How much water (in mL ) must be added to 300 mL of 0.2 M solution of $CH_{3} COOH (K_{a} = 1.8 \times 1 0^{- 5})$ . So that its percentage dissociation doubles?
[Assume degree of ionisation of acid to be negligible in both solutions]

Question

How much water (in mL ) must be added to 300 mL of 0.2 M solution of $CH_{3} COOH (K_{a} = 1.8 \times 1 0^{- 5})$ . So that its percentage dissociation doubles?

[Assume degree of ionisation of acid to be negligible in both solutions]

TestHub · Accepted Answer

Explanation:

The question asks about the volume of water added to CH₃COOH so that its degree of dissociation (D.O.D.) doubles.

Solution:

This question based on dilution Ostwald law

Solution:

α is negligible w.r.t 1

K_a= C₁α₁²+C₁α₁²

C₂= C₁ (α₁/α₂)²

0.2 Χ 1/4 = 0.05

C₁V₁ =C₂ V₂

300Χ0.2 = 0.05 Χ V₂

V₂= 1200 ml

Volume of H₂o added = 1200 - 300 =900 ml

final ans. =900

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