(a) 10 mL of 0.2 M $\mathrm{Ca(OH)_2}$ = 10 × 0.2 milli = 2 millimoles of $\mathrm{Ca(OH)_2}$ 25 mL of 0.1 M HCl = 25 × 0.1 millimoles = 2.5 millimoles of HCl $\mathrm{Ca(OH)_2} + 2\mathrm{HCl}$ → $\mathrm{CaCl_2} + 2\mathrm{H_2O}$ 1 millimole of $\mathrm{Ca(OH)_2}$ reacts with 2 millimoles of HCl ∴ 2.5 millimoles of HCl will react with 1.25 millimoles of $\mathrm{Ca(OH)_2}$ ∴ $\mathrm{Ca(OH)_2}$ left = 2 – 1.25 = 0.75 millimoles (HCl is the limiting reactant.) Total volume of the solution = 10 + 25 mL = 35 mL ∴ Molarity of Ca(OH)2 in the mixture solution = $\frac{0.75}{35}$ M = 0.0214 M ∴ $[\mathrm{OH}^-]$ = 2 × 0.0214 M = 0.0428 M = 4.28 × $10^{-2}$ M pOH = –log(4.28 × $10^{-2}$) = 2 – 0.6314 = 1.3686 ≈ 1.37 ∴ pH = 14 – 1.37 = 12.63 (b) 10 mL of 0.01 M $\mathrm{H_2SO_4}$ = 10 × 0.01 millimole = 0.1 millimole 10 mL of 0.01 M $\mathrm{Ca(OH)_2}$ = 10 × 0.01 millimole = 0.1 millimole $\mathrm{Ca(OH)_2} + \mathrm{H_2SO_4}$ → $\mathrm{CaSO_4} + 2\mathrm{H_2O}$ 1 mole of $\mathrm{Ca(OH)_2}$ reacts with 1 mole of $\mathrm{H_2SO_4}$ ∴ 0.1 millimole of $\mathrm{Ca(OH)_2}$ will react completely with 0.1 millimole of $\mathrm{H_2SO_4}$. hence, solution will be neutral with pH = 7.0 (c) 10 mL of 0.1 M $\mathrm{H_2SO_4}$ = 1 millimole 10 mL of 0.1 M KOH = 1 millimole $2\mathrm{KOH} + \mathrm{H_2SO_4}$ → $\mathrm{K_2SO_4} + 2\mathrm{H_2O}$ 1 millimole of KOH will react with 0.5 millimole of H2SO4 ∴ $\mathrm{H_2SO_4}$ left = 1 – 0.5 = 0.5 millimole Volume of reaction mixture = 10 + 10 = 20 mL ∴ Molarity of $\mathrm{H_2SO_4}$ in the mixture solution = $\frac{0.5}{20}$ = 2.5 × $10^{-2}\,\mathrm{M}$ $[\mathrm{H}^+]$ = 2 × 2.5 × $10^{-2}$ = 5 × $10^{-2}$ pH = –log(5 × $10^{-2}$) = 2 – 0.699 = 1.3