Q.21. A 40.0 ml solution of weak base, BOH is titrated with 0.1N HCl solution. The pH of the solution is found to be 10.04 and 9.14 after adding 5.0 ml and 20.0 ml of the acid respectively. Find out the dissociation constant of the base.
Ans. 1.828 × 10^{–5}
Solution. Case I. Write the concerned chemical reaction
Since the solution represents a basic buffer, following Hendersen equation can be applied.
Case II.
Again the solution is acting as basic buffer
Substituting x in (i) and solving for K_{b}
Q.22. The solubility product (K_{sp}) of Ca(OH)_{2} at 25ºC is 4.42 × 10^{–5}. A 500 ml. of saturated solution of Ca(OH)^{2} is mixed with equal volume of 0.4 M NaOH. How much Ca(OH)_{2} in milligrams is precipitated?
Ans. 743.3 mg
Solution. Let the solubility of Ca(OH)2 in pure water = S moles/litre
Then K_{sp} = [Ca^{2+}] [OH^{–}]^{2}
4.42 × 10^{–5} = S × (2S)^{2}; 4.42 × 10^{–5} = 4S^{3}
S = 2.224 × 10^{–2} = 0.0223 moles litre^{–1}
∴ No. of moles of Ca^{2+} ions in 500 ml. of solution = λ
NOTE THIS STEP : Now when 500 ml. of saturated solution is mixed with 500 ml of 0.4M NaOH, the resultant volume is 1000 ml. The molarity of OH– ions in the resultant solution would therefore be 0.2 M.
Thus, No. of moles of Ca^{2+} or Ca(OH)_{2} precipitated = 0.01115 – 0.001105 = 0.010045
Mass of Ca(OH)_{2} precipitated
= 0.010045 × 74 = 0.7433 g = 743.3 mg
[mole wt. of Ca(OH)_{2} = 74]
Q.23. 0.15 mole of CO taken in a 2.5 l flask is maintained at 750 K along with a catalyst so that the following reaction can take place :
Hydrogen is introduced until the total pressure of the system is 8.5 atmosphere at equilibrium and 0.08 mole of methanol is formed. Calculate (i) K_{p} and K_{c} and (ii) the final pressure if the same amount of CO and H_{2} as before are used, but with no catalyst so that the reaction does not take place.
Ans. (i) 0.05 atm^{–2}, 187.85 mol^{–2} l^{2}, (ii) 12.438 atm
Solution. (i)
Total moles at equilibrium can also be calculated from the following relation
∴ 0.345 = a – 0.01 [Comparing (i) and (ii)]
or a = 0.355
Thus, Moles of CO at equilibrium = 0.15 – 0.08 = 0.07
Moles of H_{2} at equilibrium = 0.355 – 0.16 = 0.195
Moles of CH_{3}OH at equilibrium = 0.08
Substituting the values in the relation,
= 187.85 mole^{–2} litre2 [∵ V = 2.5 L]
Calculation of K_{p}
K_{p} = K_{c} (RT)^{Δn }= 187.85 × (0.0821 × 750)^{–2 }= 0.05 atm^{–2}
[∵ Δn = –2]
(ii) Calculation of final pressure when there is no reaction
Moles of CO = 0.15; Moles of H_{2} = 0.355
∴ Total moles = 0.15 + 0.355 = 0.505
PV = nRT
P × 2.5 = 0.505 × 0.0821 × 750 ⇒ P = 12.438 atm.
Q.24. The pH of blood stream is maintained by a proper balance of H_{2}CO_{3} and NaHCO_{3} concentrations. What volume of 5M NaHCO_{3} solution should be mixed with a 10 ml sample of blood which is 2M in H_{2}CO_{3} in order to maintain a pH of 7.4 ? K_{a} for H_{2}CO_{3} in blood is 7.8 × 10^{–7}.
Ans. 78.36 ml
Solution. Volume of blood = 10 ml. (given)
[H_{2}CO_{3}] in blood = 2 M (given)
[NaHCO_{3}] to be added = 5 M (given)
Let volume of NaHCO_{3} added in 10 ml blood = V ml
∴ [H_{2}CO_{3}] in blood mixture
[NaHCO_{3}] in blood mixture
Q.25. An aqueous solution of a metal bromide MBr_{2} (0.05M) is saturated with H_{2}S. What is the minimum pH at which MS will precipitate?
K_{sp} for MS = 6.0 x 10 ^{-21}; concentration of saturated H_{2}S = 0.1 M
K_{1} = 10^{–7} and K_{2} = 1.3 x 10^{–13} , for H_{2}S.
