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JEE CHEMISTRY MCQ: Equilibrium

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An aqueous solution contains Ni2+,Co2+  and  Pb2+\mathrm{N}{\mathrm{i}^{2 + }},\mathrm{C}{\mathrm{o}^{2 + }}\;\mathrm{and}\;\mathrm{ P}{\mathrm{b}^{2 + }} ions at equal concentrations. The solubility product of NiS,PbS  and  CoS\mathrm{NiS},\mathrm{PbS}\;\mathrm{ and }\;\mathrm{CoS} in water at 25∘C25^\circ C are 1.4×10−24,  3.4×10−28  and  3×10−261.4 \times {10^{ - 24}},\;3.4 \times {10^{ - 28}}\;\mathrm{and}\;3 \times {10^{ - 26}}, respectively. Indicate which of these ions will be precipitated first and last when sulphide concentration is progressively increased from zero?
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Correct option: D
PbS  and  NiS\mathrm{PbS }\;\mathrm{and }\;\mathrm{NiS}
Option D (PbS  and  NiS\mathrm{PbS }\;\mathrm{and }\;\mathrm{NiS}) is correct because PbS, having the lowest solubility product, will precipitate first, and NiS, with the highest solubility product, will precipitate last.

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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