pH of a solution obtained on mixing 50 ml of 0.1 M NaCN and 50 ml of 0.2 M HCl will be (_$p^{Ka}$ for HCN = 9.40)

_{$NaC{N}_{\left(aq\right)}+HC{l}_{\left(aq\right)}\to HC{N}_{\left(aq\right)}+NaC{l}_{\left(aq\right)}$}

Excess milli mol of Hcl =50 x 0.2 -50 x 0.1 = 5

_{${\left[HCl\right]}_{excess}=5/100=0.05M$}

_{$\left[H_3{O}^{+}\right]=0.05,pH=-\log 0.05=1.30$}                       

(HCN remains paractically unionised )