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JEE CHEMISTRY MCQ: Equilibrium

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Application +4 | -1

A solution contains a mixture of

\mathrm{Ag}^{+}(0.10 \mathrm{~M})

and

\mathrm{Hg}_2^{2+}(0.10 \mathrm{~M}),

which are to be separated by selective precipitation. Calculate the maximum concentration of iodide ion at which one of them gets precipitated almost completely. What per cent of that metal ion is precipitated before the start of precipitation of second metal ion?

K_{\mathrm{sp}}(\mathrm{AgI})=8.5\times 10^{-17}

and

K_{\text {sp }}\left(\mathrm{Hg}_2 \mathrm{I}_2\right)=2.5 \times 10^{-26}.

View answer and explanation

Correct option: A

5 \times 10^{-13} \mathrm{~M}, 99.83 \%

Option A (

5 \times 10^{-13}

M, 99.83 %) is correct because it identifies the maximum iodide concentration before the second precipitation and calculates the percentage of the first metal ion precipitated.

More Questions from Equilibrium

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The indicator used in titrating oxalic acid with caustic soda solution is:

\mathrm{K}_1

\mathrm{K}_2

are equilibrium constant for reactions (i) and (ii)

\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots

\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots

K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3}

are the respective ionisation constants for the following reactions

\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

K{a_1},\,\,K{a_2}\,,K{a_3}

\mathrm{pH}

0.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4

solution is (For

\mathrm{N}_2 \mathrm{H}_4,

K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3

\log 3=0.48

{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}.

In equilibrium condition pressure of

\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.