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  • For a solution containing 0.1 M acetic acid ( ) and 0.2 M sodium acetate, the degree o

For a solution containing 0.1 M acetic acid ( \alpha = 0.4) and 0.2 M sodium acetate, the degree of association will be:

Option: 1

0.4


Option: 2

0.35


Option: 3

0.45


Option: 4

0.6


Answers (1)

best_answer

The degree of association for acetic acid is given by the formula, \alpha = \dfrac{1}{1+i}. Here, i is the Van't Hoff factor, which is 2 for acetic acid. Therefore, \alpha = \dfrac{1}{3}. Now, for sodium acetate, i = 2 (as it dissociates into Na^+ and acetate ions). Hence, its degree of association is \dfrac{1}{3}. The overall degree of association for the solution is given by the formula, \alpha = \left(\dfrac{\gamma_+ -\gamma_-}{2}\right)^2. Here,\gamma_+ \text{ and } \gamma_- are the activity coefficients for sodium acetate and acetic acid, respectively. Assuming their values to be equal to 1 (which is a good approximation for dilute solutions), we get, \alpha = \left(\dfrac{0.2 - 0.1}{2}\right)^2 = 0.01. Hence, the overall degree of association is \dfrac{1}{1+i}, where i = 2 \times 0.01 = 0.02. This gives \alpha = 0.35.

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vishal kumar

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