$pH$ 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 ${\rm{PO}}_4^{3 - }.$

NEET UG 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 A (4.67, 9.61) is correct because it is calculated using the Henderson-Hasselbalch equation for polyprotic acids.

Practice All Chapter MCQs →

Mg{(HC{O_3})_2}

MgC{O_3}

Mg{(OH)_2}

MgC{O_3}

K_{\mathrm{P}}

\mathrm{CaCO}_3

4.5 \times 10^{-2}

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}),

K_{\mathrm{P}}

2.0 \mathrm{~atm}.

\mathrm{CaO}

\mathrm{CaCO}_3

+

\leftrightharpoons

\mathrm{PC}{\mathrm{l}_5}\leftrightharpoons\mathrm{PC}{\mathrm{l}_3} + \mathrm{C}{\mathrm{l}_2}

{K_{\mathrm{eq}}} = 0.41

\mathrm{p}{K_a}