\(\\frac{N_1 V_1+N_2 V_2}{V_1+V_2}\\)
                                                                \nN = \\(\\frac{x(0.5)+x(0.3)}{x+x}\\) Given value equal
                                                                \n= \\(\\frac{0.8}{2}\\) = 0.4 = [OH–<\/sup>]
                                                                \npOH = – log [OH–<\/sup>]
                                                                \n= – log (0.4)
                                                                \n= 0.3979
                                                                \npH = 14 – 0.3979
                                                                \n= 13.6021<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 40.
                                                                \n60 m2 of 1 M HCl is mixed with 40 ml of 1 M NaOH. What is the pH of resulutant solution ?
                                                                \nSolution:
                                                                \nVA<\/sub>NA<\/sub> > VB<\/sub>NB<\/sub> \u21d2 N = \\(\\frac{V_A N_A-V_B N_B}{V_A+V_B}\\)
                                                                \n\u2234 N = \\(\\frac{60-40}{60+40}\\)
                                                                \n= \\(\\frac{20}{100}\\) = \\(\\frac{1}{5}\\) = 0.2
                                                                \npH = – log [H+<\/sup>] = – log 0.2
                                                                \n= 0.699<\/p>\n

                                                                Question 41.
                                                                \nCalculate the pH of a solution which contains 100 m2 of 0.1 M HCl and 9.9 m2 of 1.0M NaOH.
                                                                \nSolution:
                                                                \nHere VA<\/sub>NA<\/sub> > VB<\/sub>NB<\/sub>
                                                                \nN = \\(\\frac{V_A N_A-V_B N_B}{V_A+V_B}\\)
                                                                \n= \\(\\frac{10-9.9}{109.9}\\)
                                                                \n= \\(\\frac{0.1}{109.9}\\)
                                                                \n= 0.000909.
                                                                \npH = – log [H+<\/sup>]
                                                                \n= – log (0.000909)
                                                                \n= 3.0416<\/p>\n

                                                                Question 42.
                                                                \nWhat will be the resultant pH when 200 ml of an aqueons solution of HCl having pH = 2 is mixed with 300 m2 of an aqueous of NaOH having pH = 12?
                                                                \nSolution:
                                                                \nVA<\/sub> = 200 ml
                                                                \nNA<\/sub> = 10-2<\/sup>
                                                                \n[\u2235 pH = 2]<\/p>\n

                                                                VB<\/sub> = 300 ml
                                                                \nNB<\/sub> = 10-2<\/sup>
                                                                \n[\u2235 pOH = 2]
                                                                \n\u2234 VB<\/sub>NB<\/sub> > VA<\/sub>NA<\/sub>
                                                                \n\u2234 N = \\(\\frac{300 \\times 10^{-2}-200 \\times 10^{-2}}{300+200}\\)
                                                                \n= \\(\\frac{1}{500}\\) = 0.002 = [OH–<\/sup>]
                                                                \npOH = – log [OH–<\/sup>]
                                                                \n= – log [0.002]
                                                                \n= -log 2 \u00d7 10-3<\/sup>
                                                                \n= – log 2 + 3 log 10
                                                                \n= 3 – log 2
                                                                \n= 2.699
                                                                \n\u2234 pH = 14 – pOH
                                                                \n= 14 – 2.699
                                                                \n= 11.3010<\/p>\n

                                                                Question 43.
                                                                \n50 ml of 0.2 M HCl is added to 30 ml of 0.1 M KOH solution. Find pH of the solution.
                                                                \nSolution:
                                                                \nVA<\/sub> = 50 ml ; VB<\/sub> = 30 ml
                                                                \nNA<\/sub> = 0.2 N; NB<\/sub> = 0.1 ml
                                                                \nVA<\/sub>NA<\/sub> > VB<\/sub>NB<\/sub>
                                                                \n\u2234 N = \\(\\frac{V_A N_A-V_B N_B}{V_A+V_B}\\)
                                                                \n= \\(\\frac{50 \\times 0.2-30 \\times 0.1}{50+30}\\)
                                                                \n= \\(\\frac{1-0.3}{80}=\\frac{0.7}{80}\\)
                                                                \n= 0.00875
                                                                \n\u2234 pH = – log [H+<\/sup>]
                                                                \n= – log (0.00875)
                                                                \n= 1.058.<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 44.
                                                                \n40 ml of 0.2 M HNO3<\/sub> when reacted with 60 ml of 0.3 M NaOH, gave a mixed solution. What is the pH ofo the resulting solution.
                                                                \nSolution:
                                                                \nVB<\/sub>NB<\/sub> > VA<\/sub>NA<\/sub>
                                                                \n\"AP<\/p>\n

                                                                Question 45.
                                                                \n50 ml of 0.1 M H2<\/sub>SO4<\/sub> were added to 100 ml of 0.2 M HNO3<\/sub>. Then the solution is diluted to 300 ml. What is the pH of the solution.
                                                                \nSolution:
                                                                \nV1<\/sub> = 50 ml ; V2<\/sub> = 100 ml
                                                                \nN1<\/sub> = 0.1 \u00d7 2 = 0.2N ; N2<\/sub> = 02N
                                                                \n\"AP<\/p>\n

                                                                Question 46.
                                                                \nWhat is the Kw<\/sub> value in an aqueons solution of pKw<\/sub> = 13.725
                                                                \nSolution:
                                                                \nGiven pKw<\/sub> = 13.725
                                                                \npKw<\/sub> = – log Kw<\/sub>
                                                                \nKw<\/sub> = antilog of 13.725
                                                                \n= 1.884 \u00d7 10-14<\/sup><\/p>\n

                                                                Question 47.
                                                                \nThe ionic product of water at 80\u00b0C is 2.44 \u00d7 10-13<\/sup>. What are the concentrations of hydroniumion and the hydroxide in pure water at 80\u00b0C?
                                                                \nSolution:
                                                                \nKw<\/sub> = [H+<\/sup>] [OH–<\/sup>]
                                                                \nKw<\/sub> = 2.44 \u00d7 10-13<\/sup>
                                                                \nFor pure water [H+<\/sup>] = [OH–<\/sup>]
                                                                \n\u2234 [H+<\/sup>] = Kw<\/sub>
                                                                \n[H+<\/sup>] = \\(\\sqrt{2.44 \\times 10^{-13}}\\)
                                                                \n= 4.94 \u00d7 10-7<\/sup> moles \/ lit
                                                                \n\u2234 [OH–<\/sup>] = 4.94 \u00d7 10-7<\/sup> mole\/lit<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 48.
                                                                \nThe ionization constant for water is 2.9 \u00d7 10-14<\/sup> at 40\u00b0C. Calculate [H3<\/sub>O+<\/sup>], [OH]. pH and pOH for pure water at 40C.
                                                                \nSolution:
                                                                \nKw<\/sub> = [H-1<\/sup>] [OH–<\/sup>]
                                                                \n2.9 \u00d7 10-14<\/sup> = [H+<\/sup>] [OH–<\/sup>]
                                                                \nFor pure water [H+<\/sup>] = [OH–<\/sup>]
                                                                \n\u2234 [H+<\/sup>]2<\/sup> = 2.9 \u00d7 10-14<\/sup>
                                                                \n[H+<\/sup>] = 1.7 \u00d7 10-7<\/sup> = [H3<\/sub>O+<\/sup>]
                                                                \n\u2234 [OH–<\/sup>] = 1.7 \u00d7 10-7<\/sup>
                                                                \npH = – log [H+<\/sup>]
                                                                \n= -log [1.7 \u00d7 10-7<\/sup>]
                                                                \n= 7 – log 1.7 = 6.7689
                                                                \npOH = – log [OH–<\/sup>]
                                                                \n= – log [1.7 \u00d7 10-7<\/sup>]
                                                                \n= 7 – log 1.7 = 6.7689<\/p>\n

