A volume of $20 \mathrm{ml}$ of $0.8 \mathrm{M}-\mathrm{HCN}$ solution is mixed with $80 \mathrm{ml}$ of $0.4 \mathrm{M}$
- NaCN solution. Calculate the pH of the resulting solution. $K_{\mathrm{a}}$ of $\mathrm{HCN}=2.5 \times 10^{-10} .$ $(\log 2=0.3)$
(a) $9.9$
(b) $9.3$
(c) $4.1$
(d) $4.7$