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 \(1.0 \times 10^{-2} \mathrm{M} \mathrm{HCN}\)

First, we need to determine which acid among HCl, H2SO4, and HCN is the strongest. This can be done by considering their dissociation constant \(K_a\): The stronger the acid, the higher the \(K_a\) value. For the acids in question: - Hydrochloric acid (\(\mathrm{HCl}\)) is a strong acid with \(K_a\) much greater than 1 - Sulfuric acid (\(\mathrm{H_2SO_4}\)) is a strong acid with \(K_a1\) much greater than 1 for the first dissociation step and \(K_a2 = 1.0 \times 10^{-2}\) for the second dissociation step - Hydrogen cyanide (\(\mathrm{HCN}\)) is a weak acid with \(K_a = 6.2 \times 10^{-10}\) Since both \(\mathrm{HCl}\) and \(\mathrm{H_2SO_4}\) are strong acids with very high \(K_a\) values, they will dominate the dissociation process.

Now, we have to calculate the concentration of hydrogen ions (\([\mathrm{H^+}]\)) contributed by Hydrochloric acid (\(\mathrm{HCl}\)) and Sulfuric acid (\(\mathrm{H_2SO_4}\)). For \(\mathrm{HCl}\): \(1.0 \times 10^{-2}\text{ M of }\mathrm{HCl}\) will dissociate completely into hydrogen and chloride ions, thus \([\mathrm{H^+}] = 1.0 \times 10^{-2} \text{ M}\). For \(\mathrm{H_2SO_4}\): It is a diprotic acid with two dissociation steps. The first dissociation step, like HCl, is very strong and will dissociate completely, contributing \(1.0 \times 10^{-2}\text{ M}\) of \(\mathrm{H^+}\) ions. The second dissociation step, with \(K_a2 = 1.0 \times 10^{-2}\), is relatively weak, and due to the competition from the other two strong acids, can be considered negligible. Total [H+]: \([\mathrm{H^+}]_\text{total} = 1.0 \times 10^{-2} \text{ M (HCl) + } 1.0 \times 10^{-2} \text{ M (H2SO4) = 2.0 \times 10^{-2} \text{ M}\)

We now have the total contribution from the strong acids and can calculate the pH using the formula: \[ \text{pH} = -\log_{10}[\mathrm{H^+}] \] By applying the formula, we get: \[ \text{pH} = -\log_{10}(2.0 \times 10^{-2}) \] Now, we can calculate the pH of the solution:

Calculating the numerical value of the pH: \[ \text{pH} = -\log_{10}(2.0 \times 10^{-2}) \approx 1.70 \] Thus, the pH of the solution containing \(1.0 \times 10^{-2} M \mathrm{HCl}\), \(1.0 \times 10^{-2} M \mathrm{H_2SO_4},\) and \(1.0 \times 10^{-2} \mathrm{M} \mathrm{HCN}\) is approximately 1.70.