Question

The hydrolysis constant of $N{H}_{4}Cl$ is $4\times {10}^{-10}$. Find the $pH$ of 0.1 molar solution of $N{H}_{4}Cl$ at equilibrium at ${25}^{\circ }C$.

• A. 3.13
• B. 7
• C. 5.24
• D. 8.95

Answer 1

The correct option is A 3.13

Given, $K_h=4\times {10}^{-10},\left[N{H}_{4}Cl\right]=0.1M$
$N{H}_{4}Cl$ is a mixture of strong acid and weak base
we know the relation
$K_h=\frac{K_w}{K_b}$
$4\times {10}^{-10}=\frac{{10}^{-14}}{K_b}$
$K_b=2.5\times {10}^{-5}$
$p{K}_b=-log{K}_b$
$=-log2.5\times {10}^{-5}$
$=6-2\times 0.7$
$=4.6$
Also, $pH=\frac{1}{2}\times (p{K}_w-p{K}_b-logc)$
$pH=\frac{1}{2}\times (14-4.6-log0.1)$

$pH=5.2$