                                                                Question 49.
                                                                \nCalculate the pH of
                                                                \na. 0.002 M acetic acid having 2.3% dissociation.
                                                                \nb. 0.002 M NH4<\/sub>OH having 2.3% dissociation.
                                                                \nSolution:
                                                                \na) [H+<\/sup>] = C\u03b1
                                                                \n= 0.002 \u00d7 2.3 = 0.0046
                                                                \npH = – log [H+<\/sup>]
                                                                \n= – log 0.0046
                                                                \n= 4.3372<\/p>\n

                                                                b) [OH–<\/sup>] = C\u03b1
                                                                \n= 0.002 \u00d7 2.3
                                                                \n= 0.0046
                                                                \npOH = – log [OH–<\/sup>]
                                                                \n= – log 0.0046
                                                                \n= 4.3372
                                                                \npH = 14 – pOH
                                                                \n= 14 – 4.3372
                                                                \n= 9.6628<\/p>\n

                                                                Question 50.
                                                                \nCalculate Ka of acetic acid from equilibrium concentration given below
                                                                \n[H3<\/sub>O+<\/sup>] = [CH3<\/sub>COO–<\/sup>] = 1.34 \u00d7 10-3<\/sup>M, [CH3<\/sub>COOH] = 9.866 \u00d7 10-2<\/sup> M
                                                                \nSolution:
                                                                \nGiven
                                                                \n[H3<\/sub>O+<\/sup>] = [CH3<\/sub>COO–<\/sup>] = = 1.34 \u00d7 10-3<\/sup> M
                                                                \n[CH3<\/sub>COOH] = 9.866 \u00d7 10-2<\/sup> M
                                                                \nKa<\/sub> = \\(\\frac{\\left[\\mathrm{H}_3 \\mathrm{O}^{+}\\right]\\left[\\mathrm{CH}_3 \\mathrm{COO}^{-}\\right]}{\\left[\\mathrm{CH}_3 \\mathrm{COOH}\\right]}\\)
                                                                \n= \\(\\frac{1.34 \\times 10^{-3} \\times 1.34 \\times 10^{-3}}{9.866 \\times 10^{-2}}\\)
                                                                \n= 1.819 \u00d7 10-5<\/sup><\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 51.
                                                                \nCalculate pH of 0.1 M acetic acid having Ka<\/sub> =1.8 \u00d7 10-5<\/sup>.
                                                                \nSolution:
                                                                \n\"AP<\/p>\n

                                                                Question 52.
                                                                \nThe pH of 0.1 M solution of weak mono protic acid is 4.0 calculate its [H+<\/sup>] and Ka<\/sub>.
                                                                \nSolution:
                                                                \nGiven pH = 4.0
                                                                \n\u2234 [H+<\/sup>] = ?
                                                                \npH = – log[H+<\/sup>]
                                                                \n[H+<\/sup>] = 10-pH<\/sup>
                                                                \n= 10-4<\/sup>
                                                                \n\u2234 [H+<\/sup>] = C \u00d7 \u03b1
                                                                \n\u03b1 = \\(\\frac{10^{-4}}{0.1}\\)
                                                                \n= 10-3<\/sup>
                                                                \nKa<\/sub> = C \u00d7 \u03b12<\/sup>
                                                                \n= 0.1 \u00d7 (10-3<\/sup>)2<\/sup>
                                                                \n= 0.1 \u00d7 10-6<\/sup>
                                                                \n= 10-7<\/sup>.<\/p>\n

                                                                Question 53.
                                                                \nKa<\/sub> of 0.02 M CH3<\/sub> COOH is 1.8 \u00d7 10-5<\/sup>. Calculate
                                                                \na. [H3<\/sub>O+<\/sup>]
                                                                \nb. % ionization
                                                                \nc. pH
                                                                \nSolution:
                                                                \n(a), (b)
                                                                \nd2<\/sup> = \\(\\frac{K_a}{C}\\)
                                                                \n= \\(\\frac{1.8 \\times 10^{-5}}{0.02}\\)
                                                                \n\u03b12<\/sup> = 9 \u00d7 10-4<\/sup>
                                                                \n\u03b1 = 3 \u00d7 10-2<\/sup>
                                                                \n% of ionisation
                                                                \n[H3<\/sub>O+<\/sup>] = C \u00d7 \u03b1
                                                                \n= 0.02 \u00d7 3 \u00d7 10-2<\/sup> = 6 \u00d7 10-4<\/sup><\/p>\n

                                                                c) pH = -log[H3<\/sub>O+<\/sup>]
                                                                \n= -log(6 \u00d7 10-4<\/sup>) = 3.24.<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 54.
                                                                \nCalculate the pH of 0.01 M solution of CH3<\/sub> COOH. Ka<\/sub> for CHa<\/sub> OOH at 298K is 1.8 \u00d7 10-5<\/sup>.
                                                                \nSolution:
                                                                \n\u03b12<\/sup> = \\(\\frac{K_a}{C}\\)
                                                                \n= \\(\\frac{1.8 \\times 10^{-5}}{0.01}\\)
                                                                \n= 180 \u00d7 10-5<\/sup>
                                                                \n= 18 \u00d7 10-4<\/sup>
                                                                \n\u03b1 = 4.1 \u00d7 10-2<\/sup>
                                                                \n\u2234 [H3<\/sub>O+<\/sup>] = 0.01 \u00d7 4.1 \u00d7 10-2<\/sup>
                                                                \n= 4.1 \u00d7 10-4<\/sup>
                                                                \npH = – log [H3<\/sub>O+<\/sup>]
                                                                \n= -log (4.1 \u00d7 10-4<\/sup>) = 3.38.<\/p>\n

                                                                Question 55.
                                                                \nThe pH of 0.1 M solution of an organic acid is 4.0 calculate the dissociation constant of the acid.
                                                                \nSolution:
                                                                \nGiven pH = 4.0
                                                                \n\u2234 [H+<\/sup>] = ?
                                                                \npH = – log[H+<\/sup>]
                                                                \n\u2234 [H+<\/sup>] = 10-pH<\/sup> = 10-4<\/sup>
                                                                \n\u2234 [H+<\/sup>] = C \u00d7 \u03b1
                                                                \n\u03b1 = \\(\\frac{10^{-4}}{0.1}\\) = 10-3<\/sup>
                                                                \nKa<\/sub> = C \u00d7 \u03b12<\/sup>
                                                                \n= 0.1 \u00d7 (10-3<\/sup>)2<\/sup> = 0.1 \u00d7 10-6<\/sup> = 10-7<\/sup><\/p>\n

                                                                Question 56.
                                                                \nThree ionization constants for HF, HCOOH and HCN at 298 K are 6.8 \u00d7 10-4<\/sup>, 1.8 \u00d7 10-4<\/sup> and 4.8 \u00d7 10-9<\/sup> respectively. Calculate the ionization constants of the corresponding conjugate base.
                                                                \nSolution:
                                                                \ni) Consider the case of HF
                                                                \nKa<\/sub> = 6.8 \u00d7 10-4<\/sup>
                                                                \nKb<\/sub> = \\(\\frac{\\mathrm{K}_{\\mathrm{W}}}{\\mathrm{K}_{\\mathrm{a}}}=\\frac{1 \\times 10^{-14}}{6.8 \\times 10^{-4}}\\) = 0.147 \u00d7 10-10<\/sup>
                                                                \nii) Consider the case of HCOOH
                                                                \nKa = 1.8 \u00d7 10-4<\/sup>
                                                                \nKb<\/sub> = \\(\\frac{K_w}{K_a}=\\frac{1 \\times 10^{-14}}{1.8 \\times 10^{-4}}\\) = 0.56 \u00d7 10-10<\/sup><\/p>\n

                                                                iii) Consider the case of HCN
                                                                \nKa<\/sub> = 4.8 \u00d7 10-9<\/sup>
                                                                \nKb<\/sub> = \\(\\frac{\\mathrm{K}_{\\mathrm{w}}}{\\mathrm{K}_{\\mathrm{a}}}=\\frac{1 \\times 10^{-14}}{4.8 \\times 10^{-9}}\\) = 0.208 \u00d7 10-5<\/sup><\/p>\n

