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

HCl is a strong acid, meaning it completely ionizes in water. Therefore, the concentration of \(\mathrm{H}^{+}\) ions contributed by HCl in the solution will be equal to its initial concentration, which is \(1.0 \times 10^{-2} M\).

H2SO4 is a diprotic acid, meaning it can donate two protons. Its first ionization is nearly complete, so the contribution of \(\mathrm{H}^{+}\) ions from the first ionization of H2SO4 is equal to its initial concentration, which is \(1.0 \times 10^{-2} M\). Its second ionization is much weaker, and we can neglect its contribution, as it will be much smaller compared to the contributions from the other two components.

HCN is a weak acid, and its ionization needs to be considered explicitly. For the ionization of HCN in water, \[ \mathrm{HCN} \leftrightharpoons \mathrm{H^+} + \mathrm{CN^-} \] We are given the initial concentration of HCN as \(1.0 \times 10^{-2} M\). Let \(x\) be the concentration of \(\mathrm{H}^+\) produced by HCN's dissociation. Using the equilibrium expression and the \(K_\mathrm{a}\) value for HCN (which is \(6.2 \times 10^{-10}\)), we can set up the equation as follows: \[K_\mathrm{a} = \frac{[\mathrm{H^+}][\mathrm{CN^-}]}{[\mathrm{HCN}]} = \frac{x^2}{1.0 \times 10^{-2} - x}\] Due to the small value of \(K_\mathrm{a}\) and since \(x\) is very small compared to the initial concentration of HCN, we can approximate and simplify the expression to: \[ x^2 \approx 6.2 \times 10^{-10} \times 1.0 \times 10^{-2} \] Solve for \(x\) (the concentration of \(\mathrm{H}^+\) ions contributed by HCN): \[x \approx \sqrt{6.2 \times 10^{-12}}\]

Now, we add up the concentrations of \(\mathrm{H}^+\) ions from all three components: \[[\mathrm{H}^+]_{\text{total}} = [\mathrm{H}^+]_{\text{HCl}} + [\mathrm{H}^+]_{\text{H2SO4}} + [\mathrm{H}^+]_{\text{HCN}} \] \[[\mathrm{H}^+]_{\text{total}} = 1.0 \times 10^{-2} + 1.0 \times 10^{-2} + \sqrt{6.2 \times 10^{-12}}\] Finally, calculate the pH using the formula: pH = \(-\log([\mathrm{H}^+])\) \[\text{pH} = -\log(2.0 \times 10^{-2} + \sqrt{6.2 \times 10^{-12}}) \] Solving for pH, we get a value close to 1.70.