Question

An aqueous solution contains $N{i}^{2+},C{o}^{2+}$ and $P{b}^{2+}$ ions at equal concentrations. The solubility product of $NiS,PbS$ and $CoS$ in water at $25^{\circ }C$ are $1.4\times 10^{-24}$, and $3\times 10^{-26}$, respectively. Indicate which of these ions will be precipitated first and last when sulphide concentration is progressively increased from zero?

• A. NiS and PbS
• B. NiS and CoS
• C. CoS and NiS
• D. PbS and NiS

Answer 1

Answer: (D)

For precipitation, ionic product > solubility product
$\therefore$ The metal sulphide with lowest $K_{sp}(PbS)$
will precipitate out first because its ionic product will exceed its $K_{sp}$
first and the metal sulphide with lowest $K_{sp}(NiS)$ will precipitate out at last.