What is the pH of solution? (, , , ) — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Volume $= 10\text{ m}^3 = 10^4\text{ L}$. Concentration $C = \frac{0.7 \times 10^{-3}\text{ mol}}{10^4\text{ L}} = 7.0 \times 10^{-8}\text{ M}$. Because the solution is highly dilute ($C < 10^{-6}\text{ M}$), we must consider the auto-ionization of water and full dissociation of the weak acid. The $\text{H}^+$ from the acid is approximately $7 \times 10^{-8}\text{ M}$. Taking water into account: $[\text{H}^+][\text{OH}^-] = 10^{-14}$ and $[\text{H}^+] = [\text{A}^-] + [\text{OH}^-] \implies $[\text{H}^+] = 7 \times 10^{-8} + \frac{10^{-14}}{[\text{H}^+]}$. This gives $[\text{H}^+]^2 - 7 \times 10^{-8}[\text{H}^+] - 10^{-14} = 0$. Solving yields $[\text{H}^+] \approx 1.41 \times 10^{-7}\text{ M}$. $\text{pH} = -\log(1.41 \times 10^{-7}) \approx 7 - 0.15 = 6.85$. Therefore, correct answer is C.