                                                                Question 57.
                                                                \nFind the concentration of hydroxide ion in a 0.25 M solution of trimethylamine, a weak base.
                                                                \n(CH3<\/sub>)3<\/sub>N + H2<\/sub>O \u21cc (CH3<\/sub>)3<\/sub>N+<\/sup>H+OH–<\/sup>; Kb<\/sub> = 7.4 \u00d7 10-5<\/sup>.
                                                                \nSolution:
                                                                \n\u03c3\u03b1 = \\(\\sqrt{\\frac{K_b}{c}}=\\sqrt{\\frac{7.4 \\times 10^{-5}}{0.25}}\\)
                                                                \n= 1.74 \u00d7 10-3<\/sup>
                                                                \n[OH–<\/sup>] = c \u00d7 \u03b1
                                                                \n= 0.25 \u00d7 1.74 \u00d7 10-3<\/sup>
                                                                \n= 4.32 \u00d7 10-3<\/sup> M<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 58.
                                                                \nThe 0.005 M monobasic acid has a pH of 5. What is the extent of ionization ?
                                                                \nSolution:
                                                                \npH = 5
                                                                \n\u2234 [H+<\/sup>] = 10-5<\/sup>
                                                                \nConcentration (C) = 0.005 M = 5 \u00d7 10-3<\/sup> M
                                                                \n[H+<\/sup>] = C\u03b1
                                                                \n\u03b1 = degree of dissociation per mole = ?
                                                                \n\u2234 \u03b1 = \\(\\frac{\\left[\\mathrm{H}^{+}\\right]}{\\mathrm{C}}=\\frac{\\left[\\mathrm{H}^{+}\\right]}{5 \\times 10^{-3}}=\\frac{10^{-5}}{5 \\times 10^{-3}}\\)
                                                                \n= \\(\\frac{1}{5}\\) \u00d7 10-2<\/sup> = 0.2 \u00d7 10-2<\/sup>
                                                                \nPercentage of degree of dissociates
                                                                \n= 0.02 \u00d7 10-2<\/sup> \u00d7 100 = 0.2 %<\/p>\n

                                                                Question 59.
                                                                \n50 ml of 0.1 M NH4<\/sub>OH, 25m1 of 2 M NH4<\/sub>Cl were used to make a buffer. What is the pH if PKa<\/sub> is 4.8.
                                                                \nSolution:
                                                                \nThe given buffer is a base buffer.
                                                                \nFor a base buffer pOH = pKb<\/sub> + log \\(\\frac{[\\text { Salt }]}{[\\text { Base }]}\\)
                                                                \n\u2234 pOH = 4.8 + \\(\\log \\left\\{\\frac{\\frac{(25 \\times 2)}{75}}{\\frac{(50 \\times 0.1)}{75}}\\right\\}\\)
                                                                \n= 4.8 + log(10) = 4.8 + 1 = 5.8
                                                                \n\u2234 pH = (14.0 – pOH) = (14.0 – 5.8) = 8.2
                                                                \n\u2235 pH + pOH = 14.0 at room temperature.<\/p>\n

                                                                Question 60.
                                                                \nThe pH of a buffer prepared by mixing 50 ml of 0.2 M CH3<\/sub> COOH and 25 ml of CH3<\/sub> COONa is 4.8. If the PKa<\/sub> is 4.8. What is the strength of CH3<\/sub> COONa.
                                                                \nSolution:
                                                                \nThe given solution is an acid buffer.
                                                                \n\u2234 The pHof the buffer is given by,
                                                                \n\"AP<\/p>\n

                                                                Question 61.
                                                                \n50 ml of 0.1 M sodium acetate, 25ml of 0.2 M acetic acid were added together to form the buffer solution. PKa<\/sub> of CH3<\/sub> COOH is 4.8 Find the pH of the solution.
                                                                \nSolution:
                                                                \nThe buffer solution is an acid buffer.
                                                                \nHendersen\u2019s equation for an acid buffer is
                                                                \npH = pKa<\/sub> + log
                                                                \n\\(\\frac{[\\text { Salt }]}{[\\text { Base }]}=4.8+\\left\\{\\log \\frac{\\frac{(0.1 \\times 50)}{75}}{\\frac{(0.2 \\times 25)}{75}}\\right\\}\\)
                                                                \n= 4.8 + log (1)
                                                                \n\u2234 pH = 4.8.<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 62.
                                                                \nWhen 20 ml of 0.1 M NH4<\/sub>OH are added to 20ml of 1M NH4<\/sub>Cl solution, the pH of the buffer formed is 8.2. What is the pKa<\/sub> of NH4<\/sub>OH?
                                                                \nSolution:
                                                                \nFor base buffer, pH is given by
                                                                \npH = 14.0 – pOH and pOH
                                                                \n= pKb<\/sub> + \\(\\log \\frac{[\\text { Salt }]}{[\\text { Base }]}\\)
                                                                \nThe pH of the buffer = 8.2;
                                                                \n\u2234 pOH =(14 – pH) = 5.8 then
                                                                \n5.8 = pKb<\/sub> + \\(\\log \\frac{[\\text { Salt }]}{[\\text { Base }]}\\)
                                                                \n= pKb<\/sub> + \\(\\log \\left[\\frac{\\frac{20 \\times 1}{40}}{\\frac{20 \\times 0.1}{40}}\\right]\\)
                                                                \n= pKb<\/sub> + log (10) = (pKb<\/sub> + 1.0)
                                                                \n\u2234 pKb<\/sub> = (5.8 – 1) = 4.8<\/p>\n

                                                                Question 63.
                                                                \nOne litre of buffer solution contains 0.1 mole of aceticacid add 1 mole of sodium acetate. Find its pH if PKa<\/sub> of CH3<\/sub>COOH is 4.8.
                                                                \nSolution:
                                                                \nFor an acid buffer pH = PKa<\/sub> + \\(\\log \\frac{[\\text { Salt }]}{[\\text { Base }]}\\)
                                                                \npH = 4.8 + log \\(\\frac{1}{0.1}\\) = 4.8 + log 10
                                                                \npH = 4.8 + 1 \u21d2 pH = 5.8<\/p>\n

                                                                Question 64.
                                                                \n50 ml of 1 M CH3<\/sub> COOH solution, when added to 50 ml of 0.5 M NaOH gives a solution with a pH \u2018X\u2019. Find the value of \u2018X\u2019. PKa<\/sub> of acetic acid is 4.8.
                                                                \nSolution:
                                                                \nAcetic acid reacts with sodium hydroxide to form sodium acetate.
                                                                \nCH3<\/sub>COOH + NaOH \u2192 CH3<\/sub>COONa + H2<\/sub>O
                                                                \nNo.of millimoles of CH3<\/sub>COOH = 50 \u00d7 1 = 50
                                                                \nNo.of millimoles of NaOH = 50 \u00d7 0.5 = 25
                                                                \nNo.of millimoles of CH3<\/sub>. COOH remained unreacted = 50 – 25 = 25
                                                                \nTotal Vol. of solution = 50 + 50 = 100 ml
                                                                \nNo. of millimoles of CH3<\/sub>COONa formed = 25
                                                                \n\u2234 Molarity of CH3<\/sub>COONa = \\(\\frac{25}{100}\\) = .25 M
                                                                \nMolarity of CH3<\/sub>COOH = \\(\\frac{25}{100}\\) = .25 M
                                                                \npH = pKa<\/sub> + \\(\\log \\frac{[\\text { Salt }]}{[\\text { Base }]}\\)
                                                                \npH = 4.8 + log \\(\\frac{1}{0.1}\\) = 4.8 + log 10
                                                                \n= 4.8 + log \\(\\frac{.25}{.24}\\) = 4.8 + log(1)
                                                                \n\u2234 pH of the solution = 4.8 + 0 = 4.8<\/p>\n

