Question

Calculate the degree of hydrolysis and $pH$ of $0.02\, M$ ammonium cyanide $\left( NH _{4} CN \right)$ at $298\, K$. $\left[K_{a}\right.$ of $HCN =4.99 \times 10^{-9}, K_{b}$ for $NH _{4} OH =1.77 \times 10^{-5}$ ]

• A. 8.2
• B. 3.2
• C. 9.3
• D. 3.9

Answer 1

Answer: (C)

$K_{h}=\frac{10^{-14}}{4.99 \times 10^{-9} \times 1.77 \times 10^{-5}}=1.132$
It can be seen that hydrolysis constant $\left(K_{h}\right)$ is not small, and for calculating $h$, the equation used is
$h =\frac{\sqrt{K_{h}}}{1+K_{h}}=\frac{\sqrt{1.132}}{1+\sqrt{1.132}}$
$=\frac{1.06}{1+1.06} $
$=\frac{1.06}{2.06}=0.51$
Using the formula
$ pH = p K_{a}-\log h+\log (1-h) $
$=-\log \left(4.99 \times 10^{-10}\right)-\log (0.51)+\log (1-0.51)$
$=9.3$