What is the pH of solution at 25°C? — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

For highly dilute acids like $10^{-7}\text{ M HCl}$, the contribution of $\text{H}^+$ from the autoionization of water cannot be ignored. Let $[\text{H}^+]$ from water be $x$. Then total $[\text{H}^+] = 10^{-7} + x$ and $[\text{OH}^-] = x$. Using $K_w = (10^{-7} + x)(x) = 10^{-14}$. Solving the quadratic gives $x \approx 0.618 \times 10^{-7}\text{ M}$. The total $[\text{H}^+] = 1.618 \times 10^{-7}\text{ M}$. The $\text{pH} = -\log(1.618 \times 10^{-7}) = 7 - \log 1.618 \approx 6.79$.