NEET UG CHEMISTRY MCQ: Equilibrium

1 Hard

1 of 1

Analyzing +4 | -1

Acetic acid and propionic acid have

{K_a}

values

1.75 \times {10^{ - 5}}

and

1.3 \times {10^{ - 5}}

respectively at a certain temperature. An equimolar solution of a mixture of the two acids is partially neutralized by

\mathrm{NaOH}\mathrm{.}

. How is the ratio of the contents of acetate and propionate ions related to the

{K_a}

values and the molarity?

View answer and explanation

Correct option: A

\left( {\frac{\alpha }{{1 - \alpha }}} \right) = \frac{{1.75}}{{1.3}} \times \left( {\frac{\beta }{{1 - \beta }}} \right),0

where

\alpha

and

\beta

are ionised fractions of the acids

Option A (

\left( {\frac{\alpha }{{1 - \alpha }}} \right) = \frac{{1.75}}{{1.3}} \times \left( {\frac{\beta }{{1 - \beta }}} \right),0

where

\alpha

and

\beta

are ionised fractions of the acids) is correct because it relates the ratio of acetate and propionate ions to their respective

K_a

values and ionization fractions.

More Questions from Equilibrium

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}