Question

Calculate the pH of a solution prepared by mixing $100mL$ of $0.4M$ $HCl$ with $100mL$ of $0.4M$ ${NH}_3$. Hydrolysis constant of ammonium chloride is $5.6\times {10}^{-10}$.

Answer 1

Aqueous solution of $N{H}_3$ is$N{H}_{4}OH$.

$HCl+N{H}_{4}OH\rightleftharpoons N{H}_{4}Cl+H_{2}O$

$40millimoles$ $40millimoles$

$40millimoles$

$\left[N{H}_{4}Cl\right]=40/200=0.2M$

$N{H}_4^{+}+H_{2}O\rightleftharpoons H^{+}+N{H}_{4}OH$

As we know that for salt of weak base and strong acid $K_h\times K_b=K_w$

Also,

$K_b=1.78\times {10}^{-5}$

$pH=7-0.5(p{K}_b+\log C)$

$pH=7-0.5(5-\log (1.78)+\log 0.2)$

$pH=4.96$