Industrial effluent is found to be highly contaminated with silver and lead ions. A sample of water contains $8.0 \times {10^{ - 3}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{ A}}{{\text{g}}^ + }$ and $1.9 \times {10^{ - 2}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}{\text{ P}}{{\text{b}}^{2 + }}$. On the addition of chloride ions both ${\text{AgCl }}({K_{sp}} = 1.8 \times {10^{ - 10}})$ and ${\text{PbC}}{{\text{l}}_{\text{2}}}{\text{ }}({K_{sp}} = 1.7 \times {10^{ - 5}})$ precipitate from the solution. Determine the concentration of ${\text{C}}{{\text{l}}^ - }$ needed to initiate the precipitation of each salt and deduce which salt precipitates first.

${K_{{\text{sp}}}} = {\text{[A}}{{\text{g}}^ + }{\text{][C}}{{\text{l}}^ - }{\text{]}}/1.8 \times {10^{ - 10}} = 8.0 \times {10^{ - 3}} \times {\text{[C}}{{\text{l}}^ - }{\text{]}}$;

${\text{[C}}{{\text{l}}^ - }{\text{]}} = 2.3 \times {10^{ - 8}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}$;

${K_{{\text{sp}}}} = {\text{[P}}{{\text{b}}^{2 + }}{\text{][C}}{{\text{l}}^ - }{{\text{]}}^2}/1.7 \times {10^{ - 5}} = 1.9 \times {10^{ - 2}} \times {{\text{[C}}{{\text{l}}^ - }{\text{]}}^2}$;

${\text{[C}}{{\text{l}}^ - }{\text{]}} = 3.0 \times {10^{ - 2}}{\text{ mol}}\,{\text{d}}{{\text{m}}^{ - 3}}$;

AgCl will precipitate first (because it is less soluble);

Few candidates could calculate the concentrations correctly having problems with the calculations but most named AgCl as the first to precipitate.