At $25^{\circ} \mathrm{C},$ the dissociation constants of acid $\text{HA}$ and base $\mathrm{BOH}$ in aqueous solution is same. The $\mathrm{pH}$ of $0.01 \mathrm{~M}$ solution of $\mathrm{HA}$ is $5.0.$ The $\mathrm{pH}$ of $0.1 \mathrm{~M}$ solution of $\mathrm{BOH}$ is

JEE CHEMISTRY MCQ: Equilibrium

1 Medium

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Application +4 | -1

25^{\circ} \mathrm{C},

the dissociation constants of acid

\text{HA}

and base

\mathrm{BOH}

in aqueous solution is same. The

\mathrm{pH}

0.01 \mathrm{~M}

solution of

\mathrm{HA}

5.0.

The

\mathrm{pH}

0.1 \mathrm{~M}

solution of

\mathrm{BOH}

View answer and explanation

Correct option: C

9.5

Option A (

2.08 \times 10^{-13}; 12.64 \times 10^{-8}

) is correct because it provides the ionic product and degree of dissociation calculated for heavy water.

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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