Question

Calculate the pH of an aqueous solution containing $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{M}\mathbf{HCl}$, $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{M}{\mathbf{H}}_2{\mathbf{SO}}_4$, and $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{M}\mathbf{HCN}\mathbf{.}$

Short answer

pH of the given solution = 1.5

Step-by-step solution

Concentration of [H+]from HCl and H2SO4

HCl and H₂SO₄ are strong acids, so these get completely dissociated Thus,

$\left[{\mathrm{H}}^{+}\right]\mathrm{from}\mathrm{HCl}=0.01\mathrm{M}$

$\begin{array}{c}\left[{\mathrm{H}}^{+}\right]\mathrm{from}{\mathrm{H}}_2{\mathrm{SO}}_4=2\times 0.01\text{M}\\ =0.02\mathrm{M}\end{array}$

Dissociation of HCN in water

HCN dissociates in water as:

$\begin{array}{l}{\text{HCN+H}}_{\text{2}}\text{O\hspace{1em}}\rightleftharpoons {\text{\hspace{1em}H}}_3{\text{O}}^{\text{+}}{\text{+CN}}^{\text{-}}\\ 0.010.030\\ 0.01-\mathrm{x}0.03+\mathrm{x}\mathrm{x}\end{array}$

pH Calculation 

$\begin{array}{l}{\text{[H}}^{+}\text{]}=0.03+\mathrm{x}\approx 0.03\left(\mathrm{as}\mathrm{x}\mathrm{is}\mathrm{very}\mathrm{low}\right)\\ {\text{pH=-log[H}}^{\text{+}}\text{]}=-\log (0.03)=1.5\end{array}$