Question

Calculate the pH of the solution in which $0.2MN{H}_{4}Cl$ and $0.1MN{H}_3$ are present. The ${\text{pK}}_b$ of ammonia is 4.75.

Answer 1

Given: solution contains $0.2MN{H}_{4}Cl$ and $0.1MN{H}_3$, and ${\text{pK}}_b=4.75$

Dissociation of $N{H}_3$ in water is given by reaction-
$N{H}_3+H_{2}O\rightleftharpoons N{H}_4^{+}+O{H}^{-}$

Step: 1 $K_b$ of $N{H}_3$

Ionization constant of $N{H}_3$:
$K_b=antilog(-p{K}_b)$
$\Rightarrow K_b={10}^{-4.75}=1.77\times {10}^{-5}M$

Step: 2 Equilibrium concentration
$N{H}_3+H_{2}O\rightleftharpoons N{H}_4^{+}+O{H}^{-}$

	$N{H}_3$	$N{H}_4^{+}$	$O{H}^{-}$
Initial concentration	$0.10$	$0.20$	$0$
Equilibrium concentration	$0.10-x$	$0.20+x$	x

Ionization constant , $K_b$ is given as:
$K_b=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_3\right]}=\frac{(0.20+x)\left(x\right)}{0.1-x}=1.77\times {10}^{-5}$

Step:3 pH calculation

Since, $K_b$ is very small, x can be neglected in comparision to $0.1M$ and $0.2M$.
Therefore,
$\left[O{H}^{-}\right]=x=0.88\times {10}^{-5}\Rightarrow pOH=-log\left[O{H}^{-}\right]=-log(0.88\times {10}^{-5})=5.05\Rightarrow pH=14–\text{pOH}=14–5.05=8.95$

Final answer:

$8.95$.