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.
$\underset{1-\alpha }{C{H}_{3}COOH}\rightleftharpoons \underset{\alpha }{C{H}_{3}CO{O}^{-}+}\underset{\alpha +\beta }{H^{+}}$
$\underset{1-\beta }{C_2{H}_{5}COOH}\rightleftharpoons \underset{\beta }{C_2{H}_{5}CO{O}^{-}+}\underset{\alpha +\beta }{H^{+}}$
In a given mixture, the ionization of two acids can be written as : Let $\alpha ,\beta$ be degree of ionization at same concentration.
$\underset{1-\alpha }{C{H}_{3}COOH}\rightleftharpoons \underset{\alpha }{C{H}_{3}CO{O}^{-}+}\underset{\alpha +\beta }{H^{+}}$
$\underset{1-\beta }{C_2{H}_{5}COOH}\rightleftharpoons \underset{\beta }{C_2{H}_{5}CO{O}^{-}+}\underset{\alpha +\beta }{H^{+}}$
$\therefore K_{\text{Acetic acid }}=\frac{\alpha \alpha +\beta \cdot c}{1-\alpha }$
$K_{\text{Propionic acid }}=\frac{\beta \alpha +\beta \cdot c}{1-\beta }$
$\therefore \frac{K_{\text{Acetic acid }}}{K_{\text{Propionic acid }}}=\frac{\alpha }{1-\alpha }\times \frac{1-\beta }{\beta }$
Or $\frac{\alpha }{1-\alpha }=\frac{1\cdot 75}{1\cdot 3}\times \frac{\beta }{1-\beta }$