Chemistry - Ionic Equilibrium Question with Solution | TestHub

Initially: $\mathrm{HCN}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{CN}}^{-}$

${\alpha }_1=\sqrt{\frac{K_a}{c}}=\sqrt{\frac{10^{-5}}{0.1}}=10^{-2}$If V ml of HCl is added, so that

$${\alpha }_2\text{ of }\mathrm{HCN}=\frac{{\alpha }_1}{2}=\frac{10^{-2}}{2}=5\times 10^{-3}$$

$$\therefore [\mathrm{HCN}]=\frac{0.1^2\times 100^2}{(100+\mathrm{V})}=\frac{10}{(100+\mathrm{V})}$$

$$[\mathrm{HCl}]=\frac{0.1\times \mathrm{V}}{(100+\mathrm{V})}$$

$$\mathrm{HCN}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{CN}}^{-}$$At equilibrium:

$$\left[{\mathrm{H}}^{+}\right]=\frac{10}{(100+\mathrm{V})}\times 5\times 10^{-3}+\frac{0.1\mathrm{V}}{(100+\mathrm{V})}$$Therefore:

$$\left[{\mathrm{H}}^{+}\right]=\left(\frac{0.1\mathrm{V}+0.05}{100+\mathrm{V}}\right)$$and

$$\left[{\mathrm{CN}}^{-}\right]=\frac{10\times 5\times 10^{-3}}{(100+\mathrm{V})}$$

$$\therefore K_a=10^{-5}=\frac{\left(\frac{0.1\mathrm{V}+0.05}{100+\mathrm{V}}\right)\left(\frac{5\times 10^{-2}}{100+\mathrm{V}}\right)}{\frac{10}{(100+\mathrm{V})}}$$

$$10^{-4}=\left(\frac{0.1\mathrm{V}+0.05}{100+\mathrm{V}}\right)\times 5\times 10^{-2}$$

$$2\times 10^{-3}(100+\mathrm{V})=(0.1\mathrm{V}+0.05)$$

$$0.2+2\times 10^{-3}\mathrm{V}=0.1\mathrm{V}+0.05$$

$$\mathrm{V}=\frac{0.15}{0.1}=1.5\mathrm{ml}$$