The molar conductance at infinite dilution of sodium acetate, sodium sulfate and sulfuric acid solutions are
91.0 × 10−4,
259.8 × 10−4 and
859.3 × 10−4 S m2 mol−1, respectively.
Q.25 If the molar conductance of an acetic acid solution is
15.2 × 10−4 S m2 mol−1,
then the percentage (%) dissociation of acetic acid in the solution will be
(A) 3.89
(B) 2.20
(C) 1.85
(D) 1.48
Percentage Dissociation of Acetic Acid Using Molar Conductance
Problems based on molar conductance and
degree of dissociation are very common in physical chemistry.
In this question, we calculate the
percentage dissociation of acetic acid
using given molar conductance values and
Kohlrausch’s law.
Given Data
- Λ°(CH3COONa) = 91.0 × 10−4 S m2 mol−1
- Λ°(Na2SO4) = 259.8 × 10−4 S m2 mol−1
- Λ°(H2SO4) = 859.3 × 10−4 S m2 mol−1
- Λm(CH3COOH) = 15.8 × 10−4 S m2 mol−1
Concept Used: Kohlrausch’s Law
For acetic acid:
Λ°(CH3COOH) =Λ°(CH3COONa) +Λ°(H2SO4) −Λ°(Na2SO4)
Step 1: Calculate Λ° of Acetic Acid
Λ°(CH3COOH) =(91.0 + 859.3 − 259.8) × 10−4
Λ°(CH3COOH) =690.5 × 10−4 S m2 mol−1
Step 2: Calculate Degree of Dissociation (α)
α =Λm /Λ°
α =15.8 × 10−4 /690.5 × 10−4
α = 0.0229
Step 3: Calculate Percentage Dissociation
Percentage dissociation = α × 100
= 0.0229 × 100
= 2.29%
Final Answer
Correct Option: (B) 2.20%
Conclusion
Acetic acid is a weak electrolyte and shows partial dissociation.
Using molar conductance values and Kohlrausch’s law allows accurate
calculation of its percentage dissociation in solution.
