Question

The pH of 0.10 M KCN solution at 25 degre sol , for HCN , Ka =6.2*10^-10

Answer 1

Answer: pH of KCN solution will be 11.11

Explanation: Reaction of a strong base (KOH and weak acid (HCN) which leads to the formation of the salt and will have basic pH.

$KOH+HCN\rightleftharpoons KCN+H_2O$

we are getting $CN^-$ ions in the product, we need to look at the hydrolysis of $CN^-$ ions, which is

$CN^-+H_2O\rightleftharpoons HCN+ON^-$

here we are getting $OH^-$ ions, we calculate $k_b$

$k_b=\frac{[HCN][OH^-]}{[CN^-]}$           ......(1)

and $k_w=(k_a)(k_b)$

where, $k_w$ = equilibrium constant of water = $1\times10^{-14}$

$k_a$ = acid dissociation constant

$k_b$ = base dissociation constant

$k_b=\frac{k_w}{k_a}$

$k_a=6.2\times 10^{-10}$   (given)

putting values in above equation,

$k_b=\frac{1\times 10^{-14}}{6.2\times 10^{-10}}$

$k_b=1.612\times10^{-5}}$

For the hydrolysis reaction, let $[OH^-]=[HCN]=x$

$[CN^-]=0.10M$  (given)

Putting values in equation 1, we get

$1.612\times10^{-5}=\frac{(x)(x)}{0.10}$

$[OH^-]=x=1.2969\times 10^{-3}M$

$pOH=-log[OH^-]$

$pOH=-log(0.001296)$

pOH = 2.88

we can calculate pH easily by pOH by

pH = 14 - pOH

pH = 14 - 2.88

pH = 11.11