Question Details

The molar conductivity of 0.007 M acetic acid is 20 S cm2 mol⁻¹. What is the dissociation constant of acetic acid ? Choose the correct option.

[ H + = 350  S  c m 2  m o l 1 C H 3 C O O = 50  c m 2  m o l 1 ]

Options

A

1.75×10⁻⁴ mol L⁻¹

B

1.75×10⁻⁴ mol L⁻¹

C

1.75×10⁻⁵ mol L⁻¹

D

2.50×10⁻⁵ mol L⁻¹

Show Answer

Correct Answer :

Option C

1.75×10⁻⁵ mol L⁻¹

1.75×10⁻⁵ mol L⁻¹

Solution :

The correct option is 1.75×10⁻⁵ mol L⁻¹.

Step-by-step Derivation:

1. Identify the given values:
Concentration of acetic acid, C=0.007 M=7×103 mol L1
Molar conductivity of acetic acid at concentration C, m=20 S cm2 mol1
Limiting molar conductivity of hydrogen ion, H+=350 S cm2 mol1
Limiting molar conductivity of acetate ion, CH3COO=50 S cm2 mol1

2. Calculate the limiting molar conductivity of acetic acid (m):
According to Kohlrausch's law of independent migration of ions:
m(CH3COOH)=H++CH3COO
m(CH3COOH)=350+50=400 S cm2 mol1

3. Calculate the degree of dissociation (α):
The degree of dissociation is the ratio of molar conductivity at concentration C to the limiting molar conductivity:
α=mm
α=20400=0.05

4. Calculate the dissociation constant (Ka):
For a weak acid like acetic acid, the dissociation constant is given by:
Ka=Cα21α
Since α=0.05 is very small compared to 1, we can approximate 1α1:
KaCα2
Substituting the values:
Ka=0.007×(0.05)2
Ka=0.007×0.0025
Ka=1.75×105 mol L1

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