Question

when 0.1 mol ch3cooh is present in a litre solvent it shows elevation in bolling point of 0.75c. acid dissociation constant will be (kb=5k kg mol^-1​

Answer 1

Dear Student,

◆ Answer -

Ka = 0.05 KKg/mol

● Explaination -

# Given -

m = 0.1 m

∆Tb = 0.75 ℃

Kb = 5 Kkg/mol

# Solution -

Dissociation of acetic acid occurs as -

CH3COOH --> CH3COO- + H+

Elevation in boiling point is calculated as -

∆Tb = i.Kb.m

0.75 = i × 5 × 0.1

i = 0.75 / 0.5

i = 1.5

Degree of dissociation is given as -

α = (i-1) / (2-1)

α = (1.5-1) / 1

α = 0.5

Now, acid dissociation constant is given by -

Ka = α².C / (1-α)

Ka = 0.5² × 0.1 / (1-0.5)

Ka = 0.05 KKg/mol

Therefore, acid dissociation constant is 0.05 KKg/mol.