                                                                Question 65.
                                                                \nThe solubility product of AgCl is 1.6 \u00d7 10-10<\/sup> mol2<\/sup>\/L2<\/sup>. What is solubility?
                                                                \nSolution:
                                                                \nKSP<\/sub> = S2<\/sup>
                                                                \n1.6 \u00d7 10-10<\/sup> = S2<\/sup>
                                                                \n\u2234 S = \\(\\sqrt{10^{-10} \\times 1.6}\\) = 1.3 \u00d7 10-5<\/sup> moles \/litre.<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 66.
                                                                \nThe solutbility product of Zr (OH)2<\/sub> is 4.5 \u00d7 10-17<\/sup> mol3<\/sup>L-3<\/sup> what is solubility?
                                                                \nSolution:
                                                                \nZn(OH)2<\/sub> \u21cc Zn++<\/sup> + 2OH–<\/sup>
                                                                \n\u2234 KSP<\/sub> = S \u00d7 4S2<\/sup> = 4S3<\/sup>
                                                                \n\u2234 4S3<\/sup> = 4.5 \u00d7 10-17<\/sup>
                                                                \nS3<\/sup> = 4.5 \u00d7 10-18\/4<\/sup> = 11.7 \u00d7 10-18<\/sup>
                                                                \nS = \\(\\sqrt[3]{11.7} \\times \\sqrt[3]{10^{-18}}\\)
                                                                \nS = 2.6 \u00d7 10-6<\/sup><\/p>\n

                                                                Question 67.
                                                                \nThe solubility of Ag2<\/sub> CrO4<\/sub> is 1.3 \u00d7 10-4<\/sup> mol L-1<\/sup>. What is the solubility product ?
                                                                \nSolution:
                                                                \n\"AP
                                                                \nKSP<\/sub> = 4S2<\/sup> \u00d7 S = 4S3<\/sup>
                                                                \n\u2234 KSP<\/sub> = 4 \u00d7 S3<\/sup>
                                                                \n= 4 \u00d7 (1.3 \u00d7 10-4<\/sup>)3<\/sup>
                                                                \nKSP<\/sub> = 9 \u00d7 10-12<\/sup><\/p>\n

                                                                Question 68.
                                                                \nThe solubility of A2<\/sub>B = 2 \u00d7 10-3<\/sup> mol L-1<\/sup>. What is solubility product?
                                                                \nSolution:
                                                                \n\"AP
                                                                \nKSP<\/sub> = [2s]2<\/sup>[S]
                                                                \nS = 3\\(\\sqrt{\\frac{K_{S P}}{4}}\\)
                                                                \n= 3\\(\\sqrt{\\frac{2 \\times 10^{-3}}{4}}\\) = 3.2 \u00d7 10-10<\/sup><\/p>\n

                                                                Question 69.
                                                                \nThe sotubility product of a salt AB = 10-10<\/sup> mol2<\/sup> L-2<\/sup>. What is solubility?
                                                                \nSolution:
                                                                \n\"AP
                                                                \nKSP<\/sub> = s2<\/sup>
                                                                \nS = \\(\\sqrt{K_{s p}}=\\sqrt{10^{-10}}\\) = 10-5<\/sup> moles\/lit<\/p>\n

                                                                Question 70.
                                                                \nPQ and RS2<\/sub> are two sparingly soluble salts. Their solubility products are equal and each equal to 4.0 \u00d7 10-18<\/sup>. Which salt is more soluble ?
                                                                \nSolution:
                                                                \nGiven KSP<\/sub> of PQ and RS2<\/sub> salts = 4 \u00d7 10-18<\/sup>
                                                                \nS = \\(\\sqrt{\\frac{K_{\\mathrm{SP}}}{4}}\\) for RS2<\/sub>
                                                                \nS = \\(\\sqrt{K_{s p}}\\) for PQ
                                                                \n\u2234 Solubility of RS2<\/sub> is more.<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 71.
                                                                \nIn a 0.1 M solution, acetic acid is 1.34% ionized. Calculate [H+<\/sup>], [CH3<\/sub>COO–<\/sup>] and [CH3<\/sub>COOH] in the solution and Calculate Ka<\/sub> of acetic acid.
                                                                \nSolution:
                                                                \nGiven 1.34 % ionised
                                                                \n[H+<\/sup>] = C \u00d7 \u03b1
                                                                \n= 0.1 \u00d7 1.34 \u00d7 10-2<\/sup> = 1.34 \u00d7 10-3<\/sup> M
                                                                \n\u2234 [CH3<\/sub>COOH] =1.34 \u00d7 10-3<\/sup> M
                                                                \n[H+<\/sup>] = C\u03b1 = 1.34 \u00d7 10-3<\/sup>]
                                                                \n\u03b1 = \\(\\frac{1.34 \\times 10^{-3}}{0.1}\\) = 1.34 \u00d7 10-2<\/sup>]
                                                                \n[CH3<\/sub>COO–<\/sup>] (1 – \u03b1) = 0.1 (1 – 0.0134)
                                                                \n= 0.09866 M
                                                                \nKa = \u03b12<\/sup> \u00d7 C
                                                                \n= (1.34 \u00d7 10-2<\/sup>) \u00d7 0.1 = 1.79 \u00d7 10-5<\/sup><\/p>\n

                                                                Solved Problems<\/span><\/p>\n

                                                                Question 1.
                                                                \nThe following concentrations were obtained for the formation of NH3<\/sub> from N2<\/sub> and H2<\/sub> at equilibrium at 500 K. [N2<\/sub>] = 1.5 \u00d7 10-2<\/sup> M. [H2<\/sub>] = 3.0 \u00d7 10-2<\/sup> M and [NH3<\/sub>] = 1.2 \u00d7 10-2<\/sup> M. Calculate equilibrium constant.
                                                                \nSolution:
                                                                \nThe equilibrium constant for the reaction. N2<\/sub>(g) + 3 H2<\/sub> (g) \u21cc NH3<\/sub>(g) can be written as,
                                                                \n\"AP<\/p>\n

                                                                Question 2.
                                                                \nAt equilibrium, the concentrations of N2<\/sub> = 3.0 \u00d7 10-3<\/sup> M, O2<\/sub> = 4.2 \u00d7 10-3<\/sup> M and NO = 2.8 \u00d7 10-3<\/sup> M in sealed vessel at 800K. What will be Kc<\/sub> for the reaction
                                                                \nN2<\/sub>(g) + O2<\/sub> (g) \u21cc 2 NO(g)
                                                                \nSolution:
                                                                \nFor the reaction equilibrium constant, KC<\/sub> can be written as,
                                                                \nKC<\/sub> = \\(\\frac{[\\mathrm{NO}]^2}{\\left[\\mathrm{~N}_2\\right]\\left[\\mathrm{O}_2\\right]}\\)
                                                                \n= \\(\\frac{\\left(2.8 \\times 10^{-3} \\mathrm{M}\\right)^2}{\\left(3.0 \\times 10^{-3} \\mathrm{M}\\right)\\left(4.2 \\times 10^{-3} \\mathrm{M}\\right)}\\)
                                                                \n= 0.622<\/p>\n

                                                                Question 3.
                                                                \nPCl5<\/sub>, PCl3<\/sub> and Care at equilibrium at 500 K and having concentration 1.59 M PCl3<\/sub>, 159M Cl2<\/sub> and 1.41 M PCl5<\/sub>, Calculate K for the reaction
                                                                \nPCl5<\/sub> \u21cc PCl3<\/sub> + Cl2<\/sub>
                                                                \nSolution:
                                                                \nThe equilibrium constant K. for the above reaction can be written as,
                                                                \nKC<\/sub> = \\(\\frac{\\left[\\mathrm{PCl}_3\\right]\\left[\\mathrm{Cl}_2\\right]}{\\left[\\mathrm{PCl}_5\\right]}=\\frac{(1.59)^2}{(1.41)}\\) = 1.79<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 4.
                                                                \nThe value of KC<\/sub> = 4.24 of 800 K for the reaction
                                                                \nCO (g) + H2<\/sub>O (g) \u21cc CO2<\/sub> (g) + H2<\/sub> (g)
                                                                \nCalculate equilibrium concentrations of CO2<\/sub>, H2<\/sub>, CO and H2<\/sub>O at 800 K, if only CO and H2<\/sub>O are present initially at concentrations of 0.10 M each.
                                                                \nSolution:
                                                                \nFor the reaction,
                                                                \n\"AP
                                                                \nwhere x is the amount of CO2<\/sub> and H2<\/sub> at equilibrium.
                                                                \nHence, equilibrium constant can be written as,
                                                                \nKC<\/sub> = x2<\/sup>\/(0.1 – x)2<\/sup> = 4.24
                                                                \nx2<\/sup> = 4.24(0.01 + x2<\/sup> – 0.2 x)
                                                                \nx2<\/sup> = 0.0424 + 4.24 x2<\/sup> – 0.848 x
                                                                \n3.24 x2<\/sup> – 0.848 x + 0.0424 = 0
                                                                \na = 3.24, b = – 0.848, c = 0.0424
                                                                \n(for quadratic equation ax2<\/sup> + bx + c = 0
                                                                \nx = \\(\\frac{\\left(-b \\pm \\sqrt{b^2-4 a c}\\right)}{2 a}\\)
                                                                \nx = 0.848 \u00b1 \\(\\sqrt{(0.848)^2}\\) – 4(3.24) (0.04224) \/ (3.24 \u00d7 2)
                                                                \nx = (0.848 \u00b1 0.4118)\/6.48
                                                                \nx1<\/sub> = (0.848 – 0.4118)] 6.48 = 0.067
                                                                \nx2<\/sub> = (0.848 + 0.4118)\/6.48 = 0.194
                                                                \nthe value 0.194 should be neglected because it will give concentration of the reactant which is more than initial concentration. Hence the equilibrium concentrations are,
                                                                \n[CO2<\/sub>] [H2<\/sub>] = x = 0.067 M
                                                                \n[CO] = [H2<\/sub>O] = 0.1 – 0.067 = 0.033 M<\/p>\n

                                                                Question 5.
                                                                \nFor the equilibrium, 3 NOCl (g) \u21cc 2 NO(g) + C2<\/sub> (g) the value of the equilibrium constant, KC<\/sub> is 3.75 \u00d7 10-6<\/sup> at 1069 K. Calculate the Kp<\/sub> for the reaction at this temperature ?
                                                                \nSolution:
                                                                \nWe know that,
                                                                \nKP<\/sub> = KC<\/sub>(RT)\u2206n<\/sup>
                                                                \nFor the above reaction,
                                                                \n\u2206n = (2 + 1) – 2 = 1
                                                                \nKP<\/sub> = 3.75 \u00d7 10-6<\/sup> (0.0831 \u00d7 1069)
                                                                \nKP<\/sub> = 0.033<\/p>\n

                                                                Question 6.
                                                                \nThe value of KP<\/sub> for the reaction.
                                                                \nCO2<\/sub> (g) + C(s) \u21cc 2CO (g) is 3.0 at 1000 K. If initially PCO2<\/sub> = 0.48 bar and PCO<\/sub> = 0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO2<\/sub>.
                                                                \nSolution:
                                                                \nFor the reaction,
                                                                \nlet \u2018x’ be the decrease in pressure of CO2<\/sub>, then
                                                                \n\"AP
                                                                \n= (-3 \u00b1 5.66)\/8
                                                                \n= (-3 + 5.66)\/8 (as value of x cannot be negative hence we neglect that value)
                                                                \nx = 2.66\/8 = 0.33
                                                                \nThe equilibrium partial pressures are,
                                                                \nPCO<\/sub> = 2x = 2 \u00d7 0.33 = 0.66 bar
                                                                \nPCO2<\/sub> = 0.48 – x = 0.48 – 0.33 = 0.15 bar<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 7.
                                                                \nThe value of Kc for the reaction
                                                                \n2A \u21cc B + C is 2 \u00d7 10-3<\/sup>. At a given time, the composition of reaction mixture is [A] = [B] = [C] = 3 \u00d7 10-4<\/sup> M. In which direction the reaction will proceed ?
                                                                \nSolution:
                                                                \nFor the reaction the reaction quotient QC<\/sub> is given by.
                                                                \nQC<\/sub> = [B] [C] \/ [A]2<\/sup>
                                                                \nas [A] = [B] = [C] = 3 \u00d7 10-4<\/sup> M
                                                                \nQC<\/sub> = (3 \u00d7 10-4<\/sup>) (3 \u00d7 10-4<\/sup>)\/ (3 \u00d7 10-4<\/sup>)2<\/sup> = 1
                                                                \nas QC<\/sub> > KC<\/sub> so the reaction will proceed in the reverse direction.<\/p>\n

                                                                Question 8.
                                                                \n13.8 g of N2<\/sub>O4<\/sub> was placed in a 1L reaction vessel at 400 K and allowed to attain equilibrium
                                                                \nN2<\/sub>O4<\/sub>(g) \u21cc 2 NO2<\/sub>(g)
                                                                \nThe total pressure at equilbrium was found to be 9.15 bar. Calculate KC<\/sub>, KP<\/sub> and partial pressure at equilibrium.
                                                                \nSolution:
                                                                \nWe know pV = nRT
                                                                \nTotal volume (V) = 1 L
                                                                \nMolecular mass of N2<\/sub>O4<\/sub> = 92 g
                                                                \nNumber of moles = 13.8 g \/ 92 g = 0.15 of the gas (n)
                                                                \nGas constant (R) = 0.083 bar L mol-1<\/sup>K-1<\/sup>
                                                                \nTemperature (T) = 400 K
                                                                \npV = nRT
                                                                \np \u00d7 1L = 0.15 mol \u00d7 0.083 bar L mol-1<\/sup>K-1<\/sup> \u00d7 400 K
                                                                \np = 4.98 bar
                                                                \n\"AP
                                                                \nHence,
                                                                \nptotal<\/sub> at equilibrium = pN2O4<\/sub> + pNO2<\/sub>
                                                                \n9.15 = (4.98 – x) + 2x
                                                                \n9.15 = 4.98+ x
                                                                \nx = 9.15 – 4.98 = 4.17 bar
                                                                \nPartial pressures at equilibrium are,
                                                                \npN2O4<\/sub> = 4.98 – 4.17 = 0.81 bar
                                                                \npNO2<\/sub> = 2x = 2 \u00d7 4.17 = 8.34 bar
                                                                \nKp<\/sub> = (pNO2<\/sub>)2<\/sup> \/ pN2O4<\/sub>
                                                                \n= (8.34)2<\/sup> \/ 0.81 = 85.87
                                                                \nKp<\/sub> = KC<\/sub> (RT)\u2206n<\/sup>
                                                                \n85.87 = KC<\/sub>(0.083 \u00d7 400)1<\/sup>
                                                                \nKC<\/sub> = 2.586 = 2.6<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 9.
                                                                \n300 mol of PCl5<\/sub> kept in 1L closed reaction vessel was allowed to attain, equilibrium at 380K. Calculate composition of the mixture at equilibrium. Kc<\/sub> = 1.80
                                                                \nSolution:
                                                                \n\"AP<\/p>\n

                                                                Question 10.
                                                                \nThe value of \u2206G\u00b0 for the phosphorylation of glucose in glycolysis is 13.8 kJ\/mol. Find the value of Kc<\/sub> at 298 K.
                                                                \nSolution:
                                                                \n\u2206G\u00b0 = 13.8 ki\/ mol = 13.8 \u00d7 103<\/sup>J\/mol
                                                                \nAlso, \u2206G\u00b0 = – RT Inkc<\/sub>
                                                                \nHence, In Kc<\/sub> = – 13.8 \u00d7 103<\/sup> J\/mol
                                                                \n(8.314 Jmol-1<\/sup>K-1<\/sup> \u00d7 298 K)
                                                                \nIn Kc<\/sub> = 5.569
                                                                \nKc<\/sub> = e-5 569<\/sup>
                                                                \nKc<\/sub> = 3.81 \u00d7 10-3<\/sup><\/p>\n

