Question

Calculate the percent of hydrolysis in a $0.06M$ solution of $KCN$.

Use : $[K_{a\left(HCN\right)}=6\times {10}^{-10}]$.

• A. $1.667\%$
• B. $7.45\%$
• C. $34.67\%$
• D. None of these

Answer 1

The correct option is A $1.667\%$

$KCN$ is a salt of weak acid ($HCN$) and strong base ($KOH$).
Hence,
$K_h$$=\frac{K_w}{Ka}=$$\frac{{10}^{-14}}{6\times {10}^{-10}}=1.667\times {10}^{-5}$

Also,
$K_h$$=h^{2}C$

$h^2$=$\frac{1.667\times {10}^{-5}}{0.06}$=$2.778\times {10}^{-4}$

$h=\sqrt{2.778\times {10}^{-4}}=1.667\times {10}^{-2}$

Percent of hydrolysis$=h\times 100=1.667$