The conductivity of 0.001 M acetic acid is 5 × 10–5 S cm–1 and Λo is 390.5 S cm2 mol–1 then the calculated value of dissociation constant of acetic acid would be ____ × 10–6

ΛM=(K×1000)/M;α=ΛM/ΛM0Ka=cα2λM=5×10-5×10000.001=50α=50390.5=0.128Ka=α2C=(0.128)2×0.001=16×10-6

Book Online Demo

Check Your IQ

Try Test

Courses

Offline Centres

The conductivity of 0.001 M acetic acid is 5 × 10^–5 S cm^–1 and $Λ^{o}$ is 390.5 S cm² mol^–1 then the calculated value of dissociation constant of acetic acid would be ____ × 10^–6

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 18.78.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

$\begin{array}{l} Λ_{M} = (K \times 1000) / M; α = (Λ_{M}) / (Λ_{M}^{0}) \\ Ka = {cα}^{2} \end{array}$

$λ_{M} = \frac{5 \times 10^{- 5} \times 1000}{0.001} = 50$

$α = \frac{50}{390.5} = 0.128$

$K_{a} = α^{2} C = {(0.128)}^{2} \times 0.001 = 16 \times 10^{- 6}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE