Question

When 0.1 m ch3cooh is present in a solvent it shows elevation in boiling point of 0.75 acid dissociation constant is?

Answer 1

When 0.1 m CH3COOH is present in a solvent it shows elevation in boiling point of 0.75°C. Acid dissociation constant will be (Kb = 5 K ...

Answer 2

Dear Student,

◆ Answer -

Ka = 0.05

● 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

Therefore, acid dissociation constant is 0.05.

Thanks dear...