Degree of dissociation of$0.1\;{\rm{ }}N$ ${\rm{C}}{{\rm{H}}_3}{\rm{COOH}}$ is (dissociation constant $ = 1 \times {10^{ - 5}}$)

NEET UG CHEMISTRY MCQ: Equilibrium

1 Easy

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Application +4 | -1

Degree of dissociation of

0.1\;\mathrm{ }N

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

is (dissociation constant

= 1 \times {10^{ - 5}}

)

View answer and explanation

Correct option: D

{10^{ - 2}}

Option D (

{10^{ - 2}}

) is correct because it is calculated using the formula for the degree of dissociation of a weak acid.

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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}