Question

$K_a$ for $HCN$ is $5\times {10}^{-10}$ at ${25}^{o}C$. For maintaining a constant $pH=9$, the volume of $5M$ $KCN$ solution required to be added to $10ml$ of $2M$ $HCN$ solution is:[Given ${\log }_{10}5=0.69$ : antilog $(-0.3010)=0.5$]

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

Answer 1

The correct option is B $2ml$

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

$\therefore$ $V=2mL$

Hence, the correct option is $B$