What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed? — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

Let volume be V mL. Moles of $OH^-$ = 0.1V, moles of $H^+$ = 0.01V. Excess $OH^-$ = 0.09V. Total volume = 2V. $[OH^-] = \frac{0.09V}{2V} = 0.045 \text{ M}$. $pOH = -\log(0.045) \approx 1.35$. pH = $14 - 1.35 = 12.65$.