Regarding the rigorous calculation of pH for very dilute strong acid/base solutions (e.g., ), which — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

At extremely low concentrations of strong acids ($< 10^{-6} \text{ M}$), the $H^+$ coming from water is comparable and cannot be ignored. The added $H^+$ suppresses water's dissociation slightly via the common ion effect. If $x$ is the $H^+$ from water, total $H^+$ is $(x + 10^{-8})$, leading to $x(x + 10^{-8}) = 10^{-14}$, a quadratic equation. Because it's an acid, the pH must be $< 7$, so it cannot be 8.0.