Acetic acid and propionic acid have Ka values 1.75× 10- 5 and 1.3× 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 Ka values and the molarity?

Question

Acetic acid and propionic acid have Ka values 1.75× 10- 5 and 1.3× 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 Ka values and the molarity? (A) ((α /1 - α ))=(1.75/1.3)× ((β /1 - β )) , whereas α and β are ionized fractions of the acids (B) The ratio is unrelated to the Ka values (C) The ratio is unrelated to the molarity (D) The ratio is unrelated to the pH of the solution

Tardigrade Admin · Accepted Answer

In a given mixture, the ionization of two acids can be written at Let

α, β

be degree of ionization at same concentration.

C H_{3} COO H ⇌ C H_{3} CO O_{α}^{-} + α + β H^{+} C_{2} H_{5} COO H ⇌ C_{2} H_{5} CO O^{-} + H^{+} So K_{Acetic acid} = \frac{[ α ] [ α + β ] \cdot c}{[ 1 - α ]} K_{Prepionic acid} = \frac{[ β ] [ α + β ] \cdot c}{[ 1 - β ]} So \frac{K _{acetic acid}}{K _{propinoic acid}} = \frac{α}{1 - α} \times \frac{( 1 - β )}{β} or \frac{α}{1 - α} = \frac{1.75}{1.3} \times [\frac{β}{1 - β}]

$(\frac{α}{1 - α}) = \frac{1.75}{1.3} \times (\frac{β}{1 - β})$ , whereas $α$ and $β$ are ionized fractions of the acids

The ratio is unrelated to the $K_{a}$ values

The ratio is unrelated to the molarity

The ratio is unrelated to the pH of the solution

(1−αα​)=1.31.75​×(1−ββ​) , whereas α and β are ionized fractions of the acids

The ratio is unrelated to the Ka​ values

The ratio is unrelated to the molarity

The ratio is unrelated to the pH of the solution

$(\frac{α}{1 - α}) = \frac{1.75}{1.3} \times (\frac{β}{1 - β})$ , whereas $α$ and $β$ are ionized fractions of the acids

The ratio is unrelated to the $K_{a}$ values