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?

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

The ratio is unrelated to the molarity of acid.

$(\frac{α}{1 - α}) = \frac{1 .75}{1 .3} \times (\frac{β}{1 - β})$ , 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 pH of the solution

answer is A, C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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

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