                                                                Question 11.
                                                                \nHydrolysis of sucrose gives.
                                                                \nSucrose + H2<\/sub>O \u21cc Glucose + Fructose Equilibrium constatnt Kc for the reaction is 2 \u00d7 1013<\/sup> at 300K. Calculate \u2206G\u0398<\/sup> at 300K.
                                                                \nSolution:
                                                                \n\u2206G\u00b0 = -RT InKc<\/sub>
                                                                \n\u2206G\u00b0 = -8.314J mol-1<\/sup>K-1<\/sup> \u00d7 300K \u00d7 In (2 \u00d7 1013<\/sup>)
                                                                \n\u2206G\u00b0 = – 7.64 \u00d7 104J mo-1<\/sup><\/p>\n

                                                                Question 12.
                                                                \nWhat will be the conjugate bases for the following Bronsted acides: HF, H2<\/sub>SO4<\/sub> and \\(\\mathrm{HCO}_3^{-}\\) ?
                                                                \nSolution:
                                                                \nThe conjugate bases should have one proton less in each case and therefore the corresponding conjugate bases are: F–<\/sup>, \\(\\mathrm{HSO}_3^{-}\\) and CO2-<\/sup>3<\/sub> respectively.<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 13.
                                                                \nWrite the conjugate acids for the following Bronsted bases: NH–<\/sup>2<\/sub>, NH3<\/sub> and HCOO–<\/sup>.
                                                                \nSolution:
                                                                \nThe conjugate acid should have one extra proton in each case and therefore the corresponding conjugate acids are : NH3<\/sub>, NHO+<\/sup>4<\/sub> and HCOOH respectively.<\/p>\n

                                                                Question 14.
                                                                \nThe species; H2<\/sub>O, \\(\\mathrm{HCO}_{3^{\\prime}}^{-}\\), \\(\\mathrm{HCO}_{4^{\\prime}}^{-}\\) and NH3<\/sub> can act both as Bronsted acids andbases. For each case give the corresponding conjugate acid and conjugate base.
                                                                \nSolution:
                                                                \nThe answer is given in the following Table:
                                                                \n\"AP<\/p>\n

                                                                Question 15.
                                                                \nClassify the following species into Lewis acids and Lewis bases and show how these act as such :
                                                                \n(a) HO–<\/sup>
                                                                \n(b) F–<\/sup>
                                                                \n(c) H+<\/sup>
                                                                \n(d) BCl3<\/sub>
                                                                \nSolution:
                                                                \n(a) Hydroxyl ion is a Lewis base as it can donate an electron lone pair (: OH–<\/sup>).
                                                                \n(b) Flouride ion acts as a Lewis base as it can donate any one of its four electron lone pairs.
                                                                \n(c) A proton is a Lewis acid as it can accept a lone pair of electrons from bases like hydroxyl ion and fluoride ion.
                                                                \n(d) BCl2<\/sub> acts a Lewis acid as it can accept a lone pair of electrons from species like ammonia or amine molecules.<\/p>\n

                                                                Question 16.
                                                                \nThe concentration of hydrogen ion in a sample of soft drink is 3.8 \u00d7 10-3<\/sup>M. What is its pH ?
                                                                \nSolution:
                                                                \npH = – Log[3.8 \u00d7 10-3<\/sup>]
                                                                \n= -{log[3.8] + log[10-3<\/sup>}]
                                                                \n= – {(0.58) + (-3.0)} = -2.42} = 2.42
                                                                \nTherefore, the pH of the soft drink is 2.42 and it can be inferred that it is acidic.<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 17.
                                                                \nCalculate pH of a 1.0 -8<\/sup> M solution of HCl.
                                                                \nSolution:
                                                                \n2H2<\/sub>O(1) H3<\/sub>O+<\/sup> (aq) + OH–<\/sup>(aq)
                                                                \nKw<\/sub> = [OH–<\/sup>][H3<\/sub>O+<\/sup>]
                                                                \n= 10-14<\/sup>
                                                                \nLet, x = [OH–<\/sup>] = [H3<\/sub>O+<\/sup>] from H2<\/sub>O. The H3<\/sub>O+<\/sup> concentration is generated (i) from the ionization of HCl dissolved i.e., HCl (aq) + H2<\/sub>O (1) \u21cc H3<\/sub>O+<\/sup> (aq) + Cl–<\/sup>(aq), and (ii) from ionization of H2<\/sub>O. Inthese very dilute solutions. both sources of H3<\/sub>O+<\/sup> must be considered:
                                                                \n[H3<\/sub>O+<\/sup>] = 10-8<\/sup> + X
                                                                \nKw<\/sub> = (10-8<\/sup> + x)(x) = 10-14<\/sup>
                                                                \nor x2<\/sup> + 10-8<\/sup> x – 10-14<\/sup> = 0
                                                                \n[OH–<\/sup>] = x = 9.5 \u00d7 10-8<\/sup>
                                                                \nSo, pOH = 7.02 and pH = 6.98<\/p>\n

                                                                Question 18.
                                                                \nThe ionization costant of HF is 3.2 \u00d7 10-4<\/sup>. Calculate the degree of dissociation of HF in its 0.02 M solution. Calculate the concentration of all species present (H3<\/sub>O+<\/sup>.F–<\/sup> and HF) in the solution and its pH.
                                                                \nSolution:
                                                                \nThe following proton transfer reactions are possible :
                                                                \n1) HF + H2<\/sub>O \u21cc H3<\/sub>O+<\/sup>F–<\/sup>Ka<\/sub> = 3.2 \u00d7 10-4<\/sup>
                                                                \n2) H2<\/sub>O + H2<\/sub>O H3<\/sub>O+<\/sup> + OH–<\/sup> Kw<\/sub> = 1.0 \u00d7 10-14<\/sup>
                                                                \nAs ka<\/sub> >> Kw<\/sub>, [1] is the principle reaction.
                                                                \n\"AP
                                                                \nSubstituting equilibrium concentrations in the equailibrium reaction for principal reaction gives.
                                                                \nKa<\/sub> (0.02\u03b1)2<\/sup>\/ (0.02 – 0.02\u03b1)
                                                                \n= 0.02\u03b12<\/sup>\/(1 – \u03b1) = 3.2 \u00d7 10-4<\/sup>
                                                                \nWe obtain the following quadratic equation:
                                                                \n\u03b12<\/sup> + 1.6 \u00d7 10-2<\/sup>\u03b1 – 1.6 \u00d7 10-2<\/sup> = 0
                                                                \nThe quadratic equation in \u03b1 can be solved and the two values of the roots are :
                                                                \n\u03b1 = + 0.12 and – 0.12
                                                                \nThe negative root is not acceptable and hence.
                                                                \na = 0.12
                                                                \nThis means that the degree of ionization \u03b1 = 0.12, then equilibrium concentrations of other species viz., HF, F–<\/sup> and H3<\/sub>O+<\/sup> are given by:
                                                                \n[H3<\/sub>O+<\/sup>] = [F] = c\u03b1 = 0.02 \u00d7 0.12
                                                                \n= 2.4 \u00d7 10-3<\/sup>M
                                                                \n[HF] = c(1 – \u03b1) = 0.02(1 – 0.12)
                                                                \n= 17.6 \u00d7 10-3<\/sup>M
                                                                \npH = -log[H+<\/sup>] = -log (2.4 \u00d7 10-3<\/sup>) = 2.62<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 19.
                                                                \nThe pH of 0.1 M monobasic acid is 4.50. Calculate the concentration of species H+<\/sup>, A–<\/sup> and HA at equilibrium. Also, determine the value of Ka<\/sub> and pKa<\/sub> of the monobasic acid.
                                                                \nSolution:
                                                                \npH = – log [H+<\/sup>]
                                                                \nTherefore [H+<\/sup>] = 10-pH<\/sup> = 10-4.50<\/sup>
                                                                \n= 3.16 \u00d7 10-5<\/sup>
                                                                \n[H+<\/sup>] = [A–<\/sup>] = 3.16 \u00d7 10-5<\/sup>
                                                                \nThus, Ka<\/sub> = [H+<\/sup>] [A–<\/sup>] \/ [HA]
                                                                \n[HA]eqlbm<\/sub> = 0.1 – (3.16 \u00d7 10-5<\/sup>) \u2245 0.1
                                                                \nKa<\/sub> = (3.16 \u00d7 10-5<\/sup>)2<\/sup>\/0.1 = 1.0 \u00d7 10-8<\/sup>
                                                                \npKa<\/sub> = – log (10-8<\/sup>) = 8
                                                                \nAlternatively, “Percent dissociation” is another useful method for measure of strength of a weak acid and is given as : Percent dissociation
                                                                \n= [HA]dissociated<\/sub> \u00d7 100\/[HA]initial<\/sub> (7.32)<\/p>\n

