JEE CHEMISTRY MCQ: Equilibrium

1 Hard

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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 is correct because it correctly relates the ratio of ionized fractions to the ratio of the Ka values of the two acids.

More Questions from Equilibrium

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\mathrm{K}_1

\mathrm{K}_2

\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots

\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots

K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3}

\begin{array}{l} {H_2}S\leftrightharpoons{H^ + } + H{S^ - }\\ H{S^ - }\leftrightharpoons{H^ + } + {S^{2 - }}\\ {H_2}S\leftrightharpoons2{H^ + } + {S^{2 - }} \end{array}

K{a_1},\,\,K{a_2}\,,K{a_3}

\mathrm{pH}

0.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4

\mathrm{N}_2 \mathrm{H}_4,

K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3

\log 3=0.48

{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}.

\text{O}_2