To find the degree of dissociation (\( \alpha \)) and the dissociation constant (\( K_a \)) for acetic acid at 0.01 M concentration, we can use the following formulas.
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Degree of Dissociation (\( \alpha \)):
Molar conductance (\( \Lambda_m \)) of the solution can be expressed as: \[ \Lambda_m = \lambda_+ [A^+] + \lambda_- [A^-] \]
Where:
- \( \Lambda_m \) is the molar conductance of the acetic acid solution,
- \( \lambda_+ \) is the molar conductance of the ions produced. For acetic acid:
- \( \lambda_{H^+} = 390 , \text{S cm}^2/\text{mol} \) (given).
- \( \lambda_{CH_3COO^-} = 40 , \text{S cm}^2/\text{mol} \) (a general table value).
Therefore, the total molar conductance (\( \Lambda_m \)) at 0.01 M is given as: \[ \Lambda_m = 16 \times 10^{-3} , \text{S cm}^2/\text{mol} \]
Plugging in the values: \[ \Lambda_m = \alpha (\lambda_{H^+} + \lambda_{CH_3COO^-}) \cdot [C] \] Here, \( [C] = 0.01 \) M is the concentration of acetic acid.
Rearranging for \( \alpha \): \[ \alpha = \frac{\Lambda_m}{\lambda_+ + \lambda_-} \cdot \frac{1}{[C]} \]
Substituting the values: \[ \alpha = \frac{16 \times 10^{-3}}{390 + 40} \cdot \frac{1}{0.01} \]
\[ \alpha = \frac{16 \times 10^{-3}}{430} \cdot 100 \approx 0.0372 \]
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Dissociation Constant (\( K_a \)):
The dissociation constant for acetic acid can be calculated using the formula: \[ K_a = \frac{C \cdot \alpha^2}{1 - \alpha} \]
Substituting \( C = 0.01 , M \) and \( \alpha \approx 0.0372 \): \[ K_a = \frac{0.01 \cdot (0.0372)^2}{1 - 0.0372} \]
First, calculate \( \alpha^2 \): \[ (0.0372)^2 = 0.001384784 \]
Thus, \[ K_a = \frac{0.01 \cdot 0.001384784}{0.9628} \approx \frac{0.00001384784}{0.9628} \approx 0.0000144 , (or , 1.44 \times 10^{-5}) \]
Final Values:
- Degree of dissociation, \( \alpha \approx 0.0372 \) (or 3.72%)
- Dissociation constant, \( K_a \approx 1.44 \times 10^{-5} \) at 0.01 M concentration.
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