$pH$ of $0.1\;M\;{\rm{N}}{{\rm{a}}_2}{\rm{HP}}{{\rm{O}}_4}$ and $0.2\;M\;{\rm{Na}}{{\rm{H}}_2}{\rm{P}}{{\rm{O}}_4}$ solutions are respectively. ${\rm{p}}{K_a}$ for ${{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4}$ are $2.12,7.21$ and $12.0$ for respective dissociation to ${{\rm{H}}_2}{\rm{PO}}_4^ - ,{\rm{HPO}}_4^{2 - }$ and ${\rm{PO}}_4^{3 - }.$

JEE CHEMISTRY MCQ: Equilibrium

1 Medium

1 of 1

Application +4 | -1

pH

0.1\;M\;\mathrm{N}{\mathrm{a}_2}\mathrm{HP}{\mathrm{O}_4}

and

0.2\;M\;\mathrm{Na}{\mathrm{H}_2}\mathrm{P}{\mathrm{O}_4}

solutions are respectively.

\mathrm{p}{K_a}

for

{\mathrm{H}_3}\mathrm{P}{\mathrm{O}_4}

are

2.12,7.21

and

12.0

for respective dissociation to

{\mathrm{H}_2}\mathrm{PO}_4^ - ,\mathrm{HPO}_4^{2 - }

and

\mathrm{PO}_4^{3 - }.

View answer and explanation

Correct option: A

4.67,9.61

Option B (9.61, 4.67) is incorrect, the correct order is A (4.67, 9.61). The pH values are calculated using the half-sum of the relevant pKa values.

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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