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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 71 of 73
Marks: +1, -0
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, KspK_{sp} = 6.3 × 10−1810^{-18})
Solution:  
Let the concentration of each of FeSO4\mathrm{FeSO_4} and Na2S\mathrm{Na_2S} is x mol L−1x\,\mathrm{mol\,L^{-1}}. Then after mixing equal volumes :
[FeSO4][\mathrm{FeSO_4}] = x2 M\frac{x}{2}\,\mathrm{M} , [Na2S][\mathrm{Na_2S}] = x2 M\frac{x}{2}\,\mathrm{M} or [Fe2+][\mathrm{Fe}^{2+}] = x2 M\frac{x}{2}\,\mathrm{M} and [S2−][\mathrm{S}^{2-}] = x2 M\frac{x}{2}\,\mathrm{M}
FeS ⇌ Fe2++S2−\mathrm{Fe}^{2+}+\mathrm{S}^{2-}
KspK_{sp} = [Fe2+][S2−][\mathrm{Fe}^{2+}][\mathrm{S}^{2-}] ⇒ 6.3 × 10−1810^{-18} = x2×x2\frac{x}{2}\times\frac{x}{2}
or x2x^2 = 4 × 6.3 × 10−1810^{-18} = 25.2 × 10−1810^{-18} or x = 5.02 × 10−9 mol L−110^{-9}\,\mathrm{mol\,L^{-1}}
Maximum concentration of FeSO4\mathrm{FeSO_4} and Na2S\mathrm{Na_2S} which will not precipitate iron sulphide = 5.02 × 10−9 mol L−110^{-9}\,\mathrm{mol\,L^{-1}}
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