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# 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, K_sp = .

7.71     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).

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We have,
The solubility product of the $FeS = 6.3\times 10^{-18}$

Equals number of moles of ferrous sulphate and sodium sulphide are mixed in an equal volume.

Let $s$ be the concentration of ferrous sulphate and sodium sulphide. On mixing the equimolar solution, the volume of the concentration becomes half.
So, $[FeSO_4]=[Na_2S] = \frac{s}{2}M$

The ionisation of ferrous sulphide;
$FeS\rightleftharpoons Fe^{2+}+S^{2-}$
Therefore, for no precipitation, ionic product = solubility product
$\\K_{sp} = (\frac{s}{2})(\frac{s}{2})\\ 6.3\times 10^{-18} = \frac{s^2}{4}\\$
By solving the above equation, we get
$s=5.02\times 10^{-9}$

The maximum concentration of both the solution is $s=5.02\times 10^{-9}$M

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