The specific conductance of 0.01 M aqueous solution of acetic acid is 1. 63 x10-2 S m-1 and the molar conductance at infinite dilution is 390.7x10-4 S m2 mol-1.Calculate the degree of dissociation of the acetic acid solution?
0.4
0.239
0.0417
0.0239
To calculate the degree of dissociation (α) of the acetic acid solution, we'll follow these steps:
1. Given Data:
- Specific conductance (κ) = 1.63 × 10-2 S m-1
- Molar concentration (c) = 0.01 M = 0.01 × 103 mol m-3 = 10 mol m-3
- Molar conductance at infinite dilution (Λ0) = 390.7 × 10-4 S m2 mol-1
2. Calculate Molar Conductance (Λm):
Molar conductance is given by:
Λm = κ / c
= (1.63 × 10-2 S m-1) / (10 mol m-3)
= 1.63 × 10-3 S m2 mol-1
3. Calculate Degree of Dissociation (α):
The degree of dissociation is given by:
α = Λm / Λ0
= (1.63 × 10-3 S m2 mol-1) / (390.7 × 10-4 S m2 mol-1)
= (1.63 × 10-3) / (3.907 × 10-2)
= 0.0417 or 4.17%
4. Interpretation:
The low degree of dissociation (4.17%) indicates that acetic acid is a weak electrolyte, as expected, with only a small fraction of molecules dissociated into ions at this concentration.
Final Answer:
The degree of dissociation of the 0.01 M acetic acid solution is 0.0417 (or 4.17%).
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