Ans. 0.983
Solution.
Dissociation constant of H_{2}S, K = K_{1} × K_{2}
i.e.K = 1 × 10^{–7} × 1.3 × 10^{–13} = 1.3 × 10^{–20}
Now we know that
Substituting the various values in the following relation
Q.26. At temperature T, a compound AB_{2} (g) dissociates according to the reaction
with a degree of dissociation x which is small compared with unity. Deduce the expression for x in terms of the equilibrium constant K_{p} and the total pressure, P.
Ans.
Solution.
Total moles at equb.
Q.27. For the reaction
the equilibrium constant, at 25°C , is 4.0 x 10^{ -19} . Calculate the silver ion concentration in a solution which was originally 0.10 molar in KCN and 0.03 molar in AgNO_{3}.
Ans. 7.5 × 10^{–18} M
Solution. TIPS/Formulae : Consider common ion effect
Conc. of Ag+ ions = Conc. of AgNO3 = 0.03 M
Most of these Ag+ ions will be present in the form of [Ag(CN)_{2}]^{–}.
0.03 M AgNO_{3} requires 2 × 0.03 M
= 0.06 M CN^{–} to form [Ag(CN)_{2}]^{–}
∴ Conc. of free CN^{–} at equilibrium will be 0.1 – 0.06 = 0.04 M
Q.28. Calculate the pH of an aqueous solution of 1.0 M ammonium formate assuming complete dissociation.
(pK_{a} of formic acid = 3.8 and pK_{b} of ammonia = 4.8.)
Ans. 6.5
Solution. For ammonium formate which is a salt of weak acid with weak base, we know that
Q.29. What is the pH of a 0.50 M aqueous NaCN solution? pKb of CN^{–} is 4.70.
Ans. 11.5
Solution.
Now we know that
pOH = –log[OH–]
pOH = –log 3.158 × 10^{–3} = 2.5
or, pH = 14 – 2.5 = 11.5
Q.30. A sample of AgCl was tr eated with 5.0 0 mL of 1.5 M Na_{2}CO_{3} solution to give Ag_{2}CO_{3}. The remaining solution contained 0.0026 g of Cl^{–} per litre. Calculate the solubility product of AgCl (Ksp(Ag_{2}CO_{3}) = 8.2 × 10^{–12}).
Ans. 1.71 × 10^{–10}
Solution. The concerned chemical reaction is
2AgCl + Na_{2}CO_{3} → Ag_{2}CO_{3 }+ 2 NaCl
Calculation of [Ag^{+}] left in the solution :
Concentration of Cl– left = 0.0026 g/l
∴ Ksp(AgCl) = [Ag^{+}] [Cl^{–}] = (2.34 × 10^{–6}) (7.33 × 10^{–5})
= 1.71 × 10^{–10}
Q.31. An acid type in dicator, HIn differ s in colour fr om its conjugate base (In^{–}). The human eye is sensitive to colour differences only when the ratio[In^{–}]/[HIn] is greater than 10 or smaller than. 0.1. What should be the minimum change in the pH of the solution to observe a complete colour change (Ka=1.0×10^{–5})?
Ans. 2
Solution. Given K_{a} = 1 × 10^{–5}
∴ pK_{a} = 5
The two conditions when colour indicator will be visible are derived by
(i) pH = 5 + log 10 = 6
(ii) pH = 5 + log 0.1 = 4
Thus minimum change in pH = 2
Q.32. Given : and K_{sp} of AgCl = 1.8 × 10^{–10} at 298 K. If ammonia is added to a water solution containing excess of AgCl(s) only, calculate the concentration of the complex in 1.0 M aqueous ammonia.
Ans. 0.0538 M
Solution.
NOTE THIS STEP : Since the formation constant of the complex is very high, most of the [Ag^{+}] which dissolves must be converted into complex and each Ag^{+} dissolved also requires dissolution of Cl^{–}.
∴ [Cl^{–}] = [Ag (NH_{3})_{2}]^{+} and let it be c M
Equation (i) becomes
Q.33. What will be the resultant pH when 200mL of an aqueous solution of HCl (pH = 2.0) is mixed with 300 mL of an aqueous solution of NaOH (pH = 12.0) ?
Ans. 11.3010
Solution.
pH of HCl = 2, pH of NaOH = 12
∴ [HCl] = 10^{–2} M, ∴ [NaOH] = 10^{–2} M
Q.34. When 3.06 g of solid NH_{4}HS is introduced into a two litre evacuated flask at 27° C, 30% of the solid decomposes into gaseous ammonia and hydrogen sulphide. (i) Calculate Kc and Kp for the reaction at 27°C. (ii) What would happen to the equilibrium when more solid NH_{4}HS is introduced into the flask ?
