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

HCl and H2SO4 are strong acids, which means they will ionize completely in the solution. So, we need to determine their contribution to the concentration of H+ ions in the solution. For HCl, it will dissociate into H⁺ and Cl⁻, providing 1 H⁺ ion for each HCl molecule. Therefore, the H⁺ concentration from HCl is equal to the initial concentration of HCl: [H⁺] from HCl: \(1.0 \times 10^{-2}M\) H2SO4 is a dibasic acid, dissociating in two steps. In the first step, it completely ionizes, generating 1 H⁺ ion for each H2SO4 molecule: [H⁺] from first H⁺ of H2SO4: \(1.0 \times 10^{-2}M\) In the second step, the ionization is not complete, but since the concentration of the second H⁺ ion is negligible compared to the overall concentration of H⁺ ions, we can ignore the concentration of the second H⁺ ion.

HCN is a weak acid, which means it does not ionize completely. We will need to calculate the equilibrium concentration of H⁺ from the ionization of HCN using its Ka (acid dissociation constant) value. The value of Ka for HCN is \(6.2 \times 10^{-10}\). Considering the ionization equilibrium for HCN, \(HCN \rightleftharpoons H^+ + CN^-\) \(Ka = \frac{[H^+][CN^-]}{[HCN]}\) Since the initial concentration of HCl and H2SO4 is already providing a [H⁺] = \(2.0 \times 10^{-2}M\), we have to consider this when finding the concentration of H⁺ from HCN. Let x be the concentration of [H⁺] from the ionization of HCN. The equilibrium concentrations are expressed as: [H⁺] = \(2.0 \times 10^{-2} + x\) [CN⁻] = x [HCN] = \(1.0 \times 10^{-2} - x\) Now, plug these values into Ka expression: Ka = \(6.2 \times 10^{-10} = \frac{(2.0 \times 10^{-2} + x)x}{1.0 \times 10^{-2} - x}\) Due to the presence of strong acids, the value of x is small compared to \(2.0 \times 10^{-2}M\), so we can simplify the equation by ignoring x in the equation as follows: \(6.2 \times 10^{-10} ≈ \frac{(2.0 \times 10^{-2})x}{1.0 \times 10^{-2}}\)

Now, solve the equation for x: \(x ≈ 6.2 \times 10^{-10} \times \frac{1.0 \times 10^{-2}}{2.0 \times 10^{-2}}\) \(x ≈ 3.1 \times 10^{-10}\) The value of x represents the additional concentration of H⁺ ions from the ionization of HCN.

Add the H⁺ concentration from each acid to find the total H⁺ ion concentration: Total [H⁺] = [H⁺] from HCl + [H⁺] from H2SO4 + [H⁺] from HCN Total [H⁺] = \(1.0 \times 10^{-2} + 1.0 \times 10^{-2} + 3.1 \times 10^{-10}\) Total [H⁺] ≈ \(2.0 \times 10^{-2}\) Now, calculate the pH using the formula: pH = -log[H⁺] pH = -log[\(2.0 \times 10^{-2}\)] pH ≈ 1.70 So, the pH of the solution is approximately 1.70.