Question

Question 7.70 The ionization constant of benzoic acid is 6.46 × 10^(–5) and Ksp for silver benzoate is 2.5 × 10^(–13). How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Class XI Equilibrium Page 230

Answer 1

first write the equation of Ionisation of Benzoic acid and Ionisation of benzoate

C6H5COOAg <=> C6H5COO- + Ag+ ; K1 = Ksp
C6H5COO- + H+ <=> C6H5COOH; K2 = 1/Ka [where Ka is Ionisation product of acid]

Now,add both equations
C6H5COOAg + H+ <=> C6H5COOH + Ag+ ; K3 = Ksp/Ka

But K3 = [C6H5COOH][Ag+]/[H+]
Let solubility of C6H5COOAg is s
Then, K3 = s.s/[H+] = s²/[H+] =Ksp/Ka

in a buffer solution PH = 3.19
log[H+] = -3.19 = 4(bar).81
[H+] = antilog(4(bar).81) = 6.46 × 10^-4

s²/[H+] = Ksp/Ka
s = √{Ksp × [H+]/Ka}
= √{2.5 × 10^-13 × 6.46×10^-4/6.46 × 10^-5}
= 1.6 × 10^-6 M
In buffer, solubility of C6H5COOAg= 1.6 × 10^-6 M

in aqueous solution, solubility of C6H5COOAg = s.s = Ksp
s = √(Ksp)
= √{2.5 × 10^-13} = 5 × 10^-7M

hence, C6H5COOAg is 3.2 times more soluble in buffer than in pure water.

Answer 2

Answer:

Explanation:

Since pH = 3.19,  Let the solubility of C6H5COOAg be x mol/L.  Then,  Thus, the solubility of silver benzoate in a pH 3.19 solution is 1.66 × 10–6 mol/L. Now, let the solubility of C6H5COOAg be x’ mol/L.  Hence, C6H5COOAg is approximately 3.317 times more soluble in a low pH solution.