Question

The hydrolysis constant of $N{H}_{4}Cl$ is $5.6\times {10}^{-6}$. Find the $pH$ of 0.1 molar solution of $N{H}_{4}Cl$ at equilibrium. When $log1.79=0.253$.

• A. 3.13
• B. 7
• C. 5.24
• D. 8.95

Answer 1

The correct option is A 3.13

Given, $K_h=5.6\times {10}^{-6},\left[N{H}_{4}Cl\right]=0.1M$
$N{H}_{4}Cl$ is a mixture of strong acid and weak base
we know the relation
$K_h=\frac{K_w}{K_b}$
$5.6\times {10}^{-6}=\frac{{10}^{-14}}{K_b}$
$K_b=\frac{{10}^{-14}}{5.6\times {10}^{-6}}$
$K_b=1.79\times {10}^{-9}$
$p{K}_b=-log{K}_b$
$=-log1.79\times {10}^{-9}$
$=9-log1.79$
$=9-0.253$
​​​​​​​$=8.747$
Also, $pH=\frac{1}{2}\times (p{K}_w-p{K}_b-logc)$
$pH=\frac{1}{2}\times (14-8.75-log0.1)$
$pH=\frac{1}{2}\times (14-8.75+1)$
$pH=3.125$