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  4. The dissociation constant of a substituted benzoic acid at 25°C is 1.0 × 10-4. The pH of 0.01 M solution of its sodium salt is

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Q. The dissociation constant of a substituted benzoic acid at $25^{\circ}C$ is $1.0 \times 10^{-4}$. The $pH$ of $0.01 \,M$ solution of its sodium salt is

UP CPMTUP CPMT 2012Equilibrium

Solution:

The hydrolysis reaction of conjugate base of acid is
$A^{-}_{\left(aq\right)} +H_{2}O \rightarrow HO^{-}+HA$
$K_{h}=\frac{K_{w}}{K_{a}}=\frac{10^{-14}}{10^{-4}}=10^{-10}$
Since, degree of hydrolysis is negligible;
$\therefore \left[OH^{-}\right]=\sqrt{K_{h}C}$
$=\sqrt{10^{-10} \times 10^{-2}}=\sqrt{10^{-12}}=10^{-6}$
$\therefore pOH=log\,10^{-\left[OH^-\right]}$
$\therefore pOH=6$
$pH=14-6=8$