NEET UG CHEMISTRY MCQ: Equilibrium

1 Hard

1 of 1

Application +4 | -1

The dissociation constant for acetic acid and

\mathrm{HCN}

25^\circ C

are

1.5 \times {10^{ - 3}}

and

4.5 \times {10^{ - 10}}

respectively. the equilibrium constant for the equilibrium,

\mathrm{C}{\mathrm{N}^ - } + \mathrm{C}{\mathrm{H}_3}\mathrm{COOH}\;\leftrightharpoons\;\mathrm{HCN} + \mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }

View answer and explanation

Correct option: A

3.0 \times {10^4}

Option A (

3.0 \times {10^4}

) is correct because it is calculated by dividing the dissociation constant of acetic acid by the dissociation constant of HCN.

More Questions from Equilibrium

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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}