Acetic acid and propanoic acid have Ka value $1.75 \times 10^{- 5}$ and $1.3 \times 10^{- 5}$ respectively at a certain temperature, An equimolar so1ution 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?

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$(\frac{α}{1 - α}) = \frac{1 .75}{1 .3} \times (\frac{β}{1 - β})$ , where a (acetic acid) and $β$ (propanoic acid) are iomsed fractions of their acids.

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

The ratio is unrelated to the molarity of acid.

The ratio is unrelated to the pH of the solution

answer is A, C, D.

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Detailed Solution

$\underset{1 - α}{HA} + {\underset{}{OH}}^{-} \to \underset{α}{A^{+}} + H_{2} O$

$\begin{array}{r} [H^{+}] = k_{1} \times \frac{[HA]}{[A^{-}]} = 1 .75 \times 10^{- 5} \times [\frac{1 - α}{α}] \\ [H^{+}] = k_{2} \times \frac{[HPr]}{[\Pr^{-}]} = 1 .3 \times 10^{- 5} \times \frac{(1 - β)}{β} \end{array}$

$\Rightarrow [\frac{α}{1 - α}] = \frac{1 .75}{1 .3} \times [\frac{β}{1 - β}]$

The ratio is unrelated to the molarity of acid and pH of the solution.

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