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).

Asked by Pragya Singh | 1 year ago |  171

Solution :-

Assuming the maximum concentration of each solution is y mol/L

On mixing the solutions the volume of the concentration of each solution is reduced to half. After mixing the maximum concentration of each solution is

$$\dfrac{y}{2}$$ mol/L.

Thus, [FeSO4]

= [Na2S] = $$\dfrac{y}{2}$$M

So, [Fe2+]

= [FeSO4] = $$\dfrac{y}{2}$$ M

$$FeS(s) \leftrightarrow Fe^{2+}_{(aq)} + S^{2-}_{(aq)}$$

Ksp = $$[Fe^{2+}][S^{2-}]$$

$$=6.3×10^{−18}$$

$$= (\dfrac{y}{2})(\dfrac{y}{2})\dfrac{y^{2}}{4}$$

$$= 6.3\times 10^{-18}$$

Thus, y = $$5.02\times 10^{-9}$$

Thus, if the concentration of FeSO4 and Na2SOare equal to or less than that of

$$5.02\times 10^{-9}M$$, then there won’t be precipitation of FeS.

Answered by Abhisek | 1 year ago

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