Question

Calculate the pH of the resultant mixtures:

10 mL of 0.2M $Ca(OH{)}_2$ + 25 mL of 0.1M HCl

10 mL of 0.01M $H_{2}S{O}_4$ + 10 mL of 0.01M $Ca(OH{)}_2$

10 mL of 0.1M $H_{2}S{O}_4$ + 10 mL of 0.1M KOH

Answer 1

Moles of $H_3{O}^{+}=\frac{25\times 0.1}{1000}=.0025mol$
Moles of $O{H}^{-}=\frac{10\times 0.2\times 2}{1000}=.0040$ mol
Thus, excess of $O0H^{-}=.0015$ mol
$\left[O{H}^{-}\right]=\frac{.0015}{35\times {10}^{-3}}mol/L=.428pOH=-log\left[OH\right]=1.36pH=14-1.36$
=12.63 (not matched)

Moles of $H_3{O}^{+}=\frac{2\times 10\times 0.01}{1000}=.0002$ mol
Moles of $O{H}^{-}=\frac{2\times 10\times .01}{1000}=.0002$ mol
Since there is neither an excess of $H_3{O}^{+}$ or $O{H}^{-}$

The solution is neutral. Hence, pH = 7.

Moles of $H_3{O}^{+}=\frac{2\times 10\times 0.1}{1000}=.002$ mol
Moles of $O{H}^{-}=\frac{10\times 0.1}{1000}=0.001$ Mol
Excess of $H_3{O}^{+}=.001$ mol
Thus, $\left[H_3{O}^{+}\right]=\frac{.001}{20\times {10}^{-3}}=\frac{{10}^{-3}}{20\times {10}^{-3}}=.05\therefore pH=-log\left(0.025\right)=1.30$