Question

$K_a$ for $HCN$ is $5\times 10^{-10}$ at $25^{\circ }C$. For maintaining a constant $pH$ of $9$ , the vol of $5MKCN$ solution required to be added to $10ml$ of $2MHCN$ solution is ___

Answer 1

Answer: 2

$pH=p{K}_a+\log \frac{[\text{Salt}]}{[\text{ Acid }]}$
$9=-\log \left[5\times 10^{-10}\right]+\log \frac{[\text{salt}]}{[\text{ Acid }]}$
Or $\log \frac{[\text{salt}]}{[\text{ Acid }]}=9+\log \left(5\times 10^{-10}\right)$
$=9+(0.6990-10)=-0.3010=\overline{1}.6990$
$\frac{[salt]}{[\text{ Acid }]}=0.5$
Suppose vol. of $5MKCN$ to be added $=V.c.c$
Then total vol. of $HCN+KCN=(10+V)c.c$
In the final solution Vc.c of $5MKCN$
$=(10+V)c.c$ of $1MKCN$
Molarity of $KCN\left(\frac{5V}{10+V}\right)$
$10c.c$ of $2MHCN=(10+V)c.c$ of $1MHCN$
Molarity of $HCN=\left(\frac{10\times 2}{10+V}\right)$
$\frac{[\text{ Salt }]}{[\text{ Acid }]}=\left(\frac{5V}{10+V}\right)\times \frac{(10+V)}{20}=\frac{5V}{20}=0.5$
Or $V=2mol$