Question

$K_a$ for HCN is $5\times {10}^{-10}$ at $25{}^{\circ }C$. For maintaining a constant pH of 9, the volume of $5MKCN$ solution required to be added to $10mL$ of $2MHCN$ solution is $(\log 2=0.3andlog5=0.69)$

• A. 4 mL
• B. 8 mL
• C. 2 mL
• D. 10 mL

Answer 1

The correct option is C 2 mL

$k_a$ for $HCN=5\times {10}^{-10}$
volume of 5 M KCN solution required $\left(V_{mv}\right)=?$
$p{K}_a=-\log K_a=-\log (5\times {10}^{-10})=10-\log 5=9.3$
We know that $pH=p{K}_a+\log \left[\frac{salt}{acid}\right]\Rightarrow 9=9.3+\log \left[\frac{5\times V}{10\times 2}\right]\Rightarrow -0.3=\log \left[\frac{V}{4}\right]\Rightarrow 0.3=\log \left[\frac{4}{V}\right]\Rightarrow 2=\frac{4}{V}\Rightarrow V=2mL$