Question

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

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

Answer: (B)

$8$

Answer 2

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