Q. The dissociation constant of a substituted benzoic acid at $25℃\ is\ 1.0 \times 10^{- 4}$. The pH of 0.01 M solution of its sodium salt is
A) 3
B) 7
C) $8$
D) 6
Explanation
(c)
The hydrolysis reaction of conjugate base of acid is
$A^{-}(aq) + H_{2}O \longrightarrow HO^{-} + HA$
$K_{h} = \frac{K_{w}}{K_{a}} = \frac{10^{- 14}}{10^{- 4}} = 10^{- 10}$
Since, degree of hydrolysis is negligible;
$\left\lbrack OH^{-} \right\rbrack = \sqrt{K_{h}C} = 10^{- 6}.p\lbrack OH\rbrack = 6$
and pH=14-6=8
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.