# 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?

M manish

It is given that,
$pH$ of buffer solution is 3.19. So, the concentration of $H^+$ ion can be calculated as;
$[H^+ ] = anti\log(-pH)$
= antilog (-3.19)
= $6.46\times 10^{-4}M$

Ionization of benzoic acid;

$C_6H_5COOH\rightleftharpoons C_6H_5COO^-+H_3O^+$
It is given that $K_a = 6.46\times 10^{-5}$
Therefore,
$K_a =\frac{[C_H_5COO^-][H_3O^+]}{[C_H_5COOH]}$
$\frac{[C_H_5COO^-]}{[C_H_5COOH]}=\frac{[H_3O^+]}{K_a} = 10$

Let the

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