Question

The ionization constant of benzoic acid is $6.46\times {10}^{-5}$ and $K_{SP}$ for silver benzoate is $2.5\times {10}^{-13}$. How many times silver benzoate more soluble in a buffer of $pH=3.19$ as compared to its solubility in pure water?

• A. $3.317$
• B. $9.5$
• C. $1000$
• D. $7.5$

Answer 1

The correct option is A $3.317$

The reaction for the dissociation of silver benzoate in pure water is

$C_6{H}_{5}COOAg\rightleftharpoons \underset{(s-x)}{C_6{H}_{5}CO{O}^{-}}+\underset{s}{A{g}^{+}}$

Hence, the solubility product of silver benzoate is $s(s-x)=2.5\times {10}^{-13}.....\left(1\right)$

The reaction of benzoate ion with water is

$\underset{(s-x)}{C_6{H}_{5}CO{O}^{-}}+H_{2}O\rightleftharpoons \underset{x}{C_6{H}_{5}COOH}+\underset{x}{O{H}^{-}}$

The expression for the ionization constant of benzoic acid is

$\frac{x^2}{(s-x)}=\frac{{10}^{-14}}{(6.46\times {10}^{-5})}.....\left(2\right)$

In buffer solution of pH 3.19.

The reaction for the dissociation of silver benzoate is

$C_6{H}_{5}COOAg\rightleftharpoons \underset{(s^{\prime }-x^{\prime })}{C_6{H}_{5}CO{O}^{-}}+\underset{s^{\prime }}{A{g}^{+}}$

Hence, the solubility product of silver benzoate is $s(s-x)=2.5\times {10}^{-13}.....\left(3\right)$

$\underset{s^{\prime }(s^{\prime }-x^{\prime })=2.5\times {10}^{-13}}{C_6{H}_{5}CO{O}^{-}}\rightleftharpoons C_6{H}_{5}COOH$

Also,

$\frac{(s^{\prime }-x^{\prime })\times {10}^{-3.19}}{x^{\prime }}=6.46\times {10}^{-5}.....\left(4\right)$, solve for s'.

Hence, $\frac{s^{\prime }}{s}=3.317$.