Question
What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp = 6·3 × 10–18).
Solution
Let the concentration of both FeSO4 and Na2S solutions before mixing = x mol L–= xM
The solubility equilibrium of iron sulphide may be represented as,
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Now, Fe2+ ions are to be provided by FeSO4 and S2– ions and are to be provided by Na2S, as a result of dissociation.

Since equal volumes of the two solutions have been mixed, the concentration of the solutions as well as of ions on mixing will be reduced to half i.e.

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Since oh mixing, there is no precipitation of iron sulphide, therefore,
Ionic product = Solubility product

The maximum concentration of both the solutions is 5·02 × 10–9M
The solubility equilibrium of iron sulphide may be represented as,
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Now, Fe2+ ions are to be provided by FeSO4 and S2– ions and are to be provided by Na2S, as a result of dissociation.

Since equal volumes of the two solutions have been mixed, the concentration of the solutions as well as of ions on mixing will be reduced to half i.e.


Since oh mixing, there is no precipitation of iron sulphide, therefore,
Ionic product = Solubility product

The maximum concentration of both the solutions is 5·02 × 10–9M