What will be the exact resultant pH when of an aqueous solution of () is mixed with of an aqueous so — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

For $HCl$ at $pH=2$, $[H^+] = 10^{-2} \text{ M}$. Moles of $H^+ = 150 \times 10^{-2} = 1.5 \text{ mmol}$. For $NaOH$ at $pH=12$, $pOH = 2$, so $[OH^-] = 10^{-2} \text{ M}$. Moles of $OH^- = 350 \times 10^{-2} = 3.5 \text{ mmol}$. Upon mixing, $H^+$ and $OH^-$ neutralize. Remaining $OH^- = 3.5 - 1.5 = 2.0 \text{ mmol}$. Total volume is $150 + 350 = 500 \text{ mL}$. Final $[OH^-] = 2.0 / 500 = 4 \times 10^{-3} \text{ M}$. $pOH = -\log(4 \times 10^{-3}) = 3 - \log 4 = 3 - 0.602 = 2.4$. The resulting $pH = 14 - 2.4 = 11.6$.