Question

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

Answer 1

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