                                                                Question 20.
                                                                \nCalculate the pH of 0.08M solution of hypochlorous acid. HOCI. The ionization Constant of the acid is 2.5 \u00d7 10-5<\/sup>, Determine the percent dissociation of HOCl
                                                                \nSolution:
                                                                \n\"AP
                                                                \n[H+<\/sup>] = 1.41 \u00d7 10-3<\/sup> M Therefore,
                                                                \nPercent dissociation
                                                                \n= {[HOCI]dissociated<\/sub> \u00d7 100\/[HOCI]initial<\/sub>}
                                                                \n= 1.41 \u00d7 10-3<\/sup> \/0.08 = 1.76%,
                                                                \npH = -log (1.41 \u00d7 10-3<\/sup>) = 2.85<\/p>\n

                                                                Question 21.
                                                                \nThe pH of 0.004 M hydrazine solution is 9.7. Calculate is ionization constant Kb<\/sub> and pKb<\/sub>.
                                                                \nSolution:
                                                                \nNH2<\/sub>NH2<\/sub> + H2<\/sub>O \u21cc \\(\\mathrm{NH}_2 \\mathrm{NH}_3^{+}\\) +OH–<\/sup>
                                                                \nFrom the pH we can calculate the hydrogen ion concentration. Knowing hydrogen ion concentration and the ionic product of water we can calculate the concentration of hydroxyl ions. Thus we have :
                                                                \n[H+<\/sup>] = antilog (-pH)
                                                                \n= antilog (-9.7) = 1.67 \u00d7 10-10<\/sup>
                                                                \n[OH–<\/sup>] = Kw<\/sub>\/[H+<\/sup>] = 1 \u00d7 10-14<\/sup>\/ 1.67 \u00d7 10-10<\/sup>
                                                                \n= 5.98 \u00d7 10-5<\/sup>
                                                                \nThe concentration fo the corresponding hydrazinium ion is also the same as that of hydroxyl ion. The concentrations of both these ions are very small so the concentration of the undissociated base can be taken equal to 0.004 M.
                                                                \nThus,
                                                                \nKb<\/sub> = [NH2<\/sub>NH3<\/sub>+<\/sup>] [OH–<\/sup>] \/ [NH2<\/sub>NH2<\/sub>]
                                                                \n= (5.98 \u00d7 10-5<\/sup>)2<\/sup>\/0.004 = 8.96 \u00d7 10-7<\/sup>
                                                                \npKb<\/sub> = -logKb<\/sub> = -log(8.96 \u00d7 10-7<\/sup> ) = 6.04<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 22.
                                                                \nCalculate the pH of the solution in which 0.2M NH4<\/sub>Cl and 0.1 M NH3<\/sub> are present. The pKb<\/sub> of ammonia solution is 4.75.
                                                                \nSolution:
                                                                \nNH3<\/sub> + H2<\/sub>O \u21cc \\(\\mathrm{NH}_4^{+}\\) + OH–<\/sup>
                                                                \nThe ionization constant of NH3<\/sub>,
                                                                \nKb<\/sub>= antilog (-pKb<\/sub>) i.e.
                                                                \nKb<\/sub> = 10-4.75<\/sup> = 1.77 \u00d7 10-5<\/sup> M
                                                                \n3<\/sub> + H2<\/sub>O \u21cc \\(\\mathrm{NH}_4^{+}\\) + OH–<\/sup> Initial concentration (M)
                                                                \n\"AP
                                                                \nChange in concentration to reach equilibrium (M)
                                                                \n\"AP
                                                                \nKb<\/sub> = [latex]\\mathrm{NH}_4^{+}[\/latex][OH-1<\/sup>] \/ [NH3<\/sub>]
                                                                \n= (0.20 \u00d7 x) (x) \/ (0.1 – x) = 1.77 \u00d7 10-5<\/sup>
                                                                \nAs Kb<\/sub> is small, we can neglect x in comparison to 0.1 M and 0.2 M. Thus
                                                                \n[OH–<\/sup>] = x = 0.88 \u00d7 10-5<\/sup>
                                                                \nTherefore, [H+<\/sup>] = 1.12 \u00d7 10-9<\/sup>
                                                                \npH = -log [H+<\/sup>] = 8.95 .<\/p>\n

                                                                Question 23.
                                                                \nDetermine the degree of ionization and pH of a 0.05M of ammonia solution. The ionization constant of ammonia can be taken from Table 7.7. Also, calculate the ionization constant of the conjugate acid of ammonia.
                                                                \nSolution:
                                                                \nThe ionization of NH3<\/sub> in water is represented by equation :
                                                                \nNH3<\/sub> + H2<\/sub>O \u21cc \\(\\mathrm{NH}_4^{+}\\) + OH–<\/sup>
                                                                \nWe use equation (7.33) to calculate hydroxyl ion concentration,
                                                                \n[OH–<\/sup>] c \u03b1 = 0.05 \u03b1
                                                                \nKb<\/sub> = 0.05 \u03b12<\/sup>\/(1 – \u03b1)
                                                                \nThe value of \u03b1 is small, therefore the quadratic equation can be simplified by neglecting a in comparison to 1 in the denominator on right hand side of the equation,
                                                                \nThus,
                                                                \nKb<\/sub> = C \u03b12<\/sup> or \u03b1 = \\(\\sqrt{\\left(1.77 \\times 10^{-5} \/ 0.05\\right)}\\)
                                                                \n= 0.018.
                                                                \n[OH–<\/sup>] = C \u03b1 = 0.05 \u00d7 0.018 = 9.4 \u00d7 10-4<\/sup> M.
                                                                \n[H+<\/sup>] = Kw<\/sub>\/[OH–<\/sup>] = 10-14<\/sup> \/ (9.4 \u00d7 10-4<\/sup>)
                                                                \n= 1.06 \u00d7 10-11<\/sup>
                                                                \npH = -log (1.06 \u00d7 10-11<\/sup>) = 10.97
                                                                \nNow, using the relation for conjugate acid-base pair,
                                                                \nKa<\/sub> \u00d7 Kb<\/sub> = Kw<\/sub>
                                                                \nusing the value of Kb<\/sub> of NH3<\/sub> from Tabel 7.7.
                                                                \nWe can determine the concentration of con-jugate acid \\(\\mathrm{NH}_4^{+}\\)
                                                                \nKa<\/sub> \u00d7 Kw<\/sub>\/ Kb<\/sub> = 10-14<\/sup>\/ 1.77 \u00d7 10-5<\/sup>
                                                                \n= 5.64 \u00d7 10-10<\/sup>.<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 24.
                                                                \nCalculate the pH of a 0.10M ammonia solution. Calculate the pH after 50.0 mL of this solution is treated with 25.0 mL of 0.10M HCl. The dissociation constant of ammonia, Kb<\/sub> = 1.77 \u00d7 10-5<\/sup>.
                                                                \nSolution:
                                                                \nNH3<\/sub> + H2<\/sub>O \u21cc \\(\\mathrm{NH}_4^{+}\\) + OH–<\/sup>
                                                                \nKb<\/sub> = [latex]\\mathrm{NH}_4^{+}[\/latex] [OH–<\/sup>] \/ [NH3<\/sub>] = 1.77 \u00d7 10-5<\/sup>
                                                                \nBefore neutralization,
                                                                \n[latex]\\mathrm{NH}_4^{+}[\/latex] [OH–<\/sup>] = x
                                                                \n[NH3<\/sub>] = 0.10 – x – 0.10, x2<\/sup> \/ 0.10 = 1.77 \u00d7 10-5<\/sup>
                                                                \nThus, x= 1.33 \u00d7 10-3<\/sup> = [OH–<\/sup>]
                                                                \nTherefore, [H+<\/sup>] = Kw<\/sub> \/ [OH–<\/sup>] = 10-14<\/sup> \/ (1.33 \u00d7 10-3<\/sup>)= 7.51 \u00d7 10-12<\/sup>
                                                                \npH = -log (7.5 \u00d7 10-12<\/sup> ) = 11.12
                                                                \nOn addition of 25 mL of 0.1 M HCl solution (i.e., 2.5 mmol of HCl) to 50 mL of 0.1 M ammonia solution (i.e., 5 mmol of NH3<\/sub>), 2.5 mmol of ammonia molecules are neutralized. The resulting 75 mL solution contains the remaining unneutralized 2.5 mmol of NH3<\/sub> molecules and 2.5 mmol of \\(\\mathrm{NH}_4^{+}\\)
                                                                \n\"AP
                                                                \nThe resulting 75 mL of solution contains 2.5 mmol of \\(\\mathrm{NH}_4^{+}\\) ions (i.e., 0.033M) and 2.5 mmol (i.e., 0.033M) of uneutralised NH3<\/sub> molecules. This NH3<\/sub> exists as NH4<\/sub>OH in the following equilibrium :
                                                                \n\"AP
                                                                \nThe final 75 mL solution after neutralisation already contains 2.5 m mol [latex]\\mathrm{NH}_4^{+}[\/latex] ions (i.e. 0.033M), thus total concentration of [latex]\\mathrm{NH}_4^{+}[\/latex] ions is given as : [NHJ] = 0.033 + y
                                                                \nAs y is small, [NH4<\/sub>OH] \u2243 0.033 M and [latex]\\mathrm{NH}_4^{+}[\/latex] \u2243 0.033M.
                                                                \nWe know,
                                                                \nKb<\/sub> = [latex]\\mathrm{NH}_4^{+}[\/latex] [OH–<\/sup>] \/ [NH4<\/sub>OH]
                                                                \n= y(0.033)\/(0.033) 1.77 \u00d7 10-5<\/sup> M
                                                                \nThus, y = 1.77 \u00d7 10-5<\/sup> = [OH–<\/sup>]
                                                                \n[H+<\/sup>] = 10-14<\/sup>\/ 1.77 \u00d7 10-5<\/sup> = 0.56 \u00d7 10-9<\/sup>
                                                                \nHence, pH = 9.24<\/p>\n

