$0.1\;{\rm{millimole}}$ of ${\rm{CdS}}{{\rm{O}}_4}$ are present in $10\;{\rm{mL}}$ acid solution of $0.08\;N\;{\rm{HCl}}{\rm{.}}$ Now ${{\rm{H}}_2}{\rm{S}}$ is passed to precipitate all the ${\rm{C}}{{\rm{d}}^{2 + }}$ ions. The ${\rm{pH}}$ of the solution after filtering off precipitate, boiling of ${{\rm{H}}_2}{\rm{S}}$ and making the solution $100\;{\rm{mL}}$ by adding ${{\rm{H}}_2}O$ is:

JEE CHEMISTRY MCQ: Equilibrium

1 Medium

1 of 1

Application +4 | -1

0.1\;\mathrm{millimole}

\mathrm{CdS}{\mathrm{O}_4}

are present in

10\;\mathrm{mL}

acid solution of

0.08\;N\;\mathrm{HCl}\mathrm{.}

Now

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

is passed to precipitate all the

\mathrm{C}{\mathrm{d}^{2 + }}

ions. The

\mathrm{pH}

of the solution after filtering off precipitate, boiling of

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

and making the solution

100\;\mathrm{mL}

by adding

{\mathrm{H}_2}O

is:

View answer and explanation

Correct option: A

Option A (2) is correct because after the reaction, the pH is calculated based on the remaining H+ concentration in the solution.

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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