What is the maximum concentration of equimolar solutions of ferrous sulfate 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).

Given:

Ksp of FeS = 6.3 × 10–18


Let the concentration of solution of FeSO4 and Na2Sis x mol/L


On mixing the equimolar (equal moles) solutions, the volume of the concentration become half


Thus, [FeSO4] = [Na2S] = M


∴ [Fe2+] = [S2-] = M


Ionization of ferrous sulphide:


FeS Fe2+ +S2-


As we know that,


Ksp = [A+] [B-]


Where A and B are the ions


In the above reaction,


⇒[A+] = Fe2+


⇒[B-] = S2-


∴ Ksp = [Fe2+][S2-]


As Ksp = 6.3 × 10-18 (given)


⇒6.3×10-18 =


⇒ =6.3×10-18


Thus, x = 5.02×10-9


As we know that precipitation only occur when Ksp < [A+][B-]


Thus, if the concentration of ferrous sulphate and sodium sulhide are equal to or less than that of 5.02×10-9, then there no precipitation of ferrous sulphide takes place.


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