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:
$\underset{1-\alpha}{ C H _{3} COOH } \rightleftharpoons \underset{\alpha}{ CH _{3} COO ^{-}}+\underset{\alpha+\beta}{ H ^{+}}$
The equilibrium reaction for the ionization of propanoic acid is given below:
$\underset{1-\beta}{ C _{2} H _{5} COOH } \rightleftharpoons \underset{\beta}{ C _{2} H _{5} COO ^{-}}+\underset{\alpha+\beta}{ H ^{+}}$
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, $\frac{\alpha}{1-\alpha}=\frac{1.75}{1.3} \times\left[\frac{\beta}{1-\beta}\right]$