All Exams Test series for 1 year @ ₹349 only
Question

What is the ionization constant of acetic acid (\(CH_3COOH\))?

This question was previously asked in
RPF SI 2018 Question Paper PDF (10-Jan-2019) (Shift 1)
The correct answer is

\(1.74 \times 10^{-5}\)

Acetic acid is a weak acid that ionizes only partially: \(CH_3COOH \rightleftharpoons CH_3COO^- + H^+\).

Its ionization constant is \(K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]}\).

At 298 K, the standard tabulated value of \(K_a\) for acetic acid is \(1.74 \times 10^{-5}\) (\(pK_a \approx 4.76\)), showing it is a weak acid.

Hence, \(1.74 \times 10^{-5}\).

Was this answer helpful?

Similar Questions

  1. What is the ionization constant of formic acid (HCOOH)?


Important Questions from Equilibrium

  1. What is the pH of a solution prepared by dissolving $0.0025$ moles of $HNO_3$ in $250\,mL$ of water?

  2. Which species acts as acid like behaviour in the following reaction?

    \(HPO_4^{2-}+NH_4^{1+}\rightarrow H_2PO_4^{1-}+NH_3\)

  3. Aqueous solution of CH 3COONa is:

  4. pH of a neutral solution is ___________.

  5. For the reaction N 2(g) + 3H 2(g) \(\rightleftharpoons\)  2NH 3(g) ΔH = -ve : 

Need Expert Advice?
Upcoming Exams
RRB JE
October 27, 2026
RRB ALP
November 03, 2026
Test Series
RPF SI img
Railways
RPSF 2025 Mock Test Series
391 Tests 13 Tests Free
2162 Attempts
4.3(314)
English, Hindi, Bengali +7 More

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App