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

• A. $\left(\frac{\alpha }{1 - \alpha }\right)=\frac{1.75}{1.3}\times \left(\frac{\beta }{1 - \beta }\right),$ where α 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 as : Let $\alpha ,\beta $ be degree of ionization at same concentration.
$\underset{1 - \alpha }{C H_{3} C O O H}\rightleftharpoons\underset{\alpha }{C H_{3} C O O^{-} +}\underset{\alpha + \beta }{H^{+}}$
$\underset{1 - \beta }{C_{2} H_{5} C O O H}\rightleftharpoons\underset{\beta }{C_{2} H_{5} C O O^{-} +}\underset{\alpha + \beta }{H^{+}}$
In a given mixture, the ionization of two acids can be written as : Let $\alpha ,\beta $ be degree of ionization at same concentration.
$\underset{1 - \alpha }{C H_{3} C O O H}\rightleftharpoons\underset{\alpha }{C H_{3} C O O^{-} +}\underset{\alpha + \beta }{H^{+}}$
$\underset{1 - \beta }{C_{2} H_{5} C O O H}\rightleftharpoons\underset{\beta }{C_{2} H_{5} C O O^{-} +}\underset{\alpha + \beta }{H^{+}}$
$\therefore K _{\text {Acetic acid }}=\frac{\alpha \alpha+\beta \cdot c}{1-\alpha}$
$K _{\text {Propionic acid }}=\frac{\beta \alpha+\beta \cdot c}{1-\beta}$
$\therefore \frac{K_{\text {Acetic acid }}}{K_{\text {Propionic acid }}}=\frac{\alpha}{1-\alpha} \times \frac{1-\beta}{\beta}$
Or $\frac{\alpha}{1-\alpha}=\frac{1 \cdot 75}{1 \cdot 3} \times \frac{\beta}{1-\beta}$