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How much water must be added in 900ml of 0.1M CH3COOH solution to triple its degree of dissociation(Assume a

How much water must be added in 900ml of 0.1M CH3COOH solution to triple its degree of dissociation(Assume a

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Grade:12th pass

1 Answers

Arun
25750 Points
3 years ago
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.
 
 

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