Question

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 neutralised by NaOH. How is the ratio of the contents of acetate and propionate ions related to the $K_{a}$ values and the molarity?

• A. $\left(\frac{\alpha }{1 - \alpha }\right)=\frac{1.75}{1.3}\times \left(\frac{\beta }{1 - \beta }\right)$ , whereas $\alpha $ and $\beta $ are ionized fractions of the acids
• B. The ratio is unrelated to the $K_{a}$ values
• C. The ratio is unrelated to the molarity
• D. The ratio is unrelated to the pH of the solution

Answer 1

Answer: (A)

In a given mixture, the ionization of two acids can be written at Let $\alpha, \beta$ be degree of ionization at same concentration.
$ \begin{array}{l} CH _3 COOH \rightleftharpoons CH _3 COO _\alpha^{-}+\underset{\alpha+\beta}{ H ^{+}} \\ C _2 H _5 COOH \rightleftharpoons C _2 H _5 COO ^{-}+ H ^{+} \\ \text {So } K _{\text {Acetic acid }}=\frac{[\alpha][\alpha+\beta] \cdot c }{[1-\alpha]} \\ K _{\text {Prepionic acid }}=\frac{[\beta][\alpha+\beta] \cdot c }{[1-\beta]} \\ \text { So } \frac{ K _{\text {acetic acid }}}{ K _{\text {propinoic acid }}}=\frac{\alpha}{1-\alpha} \times \frac{(1-\beta)}{\beta} \\ \text { or } \frac{\alpha}{1-\alpha}=\frac{1.75}{1.3} \times\left[\frac{\beta}{1-\beta}\right] \end{array} $