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

💡 Solution & Explanation

Let the volume of each solution be $V \text{ mL}$. Millimoles of $OH^- = 0.1V$. Millimoles of $H^+ = 0.01V$. The $H^+$ neutralizes an equal amount of $OH^-$. The remaining unreacted $OH^- = 0.1V - 0.01V = 0.09V \text{ mmol}$. The total volume of the resulting mixture is $2V$. The final concentration $[OH^-] = \frac{0.09V}{2V} = 0.045 \text{ M} = 4.5 \times 10^{-2} \text{ M}$. The $pOH = -\log(4.5 \times 10^{-2}) = 2 - \log 4.5 = 2 - 0.65 = 1.35$. The resulting $pH = 14 - pOH = 14 - 1.35 = 12.65$.