What is the exact pH of the resulting solution when equal volumes of and are intimately mixed? (Expr — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Let the volume of each solution be $V$. Millimoles of $OH^- = 0.1V$. Millimoles of $H^+ = 0.01V$. The strong acid completely neutralizes an equal amount of strong base. Remaining $OH^- = 0.1V - 0.01V = 0.09V \text{ mmol}$. The total volume is $V + V = 2V$. The resulting hydroxide concentration $[OH^-] = \frac{0.09V}{2V} = 0.045 \text{ M} = 4.5 \times 10^{-2} \text{ M}$. The $pOH = -\log(4.5 \times 10^{-2}) = 2 - \log 4.5$. Since $\log 4.5 \approx 0.65$, $pOH = 2 - 0.65 = 1.35$. The $pH = 14 - 1.35 = 12.65$.