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 $\alpha$ 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)

The equilibrium reaction for the ionization of acetic acid is as given below:
$\underset{1-\alpha }{C{H}_{3}COOH}\rightleftharpoons \underset{\alpha }{C{H}_{3}CO{O}^{-}}+\underset{\alpha +\beta }{H^{+}}$
The equilibrium reaction for the ionization of propanoic acid is given below:
$\underset{1-\beta }{C_2{H}_{5}COOH}\rightleftharpoons \underset{\beta }{C_2{H}_{5}CO{O}^{-}}+\underset{\alpha +\beta }{H^{+}}$
Let, represent the degree of ionization at same concentrations of acetic acid and propanioc acid respectively.
The expression for the dissociation constant for acetic acid is as given below:
$K_{\text{Acetic acid }}=\frac{[\alpha ][\alpha +\beta ]\cdot c}{[1-\alpha ]}$
The expression for the dissociation constant for propanoic acid is as given below:
$K_{\text{Propionic acid }}=\frac{[\beta ][\alpha +\beta ]\cdot c}{[1-\beta ]}$
The ration of the acid dissociation constants is calculated as shown below:
$\frac{K_{\text{Acetic acid }}}{K_{\text{Propionic acid }}}=\frac{\alpha }{1-\alpha }\times \frac{(1-\beta )}{\beta }$
Hence, $\frac{\alpha }{1-\alpha }=\frac{1.75}{1.3}\times \left[\frac{\beta }{1-\beta }\right]$