Question

$K_a$ for $HCN$ is $5\times {10}^{-10}$ at ${25}^{o}C$. For maintaining a constant $pH$ of $9$, the volume of $5M$ $KCN$ solution required to be added to $10mL$ of $2M$ $HCN$ solution is

• A. $4$ mL
• B. $7.95$ mL
• C. $2$ mL
• D. $9.3$ mL

Answer 1

Answer: (C)

$2$ mL

$HCN$ and $KCN$ will form Buffer solution,

$pH=p{K}_a+log\frac{\left[salt\right]}{\left[acid\right]}$

$pH=-\log \left(5\right)\times {10}^{-10}+\log \left[\frac{5\times V}{V+10}/\frac{10\times 2}{V+10}\right]$

or $9=-\log \left(5\right)\times {10}^{-10}+\log \left(\frac{V}{4}\right)$

$\therefore$ $V=2$ mL