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JEE CHEMISTRY MCQ: Equilibrium

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Analyzing +4 | -1
A buffer solution is 0.25 M−CH3COOH+0.15 M−CH3COONa,0.25 \mathrm{~M}-\mathrm{CH}_3 \mathrm{COOH}+0.15 \mathrm{~M}-\mathrm{CH}_3 \mathrm{COONa}, saturated in H2S(0.1 M)\mathrm{H}_2 \mathrm{S}(0.1 \mathrm{~M}) and has [Mn2+]=0.04 M,Ka(CH3COOH)=2.0×10−5,\left[\mathrm{Mn}^{2+}\right]=0.04 \mathrm{~M}, K_{\mathrm{a}}\left(\mathrm{CH}_3 \mathrm{COOH}\right)=2.0 \times 10^{-5}, Ka(H2S)=1.0×10−21K_{\mathrm{a}}\left(\mathrm{H}_2 \mathrm{S}\right)=1.0 \times 10^{-21} and Ksp(MnS)=2.5×10−13.K_{\mathrm{sp}}(\mathrm{MnS})=2.5 \times 10^{-13}.
Which buffer component should be increased in concentration and to which minimum value to just start precipitation of MnS?\mathrm{MnS}?
View answer and explanation
Correct option: C
CH3COONa,1.25 M\mathrm{CH}_3 \mathrm{COONa}, 1.25 \mathrm{~M}
Option C (CH3COONa, 1.25 M) is correct because increasing the concentration of CH3COONa to 1.25 M will decrease the [H+] enough to start the precipitation of MnS.

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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