Chemistry - Ionic Equilibrium Question with Solution | TestHub

Suppose, volume of KCN added = x ml
Total volume of solution after mixing = (10 + x) ml
Molarity of HCN in the final solution
$=\frac{{\mathrm{M}}_1{\mathrm{V}}_1}{{\mathrm{V}}_2}=\frac{2\times 10}{(10+\mathrm{x})}=\frac{20}{(10+\mathrm{x})}\mathrm{M}$
Molarity of HCN in the final solution
$=\frac{{\mathrm{M}}_1{\mathrm{V}}_1}{{\mathrm{V}}_2}=\frac{5\times \mathrm{x}}{(10+\mathrm{x})}=\frac{5\mathrm{x}}{(10+\mathrm{x})}\mathrm{M}$
pH$=-\log {\mathrm{k}}_{\mathrm{a}}+\log \frac{\left[\mathrm{K}\mathrm{C}\mathrm{N}\right]}{\left[\mathrm{H}\mathrm{C}\mathrm{N}\right]}$
$\text{9}\text{=}-\text{log}{\text{(5×10}}^{-\text{10}}\text{)}\text{+}\text{log}\frac{\frac{\text{5x}}{\left(\text{10}\text{+}\text{x}\right)}}{\frac{\text{20}}{\left(\text{10}\text{+}\text{x}\right)}}$
On solving,$\text{x}\text{=}\text{2}\text{mL}$