                                                                Question 25.
                                                                \nThe pKa<\/sub> of acetic acid and pKa<\/sub> of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of am-monium acetate solution.
                                                                \nSolution:
                                                                \npH = 7 + 1\/2 [pKa<\/sub> – pKb<\/sub>]
                                                                \n= 7 + 1\/2 [4.76 – 4.75]
                                                                \n= 7 + 1\/2 [0.01] = 7 + 0.005 = 7.005<\/p>\n

                                                                Question 26.
                                                                \nCalculate the solubility of A2<\/sub>X3<\/sub> in pure water, assuming that neither kind of ion reacts with water. The solubility product of A2<\/sub>X3<\/sub>, Ksp<\/sub> = 1.1 \u00d7 10-23<\/sup>.
                                                                \nSolution:
                                                                \nA2<\/sub>X3<\/sub> \u21cc 2A3+<\/sup> + 3X2-<\/sup>
                                                                \nKsp<\/sub> = [A3+<\/sup>]2<\/sup> [X2-<\/sup>]3<\/sup> = 1.1 \u00d7 10-23<\/sup>
                                                                \nIf S = solubility of A2<\/sub>X3<\/sub>, then
                                                                \n[A3+<\/sup>] = 2S; [X2-<\/sup>] = 3S
                                                                \ntherefore, Ksp<\/sub> = (2S)2<\/sup> (3S)3<\/sup> = 108S5<\/sup>
                                                                \n= 1.1 \u00d7 10-23<\/sup>
                                                                \nthus, S5<\/sup> = 1 \u00d7 10-23<\/sup>
                                                                \nS = 1.0 \u00d7 10-5<\/sup> mol\/L.<\/p>\n

                                                                \"AP<\/p>\n

                                                                Question 27.
                                                                \nThe values of Ksp<\/sub> of two sparingly soluble salts Ni(OH)2<\/sub> and AgCN are 2.0 \u00d7 10-15<\/sup> and 6 \u00d7 10-17<\/sup> respectively. Which salt is more soluble ? Explain.
                                                                \nSolution:
                                                                \nAgCN \u21cc Ag+<\/sup> + CN–<\/sup>
                                                                \nKsp<\/sub> = [Ag+<\/sup>] [CN–<\/sup>] = 6 \u00d7 10-17<\/sup>
                                                                \nNi(OH)2<\/sub> \u21cc Ni2<\/sup> + 2OH–<\/sup>
                                                                \nKsp<\/sub> = [Ni2+<\/sup>] [OH–<\/sup>]2<\/sup> = 2 \u00d7 10-15<\/sup>
                                                                \nLet [Ag+<\/sup>] = S1<\/sub> then [CN–<\/sup>] = S1<\/sub>
                                                                \nLet [Ni2+<\/sup>] = S2<\/sub> then [OH–<\/sup>] = 2S2<\/sub>
                                                                \nS1<\/sub>2<\/sup> = 6 \u00d7 10-17<\/sup> = S1<\/sub> = 7.8 \u00d7 10-9<\/sup>
                                                                \n(S2<\/sub>) (2S2<\/sub>)2<\/sup> = 2 \u00d7 10-15<\/sup>, S2<\/sub> = 0.58 \u00d7 10-4<\/sup>
                                                                \nNi(OH)2<\/sub> is more soluble than AgCN.<\/p>\n

                                                                Question 28.
                                                                \nCalculate the molar solubility of Ni(OH)2<\/sub> in 0.10 M NaOH. The ionic product of Ni(OH)2<\/sub> is 2.0 \u00d7 10-15<\/sup>.
                                                                \nSolution:
                                                                \nLet the solubility of Ni(OH)2<\/sub> be equal to S. Dissolution of S mol\/L of Ni(OH)2<\/sub> provides S mol\/L of Ni2+<\/sup> and 2S mol\/L of OH–<\/sup>, but the total concentration of OH–<\/sup> = (0.10 + 2S) mol\/L because the solution already contains 0.10 mol\/L of OH–<\/sup> from NaOH.
                                                                \nKsp<\/sub> = 2.0 \u00d7 10-15<\/sup> = [Ni2+<\/sup>] [OH–<\/sup>]2<\/sup>
                                                                \n(S) (0.10 + 2S)2<\/sup> As Ksp<\/sub> is small, 2S < < 0.10,
                                                                \nthus, (0.10 + 2S) \u2248 0.10
                                                                \nHence,
                                                                \n2.0 \u00d7 10-15<\/sup> S(0.10)2<\/sup>
                                                                \nS = 2.0 \u00d7 10-13<\/sup> M = [Ni2+<\/sup>]<\/p>\n","protected":false},"excerpt":{"rendered":"

                                                                Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 7th Lesson Chemical Equilibrium and Acids-Bases Textbook Questions and Answers. AP Inter 1st Year Chemistry Study Material 7th Lesson Chemical Equilibrium and Acids-Bases Very Short Answer Questions Question 1. State law of chemical equilibrium. Answer: The ratio of product of molar concentrations of the products …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35430"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35430"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35430\/revisions"}],"predecessor-version":[{"id":35431,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35430\/revisions\/35431"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35430"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35430"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35430"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}