Question

How much water must be added to 300 ml of 0.2 M solution of acetic acid for the degree of dissociation of the acid to double Ka=1.8*10 raised -5

Answer 1

300 ml of water we have to use o.2 m solution of acetic acid

Answer 2

Hello dear,

● Answer:- 900 ml

● Explanation -
# Given -
Ka = 1.8×10^-5
C1 = 0.02 M
V1 = 300 ml

# Solution -
Dissociation of acetic acid occurs as -
CH3COOH <--> CH3COO- + H+

Initial degree of ionization is calculated by -
α = √(Ka/C1)
α = √(1.8×10^-5 / 0.2)
α = 9.5×10^-3

Later degree of dissociation is doubled, 2α = 19×10^-3

Later conc is calculated by -
C = Ka / (2α)^2
C = 1.8×10^-5 / (19×10^-3)
C = 0.05 M

New volume of solution -
V2 = C1V1 / C2
V2 = 300×0.2 / 0.05
V2 = 1200 ml

Water to be added -
∆V = V2 - V1
∆V = 1200 - 300
∆V = 900 ml

Therefore, total 900 ml water need to be added.

Hope this helps you...