Calculate the $pH$ of the resultant mixtures:
$(a)$ $10 \, mL$ of $0.2 \, M \, Ca(OH)_{2} + 25 \, mL$ of $0.1 \, M \, HCl$
$(b)$ $10 \, mL$ of $0.01 \, M \, H_{2}SO_{4} + 10 \, mL$ of $0.01 \, M \, Ca(OH)_{2}$
$(c)$ $10 \, mL$ of $0.1 \, M \, H_{2}SO_{4} + 10 \, mL$ of $0.1 \, M \, KOH$

$(a)$ Moles of $H_{3}O^{+} = \frac{25 \times 0.1}{1000} = 0.0025 \, mol$. Moles of $OH^{-} = \frac{10 \times 0.2 \times 2}{1000} = 0.0040 \, mol$. Excess $OH^{-} = 0.0040 - 0.0025 = 0.0015 \, mol$. Total volume = $35 \, mL = 0.035 \, L$. $[OH^{-}] = \frac{0.0015}{0.035} \approx 0.0428 \, M$. $pOH = -\log(0.0428) \approx 1.37$. $pH = 14 - 1.37 = 12.63$.
$(b)$ Moles of $H_{3}O^{+} = \frac{2 \times 10 \times 0.01}{1000} = 0.0002 \, mol$. Moles of $OH^{-} = \frac{2 \times 10 \times 0.01}{1000} = 0.0002 \, mol$. Since moles of $H_{3}O^{+}$ and $OH^{-}$ are equal,the solution is neutral. $pH = 7$.
$(c)$ Moles of $H_{3}O^{+} = \frac{2 \times 10 \times 0.1}{1000} = 0.002 \, mol$. Moles of $OH^{-} = \frac{10 \times 0.1}{1000} = 0.001 \, mol$. Excess $H_{3}O^{+} = 0.002 - 0.001 = 0.001 \, mol$. Total volume = $20 \, mL = 0.02 \, L$. $[H_{3}O^{+}] = \frac{0.001}{0.02} = 0.05 \, M$. $pH = -\log(0.05) \approx 1.30$.