At $30^\circ C\;{\rm{the}}\;{\rm{ solubility}}\;{\rm{ of}}\,{\rm{ A}}{{\rm{g}}_2}{\rm{C}}{{\rm{O}}_3}\;\left( {{K_{{\rm{sp}}}} = 8 \times {{10}^{ - 12}}} \right)$ would be greatest in $1\;L$ of

$0.05\;{\rm{M }}\;{\rm{N}}{{\rm{H}}_3}$

$0.05\;{\rm{M}}\;{\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3}$

$0.05\;{\rm{M}}\;{\rm{ AgN}}{{\rm{O}}_3}$

JEE CHEMISTRY MCQ: Equilibrium

1 Easy

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Analyzing +4 | -1

30^\circ C\;\mathrm{the}\;\mathrm{ solubility}\;\mathrm{ of}\,\mathrm{ A}{\mathrm{g}_2}\mathrm{C}{\mathrm{O}_3}\;\left( {{K_{\mathrm{sp}}} = 8 \times {{10}^{ - 12}}} \right)

would be greatest in

1\;L

View answer and explanation

Correct option: D

0.05\;\mathrm{M }\;\mathrm{N}{\mathrm{H}_3}

The correct answer is D (

0.05\;\mathrm{M }\;\mathrm{N}{\mathrm{H}_3}

) because the solubility of

Ag_2CO_3

is greatest in

NH_3

due to complex formation, which reduces the concentration of

Ag^+

ions.

Practice All Chapter MCQs →

\mathrm{K}_1

\mathrm{K}_2

\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}

\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}

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

\mathrm{pH}

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

\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}}.

\text{O}_2