Calculate the pH of the resultant mixtures:
a) $10 \mathrm{~mL}$ of $0.2 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}+25 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCl}$
b) $10 \mathrm{~mL}$ of $0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}+10 \mathrm{~mL}$ of $0.01 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}$
c) $10 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}+10 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{KOH}$
(a) Moles of $\mathrm{H}_{3} \mathrm{O}^{+}=\frac{25 \times 0.1}{1000}=.0025 \mathrm{~mol}$
Moles of $\mathrm{OH}^{-}=\frac{10 \times 0.2 \times 2}{1000}=.0040 \mathrm{~mol}$
Thus, excess of $\mathrm{OH}^{-}=0015 \mathrm{~mol}$
$\left[\mathrm{OH}^{-}\right]=\frac{.0015}{35 \times 10^{-3}} \mathrm{~mol} / \mathrm{L}=.0428$
$\mathrm{pOH}=-\log [\mathrm{OH}]$
$=1.36$
$\mathrm{pH}=14-1.36$
$=12.63$
(b) Moles of $\mathrm{H}_{3} \mathrm{O}^{+}=\frac{2 \times 10 \times 0.01}{1000}=.0002 \mathrm{~mol}$
Moles of $\mathrm{OH}^{-}=\frac{2 \times 10 \times .01}{1000}=.0002 \mathrm{~mol}$
Since there is neither an excess of $\mathrm{H}_{3} \mathrm{O}^{+}$or $\mathrm{OH}^{-}$, the solution is neutral. Hence, $\mathrm{pH}=7$.
(c) Moles of $\mathrm{H}_{3} \mathrm{O}^{+}=\frac{2 \times 10 \times 0.1}{1000}=.002 \mathrm{~mol}$
Moles of $\mathrm{OH}^{-}=\frac{10 \times 0.1}{1000}=0.001 \mathrm{~mol}$
Excess of $\mathrm{H}_{3} \mathrm{O}^{+}=.001 \mathrm{~mol}$
Thus, $\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{.001}{20 \times 10^{-3}}=\frac{10^{-3}}{20 \times 10^{-3}}=.05$
$\therefore \mathrm{pH}=-\log (0.05)$
$=1.30$