At $90^\circ C$, pure water has $\left[ {{{\rm{H}}^ + }} \right] = {10^{ - 6}}$ M, if $100\;mL$ of $0.2\;M\;\;HCl$ is added to 200 mL of $0.1\;M{\rm{ }}\;KOH$ at $90^\circ C$ then pH of the resulting solution will be

JEE CHEMISTRY MCQ: Equilibrium

1 Easy

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Application +4 | -1

90^\circ C

, pure water has

\left[ {{\mathrm{H}^ + }} \right] = {10^{ - 6}}

M, if

100\;mL

0.2\;M\;\;HCl

is added to 200 mL of

0.1\;M\mathrm{ }\;KOH

90^\circ C

then pH of the resulting solution will be

View answer and explanation

Correct option: D

Option D (6) is correct because at

90^{\circ}C

, pure water is neutral with

[H^+] = 10^{-6}

M, and the addition of equal equivalents of HCl and KOH results in a neutral 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