An aqueous solution contains $Ni^{2 +},Co^{2 +}\ and\ Pb^{2 +}$ ions at equal concentrations. The solubility product of $NiS,PbS\ and\ CoS$ in water at $25℃$ are $1.4 imes 10^{- 24},3.4 imes 10^{- 28}\ and\ 3 imes 10^{- 26}$, respectively. Indicate which of these ions will be precipitated first and last when sulphide concentration is progressively increased from zero?

Question

An aqueous solution contains $Ni^{2 +},Co^{2 +}\ and\ Pb^{2 +}$ ions at equal concentrations. The solubility product of $NiS,PbS\ and\ CoS$ in water at $25℃$ are $1.4 	imes 10^{- 24},3.4 	imes 10^{- 28}\ and\ 3 	imes 10^{- 26}$, respectively. Indicate which of these ions will be precipitated first and last when sulphide concentration is progressively increased from zero?

Dr. Abubacker · Accepted Answer

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

Explanation

Reviewed by Dr. Abubacker

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