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 $10$ ml of $2M$ $HCN$ solution is:

• A. $2$ ml
• B. $4$ ml
• C. $7$ ml
• D. $9$ ml

Answer 1

Answer: (A)

$2$ ml

$HCN$ and $KCN$ will form buffer solution.

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

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

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

$\therefore$ $V=2$ ml.