Calculate the \(\mathrm{pH}\) of an aqueous solution containing \(1.0 \times 10^{-2} M\) \(\mathrm{HCl}, 1.0 \times 10^{-2} \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\), and \(1.0 \times 10^{-2} \mathrm{M} \mathrm{HCN}\).

Since \(\mathrm{HCl}\) is a strong acid, we assume it fully dissociates in water: \(\mathrm{HCl} \rightarrow \mathrm{H}^{+} + \mathrm{Cl}^{-}\) The initial concentration of \(\mathrm{HCl}\) is \(1.0 \times 10^{-2} M\). As it fully dissociates, the \(\mathrm{H}^{+}\) concentration coming from \(\mathrm{HCl}\) is the same as the initial concentration of \(\mathrm{HCl}\): \[[\mathrm{H}^{+}]_{\mathrm{HCl}}=1.0 \times 10^{-2} M\]

For the first dissociation of \(\mathrm{H}_{2}\mathrm{SO}_{4}\), which is a strong acid, it dissociates as follows: \(\mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow \mathrm{H}^{+} + \mathrm{HSO}_{4}^{-}\) The initial concentration of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) is \(1.0 \times 10^{-2} M\). As only the first dissociation is strong, the \(\mathrm{H}^{+}\) concentration coming from \(\mathrm{H}_{2}\mathrm{SO}_{4}\) is also the same as the initial concentration of \(\mathrm{H}_{2}\mathrm{SO}_{4}\): \[[\mathrm{H}^{+}]_{\mathrm{H}_{2}\mathrm{SO}_{4}}=1.0 \times 10^{-2} M\]

Now for \(\mathrm{HCN}\), a weak acid, we must use its \(K_a\) value to find the \(\mathrm{H}^{+}\) contribution. The acid dissociation for \(\mathrm{HCN}\) is: \(\mathrm{HCN} \rightleftharpoons \mathrm{H}^{+} + \mathrm{CN}^{-}\) For this equation, the \(K_a\) expression is: \[K_a = \frac{[\mathrm{H}^{+}][\mathrm{CN}^{-}]}{[\mathrm{HCN}]}\] Given \(K_a=6.2 \times 10^{-10}\) for \(\mathrm{HCN}\) and the initial concentration, \(1.0 \times 10^{-2} M\), we can set up an ICE (Initial, Change, Equilibrium) table: | | \(\mathrm{H}^{+}\) | \(\mathrm{CN}^{-}\) | \(\mathrm{HCN}\) | |---------|--------------|----------------|-------------| | Initial | \(x\) | \(0\) | \(1.0 \times 10^{-2}\) | | Change | +\(x\) | +\(x\) | -\(x\) | | Eqm | \(x\) | \(x\) | \(1.0 \times 10^{-2} - x\) | Now we can substitute the values into the \(K_a\) expression and solve for \(x\), which is the \(\mathrm{H}^{+}\) concentration from \(\mathrm{HCN}\): \[6.2 \times 10^{-10} = \frac{x \times x}{1.0 \times 10^{-2} - x}\] As \(K_a\) for \(\mathrm{HCN}\) is very small, the dissociation of \(\mathrm{HCN}\) is very low, and we can make the assumption that \(x \ll 1.0 \times 10^{-2}\). This simplifies the equation to: \[6.2 \times 10^{-10} \approx \frac{x^{2}}{1.0 \times 10^{-2}}\] Solving for \(x\), we find: \[[\mathrm{H}^{+}]_{\mathrm{HCN}} \approx 2.5 \times 10^{-6} M\]

Now that we have the \(\mathrm{H}^{+}\) concentrations coming from each acid, we can find the total \(\mathrm{H}^{+}\) concentration by adding all three contributions: \[[\mathrm{H}^{+}]_{\text{total}}=[\mathrm{H}^{+}]_{\mathrm{HCl}}+[\mathrm{H}^{+}]_{\mathrm{H}_{2}\mathrm{SO}_{4}}+[\mathrm{H}^{+}]_{\mathrm{HCN}}\] \[[\mathrm{H}^{+}]_{\text{total}}=1.0 \times 10^{-2} M + 1.0 \times 10^{-2} M + 2.5 \times 10^{-6} M \approx 2.0 \times 10^{-2} M\] Finally, we can use the pH formula to calculate the pH of the solution: \[pH = -\log_{10}[\mathrm{H}^{+}]\] \[pH = -\log_{10}(2.0 \times 10^{-2} M) \approx 1.70\] So, the pH of the aqueous solution containing \(1.0 \times 10^{-2} M\) \(\mathrm{HCl}\), \(1.0 \times 10^{-2} \mathrm{M} \mathrm{H}_{2}\mathrm{SO}_{4}\), and \(1.0 \times 10^{-2} \mathrm{M} \mathrm{HCN}\) is approximately 1.70.