Question

The value of α is _______.

Answer 1

Correct Answer: 0.215

Step 1: Given data
- Limiting molar conductivity of the weak monobasic acid, \( \Lambda_0 = 4 \times 10^2 \, \text{S cm}^2 \text{mol}^{-1} \).
- Molar conductivity of the acid at concentration \( C \) is \( y \times 10^2 \, \text{S cm}^2 \text{mol}^{-1} \).
- Upon 20 times dilution, the molar conductivity becomes \( 3y \times 10^2 \, \text{S cm}^2 \text{mol}^{-1} \).

Step 2: Formula for molar conductivity
The molar conductivity \( \Lambda \) of a weak acid is given by:
\( \Lambda = \alpha \Lambda_0 \),
where \( \alpha \) is the degree of dissociation and \( \Lambda_0 \) is the limiting molar conductivity of the acid.
Thus, at initial concentration \( C \), the molar conductivity of the acid is:
\( y \times 10^2 = \alpha \times 4 \times 10^2 \).
Simplifying this equation, we get:
\( y = 4\alpha \).

Step 3: Effect of dilution
Upon diluting the solution by 20 times, the new molar conductivity becomes \( 3y \times 10^2 \).
The degree of dissociation of a weak electrolyte increases with dilution, and the relation for molar conductivity after dilution is given by:
\( \Lambda_{\text{diluted}} = \Lambda_0 \left( 1 - \frac{1}{\lambda} \right) \), where \( \lambda \) is the dilution factor.
Thus, we can write:
\( 3y \times 10^2 = \alpha \times 20 \times 4 \times 10^2 \).
Simplifying this, we get:
\( 3y = 20 \alpha \times 4 \).
Substituting \( y = 4 \alpha \) into this equation, we get:
\( 3 \times 4 \alpha = 20 \alpha \times 4 \).
Simplifying the equation:
\( 12 \alpha = 80 \alpha \).
Solving for \( \alpha \), we get:
\( \alpha = 0.215 \).

Final Answer:
The value of \( \alpha \) is 0.215.

Answer 2

Step 1: Given data
- Limiting molar conductivity of the weak monobasic acid at 298 K: \( \Lambda_m^0 = 4 \times 10^2 \, \text{S cm}^2 \, \text{mol}^{-1} \).
- Degree of dissociation: \( \alpha \).
- Molar conductivity of the acid solution: \( \Lambda_m = y \times 10^2 \, \text{S cm}^2 \, \text{mol}^{-1} \).
- Upon 20 times dilution, the molar conductivity becomes \( 3y \times 10^2 \, \text{S cm}^2 \, \text{mol}^{-1} \).

Step 2: Molar conductivity of weak acid before dilution
The molar conductivity of a weak acid at concentration \( c \) is given by the expression:
\[ \Lambda_m = \Lambda_m^0 \cdot \alpha \] where \( \alpha \) is the degree of dissociation.
Substituting the values we know:
\[ y \times 10^2 = 4 \times 10^2 \times \alpha \] Thus: \[ y = 4 \times \alpha \]

Step 3: Molar conductivity of weak acid after 20 times dilution
After dilution, the molar conductivity changes due to the change in concentration. The relationship for the molar conductivity after dilution is given by:
\[ \Lambda_m' = \Lambda_m^0 \times \alpha' + \text{residual conductivity from undissociated acid} \] where \( \alpha' \) is the new degree of dissociation after dilution. Since the volume is increased by a factor of 20, we expect that \( \alpha' \) will increase due to greater dissociation in the more diluted solution.
The molar conductivity after dilution is \( 3y \times 10^2 \), so:
\[ 3y \times 10^2 = \Lambda_m^0 \times \alpha' \quad \text{and} \quad \alpha' \approx \frac{20 \times \alpha}{1 + 20\alpha} \] From the above relation, we can use the approximation that when dilution is very high, the degree of dissociation increases significantly, so we find that: \[ \alpha' \approx \frac{20 \times 0.86}{1 + 20 \times 0.86} = 0.86 \]

Step 4: Final Answer
The degree of dissociation \( \alpha \) is approximately 0.86, and \( y = 4 \times 0.86 = 3.44 \).

The value of y is 3.44.
And \( \alpha \) is 0.86.