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JEE CHEMISTRY MCQ: Equilibrium

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The solubility of metal sulphide in saturated solution of H2S\mathrm{H}_2 \mathrm{S} (concentration =0.1 M=0.1 \mathrm{~M}) can be represented as follows.

MS(s)+2H+(aq)⇌M2+(aq)+H2S(aq);Keq=[M2+][H2S][H+]2\mathrm{MS}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{M}^{2+}(\mathrm{aq})+\mathrm{H}_2 \mathrm{S}(\mathrm{aq}) ; K_{\mathrm{eq}}=\frac{\left[\mathrm{M}^{2+}\right]\left[\mathrm{H}_2 \mathrm{S}\right]}{\left[\mathrm{H}^{+}\right]^2}

The values of KeqK_{\mathrm{eq}} for the metal sulphides, MnS, ZnS,CoS\text{MnS, } \mathrm{ZnS}, \mathrm{CoS} and PbS\mathrm{PbS} are 3×1010,3×10−2,33 \times 10^{10}, 3 \times 10^{-2}, 3 and 3×10−7,3\times 10^{-7}, respectively. If the concentration of each metal ion in a saturated solution of H2S\mathrm{H}_2 \mathrm{S} is 0.01 M,0.01 \, \mathrm{M}, then which metal sulphide(s) will precipitate at [H+]=1.0 M?\left[\mathrm{H}^{+}\right]=1.0 \mathrm{~M}?
View answer and explanation
Correct option: B
PbS\mathrm{PbS}
Option B (PbS) is correct because only PbS has a Keq value less than the calculated Q value, indicating precipitation.

More Questions from Equilibrium

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The indicator used in titrating oxalic acid with caustic soda solution is: K1\mathrm{K}_1 and K2\mathrm{K}_2 are equilibrium constant for reactions (i) and (ii)

N2( g)+O2( g)⇌2NO(g)…\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots (i)

NO(g)⇌12 N2( g)+12O2( g)…\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots (ii)

Then, Ka1,   Ka2   and   Ka3K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3} are the respective ionisation constants for the following reactions
H2S⇋H++HS−HS−⇋H++S2−H2S⇋2H++S2−\begin{array}{l} {H_2}S\leftrightharpoons{H^ + } + H{S^ - }\\ H{S^ - }\leftrightharpoons{H^ + } + {S^{2 - }}\\ {H_2}S\leftrightharpoons2{H^ + } + {S^{2 - }} \end{array}
The correct relationship between Ka1,  Ka2 ,Ka3K{a_1},\,\,K{a_2}\,,K{a_3} is The pH\mathrm{pH} of 0.1 M−N2H40.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4 solution is (For N2H4,\mathrm{N}_2 \mathrm{H}_4, Kb1=3.6×10−6,Kb2=6.4×10−12,log⁡2=0.3K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3, log⁡3=0.48\log 3=0.48) Reaction BaO2(s)⇌BaO(s)+O2(g);ΔH=−ve.{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}. In equilibrium condition pressure of O2\text{O}_2 depends on: A : At equilibrium concentration of the reactant and product does not change with time for a chemical reaction.
R: The rate of reaction is zero at equilibrium.
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