Ans. 8.1 × 10^{–5} mol^{2} l^{–2}, 4.90 × 10^{–2} atm^{2}
Solution.
NOTE : Addition of more NH_{4}HS on this equilibrium will cause no effect because concentration of NH_{4}HS is not involved in formula of K_{p} or K_{c}.
Q.35. The solubility of Pb(OH)_{2} in water is 6.7×10^{–6} M. Calculate the solubility of Pb(OH)_{2} in a buffer solution of pH = 8.
Ans. 1.203 × 10^{–3} mol litre^{–1}
Solution.
∴ K_{sp} = [Pb^{2+}][OH^{-}]^{2} = (6.7 × 10^{– 6}) (2 × 6.7 × 10^{– 6})^{2}
= 1.203 ×10^{–15}
The buffer solution pH = 8 (given)
∴ pOH = 6 or [OH^{–}] = 10^{–6}
Thus in this buffer we have, [Pb^{2+}][OH^{–}]^{2} = 1.203 x 10^{–15}
or [Pb^{2+}] × [10^{–6}]^{2} = 1.203 x 10^{–15}
∴ [Pb^{2+}] = 1.203 x 10^{–3} mol litre^{–1}
Q.36. The average concentration of SO_{2} in the atmosphere over a city on a certain day is 10 ppm, when the average temperature is 298 K. Given that the solubility of SO_{2} in water at 298 K is 1.3653 moles litre^{–1} and the pKa of H_{2}SO_{3} is 1.92, estimate the pH of rain on that day.
Ans. 4.865
Solution. Amount of SO_{2} in atmosphere =
Molar concentration of SO_{2} present in water = Amount of SO_{2} × Solubility of SO_{2} in water = 10 × 10^{–6 }× 1.3653 mole L^{–1} = 1.3653 × 10^{–5} M
Writing the concerned chemical equation
(pK_{a} = 1.92, ∴ K_{a} = 10^{–1.92})
x^{2} = 1.2 × 10^{–2} (1.3653 × 10^{-5} – x) On solving, x = 1.364 × 10^{–5}
Therefore, pH = – log (1.364 × 10^{–5}) = 4.865
Q.37. 500 mL of 0.2 M aqueous solution of acetic acid is mixed with 500 mL of 0.2 M HCl at 25°C.
(i) Calculate the degree of dissociation of acetic acid in the resulting solution and pH of the solution.
(ii) If 6 g of N aOH is a dded to th e above soluti on , determine the final pH. [Assume there is no change in volume on mixing; K_{a} of acetic acid is 1.75 × 10^{-5} mol L^{-1}].
Ans. 1.75 × 10^{–4}, 1, 4.75
Solution. (i) The volume being doubled by mixing the two solutions, the molarity of each component will be halved i.e.
[CH_{3}COOH] = 0.1 M, [HCl] = 0.1 M.
NOTE :
HCl being a strong acid will remain completely ionised and hence H^{+} ion concentration furnished by it will be 0.1 M. This would exert common ion effect on the dissociation of acetic acid, (a weak acid.)
Since α is very very small, Ca^{2} can be neglected and 1 – α can be taken as unity
Cα is negligible as compared to 0.1.
(ii)
0.1 mole of NaOH will be consumed by 0.1 mole of HCl.
Thus, 0.05 mole of NaOH will react with acetic acid according to the equation.
Thus, solution of acetic acid and sodium acetate will become acidic buffer. So pH of the buffer will be
Q.38. Match the following if the molecular weights of X, Y and Z are same.
| Boiling Point | ^{K}b |
X | 100 | 0.63 |
V | 27 | 0.53 |
z | 253 | 0.98 |
Ans. x = 0.63, y = 0.53, z = 0.98
Solution. TIPS/Formulae :
Higher the value of dipole-dipole interaction higher is b.p.
Higher value of K_{b} of a solvent suggests larger polarity of solvent molecules which in turn leads to higher dipole – dipole interaction implies higher boiling point due to dipole – dipole interaction. Therefore, the correct order of K_{b} values of the three given solvents is
or K_{b} ∝T_{b} (b.pt.)
Solvents | Boiling point | K_{b} values |
X | 100°C | 0.63 |
Y | 27°C | 0.53 |
Z | 283°C | 0.98 |