Question

If the hydrolysis constant of $N{H}_{4}Cl$ is $2\times {10}^{-5}$ 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, $K_h=2\times {10}^{-5},\left[N{H}_{4}Cl\right]=0.5M$
$N{H}_{4}Cl$ is a mixture of strong acid and weak base
we know the relation
$K_h=\frac{K_w}{K_b}$
$2\times {10}^{-5}=\frac{{10}^{-14}}{K_b}$
$K_b=\frac{{10}^{-14}}{2\times {10}^{-5}}$
$K_b=5.0\times {10}^{-10}$
$p{K}_b=-{\log }_{10}{K}_b$
$=-\log (5.0\times {10}^{-10})$
$=10-\log 5$
$=10-0.7$
$=9.3$
Also, $pH=\frac{1}{2}\times (p{K}_w-p{K}_b-\log C)$
$pH=\frac{1}{2}\times [14-9.3-\log 0.5]$
$pH=\frac{1}{2}\times [14-9.3+(1-0.7\left)\right]$
$pH=2.5$