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JEE Advanced 2020 · Paper 2, Chemistry question 18

An acidified solution of 0.05 M ZnX2+\ce{Zn^{2+}} is saturated with 0.1 M HX2S\ce{H2S}. What is the minimum molar concentration (M) of HX+\ce{H+} required to prevent the precipitation of ZnS?

Use Ksp(ZnS)=1.25×10−22K_{sp}(\ce{ZnS}) = 1.25\times10^{-22} and overall dissociation constant of HX2S\ce{H2S}, KNET=K1K2=1×10−21K_{\text{NET}} = K_1K_2 = 1\times10^{-21}.

Answer and worked solution

Official answer: 0.2

Largest allowed sulfide. ZnS must not precipitate, so
[SX2−]≤Ksp[ZnX2+]=1.25×10−220.05=2.5×10−21 M.[\ce{S^{2-}}]\le\frac{K_{sp}}{[\ce{Zn^{2+}}]}=\frac{1.25\times10^{-22}}{0.05}=2.5\times10^{-21}\text{ M}.

Acid needed. From HX2S⇌2 HX++SX2−\ce{H2S <=> 2H+ + S^{2-}}, [SX2−]=KNET[HX2S][HX+]2[\ce{S^{2-}}]=\dfrac{K_{NET}[\ce{H2S}]}{[\ce{H+}]^2}. So
[HX+]2=10−21×0.12.5×10−21=0.04  ⟹  [HX+]=0.20 M.[\ce{H+}]^2=\frac{10^{-21}\times0.1}{2.5\times10^{-21}}=0.04\implies[\ce{H+}]=0.20\text{ M}.
Answer: 0.20

Topics:Solubility productCommon ion effect
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