What will be the resultant pH when 200 ml of an aqueous solution of HCl (pH = 2.0) is mixed with 300 — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

For HCl: $pH = 2.0 \implies [H^+] = 10^{-2} \text{ M}$. Moles of $H^+ = 200 \times 10^{-3} \text{ L} \times 10^{-2} \text{ M} = 2 \text{ mmol}$. For NaOH: $pH = 12.0 \implies pOH = 2.0 \implies [OH^-] = 10^{-2} \text{ M}$. Moles of $OH^- = 300 \times 10^{-3} \text{ L} \times 10^{-2} \text{ M} = 3 \text{ mmol}$. Upon mixing, $OH^-$ is in excess: $3 - 2 = 1 \text{ mmol}$. Total volume = $500 \text{ ml}$. Final $[OH^-] = 1 \text{ mmol} / 500 \text{ ml} = 2 \times 10^{-3} \text{ M}$. $pOH = -\log(2 \times 10^{-3}) = 3 - \log 2 = 2.7$. $pH = 14 - 2.7 = 11.3$. Therefore, correct answer is B.