$pH$ of the aq. solution of $0.05 M$ $\mathrm{Ca}(\mathrm{CN}{)}_2\text{ at }{25}^{\circ }\mathrm{C}$ is: $\left[p{K}_b\left({\mathrm{CN}}^{-}\right)=8.26\right]$

$\frac{\mathrm{Ca}(\mathrm{OH}{)}_2}{\mathrm{SB}}+\frac{2\mathrm{HCN}}{W\mathrm{A}}\to \frac{\mathrm{Ca}(\mathrm{CN}{)}_2}{+2H_2\mathrm{O} \left(\mathrm{salt}\right)}$

$p^H=7+\frac{1}{2}\left(p^{ka}+\log c\right)$

$\text{ * At }{25}^{\circ }\mathrm{C}\text{; }$

$K_a\cdot K_b=K_w$

$p^{Ka}+p^{K_b}=p^{K_w}$

$p^{ka}+8\cdot 26=14$

$p{k}^a=5.74$

$=7+\frac{1}{2}\left(5.74+\log \left(5\times {10}^{-2}\right)\right)$

$7+\frac{1}{2}(5.74-1.33)$

$7+\frac{1}{2}\left(4.41\right)$

$=9.37$