Heavy metal ions can be removed by adding hydroxide ions. When hydroxide ions are added to a solution containing nickel ions, a precipitate of nickel(II) hydroxide, \({\text{Ni(OH}}{{\text{)}}_{\text{2}}}\), is formed. The solubility product of nickel(II) hydroxide is \(6.50 \times {10^{ - 18}}\) at 298 K. Determine the mass of nickel ions that remains in one litre (\({\text{1.00 d}}{{\text{m}}^{\text{3}}}\)) of water at 298 K with a pH of 7 after the precipitation reaction has occurred.

\({K_{{\text{sp}}}} = {\text{[N}}{{\text{i}}^{2 + }}{\text{]}} \times {{\text{[O}}{{\text{H}}^ - }{\text{]}}^2}\);

\({\text{[O}}{{\text{H}}^ - }{\text{]}} = 2{\text{[N}}{{\text{i}}^{2 + }}{\text{] hence }}{K_{{\text{sp}}}} = 4{{\text{[N}}{{\text{i}}^{2 + }}{\text{]}}^3}\);

\({\text{[N}}{{\text{i}}^{2 + }}{\text{]}} = {\left( {\frac{{6.50 \times {{10}^{ - 18}}}}{4}} \right)^{\frac{1}{2}}} = 1.18 \times {10^{ - 6}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}\);

Mass of \({\text{N}}{{\text{i}}^{2 + }}\) in \({\text{1 d}}{{\text{m}}^3} = 58.71 \times 1.18 \times {10^{ - 6}} = 6.90 \times {10^{ - 5}}{\text{ g}}\);

Award [4] for correct final answer.

Accept

K_sp = [Ni²⁺] \( \times \) [OH^–]²;

pH = pOH = 7 \( \to \) [OH^–] = 10^–7 mol\(\,\)dm^–3;

[Ni²⁺] = \(\frac{{6.50 \times 1{0^{ - 18}}}}{{{{(1{0^{ - 7}})}^2}}}\) = 6.50 \( \times \) 10^–4 mol\(\,\)dm^–3;

Mass of Ni²⁺ in 1 dm³ = 6.50 \( \times \) 10^–4 \( \times \) 58.71 = 3.82 \( \times \) 10^–2 g;

Award [4] for correct final answer.

Part (d) was poorly answered. Only very few candidates scored full marks, generally giving the alternative calculation. Some managed to score partial and EFC marks.