Question

Calculate the pH of the resultant mixtures:
a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl
b) 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2
c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH

Answer 1

(a) Moles of H3O+ = $\frac{25 × 0.1}{1000}$ = .0025 mol
Moles of OH- = $\frac{10 × 0.2 × 2}{1000}$ = .0040 mol
Thus, an excess of OH- = .0015 mol
[OH-] =$\frac{ .0015}{35 × 10^{-3}}$ mol/L = .0428
pOH = -log[OH] = 1.36
pH = 14 - 1.36 = 12.63 (not matched)

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(b) Moles of H3O+ =$\frac{10 × 0.01 × 2}{1000}$ = .0002 mol
Moles of OH- = $\frac{10 × 0.1 × 2}{1000}$ = .0002 mol
Since there is neither an excess of H3O+ or OH-, the solution is neutral. Hence, pH = 7.

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(c) Moles of H3O+ = $\frac{10 × 0.1 × 2}{1000}$= .002 mol
Moles of OH- = $\frac{10 × 0.1 }{1000}$ = 0.001 mol
Excess of H3O+ = .001 mol
Thus, [H3O+] = $\frac{.001}{20 × 10^{-3}}$ = $\frac{10^{-3}}{20 × 10{-3}}$ = 0.5
∴ pH = -log(0.05) = 1.30