Statement (S): The dissociation constant of water at $60^{\circ} \mathrm{C}$ is $10^{-13}$. Explanation (E): The $\mathrm{pH}$ of water is 6.5 and that it behaves as acid at $60^{\circ} \mathrm{C}$.

Both S and E are correct and E is correct explanation for S

Both S and E are correct but E is not correct explanation for S

JEE CHEMISTRY MCQ: Equilibrium

1 Medium

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Analyzing +4 | -1

Statement (S): The dissociation constant of water at

60^{\circ} \mathrm{C}

10^{-13}

.

Explanation (E): The

\mathrm{pH}

of water is 6.5 and that it behaves as acid at

60^{\circ} \mathrm{C}

View answer and explanation

Correct option: A

S is correct but E is wrong

Option A (S is correct but E is wrong) is correct because while the dissociation constant of water at 60°C is indeed around

10^{-13}

, the pH of 6.5 at that temperature still indicates neutrality, not acidity.

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