Find change in $\mathrm{pH}$ of water when $1 \mathrm{~mL}$ of $10^{-3} \mathrm{M}$ $\mathrm{KOH}$ is added to $99 \mathrm{~mL}$ of $\mathrm{H}_2 \mathrm{O}$ at $25^{\circ} \mathrm{C}$.

JEE CHEMISTRY MCQ: Equilibrium

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Application +4 | -1

Find change in

\mathrm{pH}

of water when

1 \mathrm{~mL}

10^{-3} \mathrm{M}

\mathrm{KOH}

is added to

99 \mathrm{~mL}

\mathrm{H}_2 \mathrm{O}

25^{\circ} \mathrm{C}

View answer and explanation

Correct option: A

Option A (2) is correct because the addition of KOH to water increases the pH from 7 to 9, resulting in a change of 2 pH units.

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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