JEE CHEMISTRY MCQ: Equilibrium

1 Medium

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Analyzing +4 | -1

\mathrm{H}_2 \mathrm{~S}

is passed into

1 \mathrm{~dm}^3

of a solution containing 0.1 mole of

\mathrm{Zn}^{2+}

and 0.01 mole of the

\mathrm{Cu}^{2+}

till the sulphide ion concentration reaches

8.1 \times 10^{-19}

mole. Which one of the following statements is true? [

K_{\mathrm{sp}}

of Zns and CuS are

3 \times 10^{-22}

and

8 \times 10^{-36}

, respectively].

View answer and explanation

Correct option: B

Both CuS and ZnS precipitate

Option B (Both CuS and ZnS precipitate) is correct because the ion product for both CuS and ZnS exceeds their respective

K_{sp}

values, indicating that both will precipitate.

More Questions from Equilibrium

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\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