The degree of hydrolysis in hydrolytic equilibrium ${A^ - } + {{\rm{H}}_2}{\rm{O}}\leftrightharpoons{\rm{H}}A + {\rm{O}}{{\rm{H}}^ - }$ at salt concentration of $0.001\;M$ is $({K_a} = 1 \times {10^{ - 5}})$

NEET UG CHEMISTRY MCQ: Equilibrium

1 Hard

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Application +4 | -1

The degree of hydrolysis in hydrolytic equilibrium

{A^ - } + {\mathrm{H}_2}\mathrm{O}\leftrightharpoons\mathrm{H}A + \mathrm{O}{\mathrm{H}^ - }

at salt concentration of

0.001\;M

({K_a} = 1 \times {10^{ - 5}})

View answer and explanation

Correct option: A

1 \times {10^{ - 3}}

Option A (

1 \times {10^{ - 3}}

) is correct because the degree of hydrolysis can be calculated using the formula h = sqrt(Kh/C) where Kh = Kw/Ka.

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}