Find the resulting $pH$ of the mixture of $200 \ mL$ of $HCl$ $(pH = 2)$ and $300 \ mL$ of $NaOH$ $(pH = 12.0)$.
(N/A) For $HCl$: $pH = 2$,so $[H^+] = 10^{-2} \ M$. Moles of $H^+ = 10^{-2} \ mol/L \times 0.2 \ L = 2 \times 10^{-3} \ mol$.
For $NaOH$: $pH = 12$,so $pOH = 14 - 12 = 2$. Thus,$[OH^-] = 10^{-2} \ M$. Moles of $OH^- = 10^{-2} \ mol/L \times 0.3 \ L = 3 \times 10^{-3} \ mol$.
Since $n(OH^-) > n(H^+)$,the mixture is basic.
Remaining moles of $OH^- = 3 \times 10^{-3} - 2 \times 10^{-3} = 1 \times 10^{-3} \ mol$.
Total volume $= 200 \ mL + 300 \ mL = 500 \ mL = 0.5 \ L$.
$[OH^-]_{result} = \frac{1 \times 10^{-3} \ mol}{0.5 \ L} = 2 \times 10^{-3} \ M$.
$pOH = -\log(2 \times 10^{-3}) = 3 - \log(2) = 3 - 0.301 = 2.699$.
$pH = 14 - 2.699 = 11.301 \approx 11.3$.
For $NaOH$: $pH = 12$,so $pOH = 14 - 12 = 2$. Thus,$[OH^-] = 10^{-2} \ M$. Moles of $OH^- = 10^{-2} \ mol/L \times 0.3 \ L = 3 \times 10^{-3} \ mol$.
Since $n(OH^-) > n(H^+)$,the mixture is basic.
Remaining moles of $OH^- = 3 \times 10^{-3} - 2 \times 10^{-3} = 1 \times 10^{-3} \ mol$.
Total volume $= 200 \ mL + 300 \ mL = 500 \ mL = 0.5 \ L$.
$[OH^-]_{result} = \frac{1 \times 10^{-3} \ mol}{0.5 \ L} = 2 \times 10^{-3} \ M$.
$pOH = -\log(2 \times 10^{-3}) = 3 - \log(2) = 3 - 0.301 = 2.699$.
$pH = 14 - 2.699 = 11.301 \approx 11.3$.
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