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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 70 of 73
Marks: +1, -0
The ionization constant of benzoic acid is 6.46 × 10−510^{-5} and Ksp for silver benzoate is 2.5 × 10−1310^{-13}. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?
Solution:  
Suppose S is the molar solubility of silver benzoate in water, then
C6H5COOAg(s)\mathrm{C_6H_5COOAg}_{(s)} ⇌ C6H5COO−(aq)+Ag+(aq)\mathrm{C_6H_5COO^{-}}_{(aq)} + \mathrm{Ag^{+}}_{(aq)}
KspK_{sp} = S2S^2 ∴ S = 2.5×10−13\sqrt{2.5 \times 10^{-13}} = 5.0 × 10−7 M10^{-7}\,\mathrm{M}
If the solubility of salt of weak acid of ionization constant KaK_a is S′, then KspK_{sp},
KaK_a and S′ are related to each other at pH = 3.19.
∴ [H+][\mathrm{H}^{+}] = 6.46 × 10−4 M10^{-4}\,\mathrm{M}
KspK_{sp} = S′2[KaKa+[H+]]S'^2 \left[ \frac{K_a}{K_a + [\mathrm{H}^{+}]} \right] , S′S' =
[2.5×10−136.46×10−56.46×10−5+6.46×10−4]1/2\left[ \frac{2.5 \times 10^{-13}}{ \frac{6.46 \times 10^{-5}}{6.46 \times 10^{-5} + 6.46 \times 10^{-4}} } \right]^{1/2}
S′S' = [2.5×10−13×7.106×10−46.46×10−5]1/2\left[ \frac{2.5 \times 10^{-13} \times 7.106 \times 10^{-4}}{6.46 \times 10^{-5}} \right]^{1/2} = (2.75×10−12)1/2(2.75 \times 10^{-12})^{1/2} = 1.658 × 10−6 M10^{-6}\,\mathrm{M}
∴ The ratio S′S\frac{S'}{S} = 1.658×10−65.0×10−7\frac{1.658 \times 10^{-6}}{5.0 \times 10^{-7}} = 3.32
Silver benzoate is 3.32 times more soluble in buffer of pH 3.19 than in pure water.
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