Question

If the hydrolysis constant of $N{H}_{4}Cl$ is $4\times {10}^{-10}$ at $298K$. Find the $pH$ of $0.5M$ solution of $N{H}_{4}Cl$ at equilibrium.
Given : $\log 5=0.7$

• A. 8.22
• B. 2.5
• C. 7.34
• D. 4.85

Answer 1

The correct option is D 4.85

Given, $K_h=4\times {10}^{-10},\left[N{H}_{4}Cl\right]=0.5M$
$N{H}_{4}Cl$ is a mixture of strong acid and weak base
we know the relation
$K_h=\frac{K_w}{K_b}$
$4\times {10}^{-10}=\frac{{10}^{-14}}{K_b}$
$K_b=\frac{{10}^{-14}}{4\times {10}^{-10}}$
$K_b=2.5\times {10}^{-5}$
$p{K}_b=-{\log }_{10}{K}_b$
$=-\log (2.5\times {10}^{-5})$
$=6-\log 25$
$=6-2\times \log 5$
$=4.6$
Also, $pH=\frac{1}{2}\times (p{K}_w-p{K}_b-\log C)$
$pH=\frac{1}{2}\times [14-4.6-\log 0.5]$
$pH=\frac{1}{2}\times [14-4.6+(1-0.7\left)\right]$
$pH=4.85$