The resistance of a 0.10 M weak acid HA in a conductivity cell is `2.0xx10^(3)ohm`. The cell constat of the cell is 0.78 `cm^(-1)` and `wedge_(0)` of the acid is 390 S `cm^(2)mol^(-1)`.
Consider the following statements:
1. pH of the acid solution=3
2. `pK_(a)` of the acid=5
3. Degree of dissociation the acid=0.01
Which of the statements given above are correct?

Options

A. 1 ad 2 only
B. 1 and 3 only
C. 2 and 3 only
D. 1,2 and 3

Explanation

To solve the problem, we will analyze the given data and verify the statements one by one.

### Given Data:
- Concentration of weak acid (HA), \( C = 0.10 \, \text{M} \)
- Resistance of the solution, \( R = 2.0 \times 10^{3} \, \Omega \)
- Cell constant, \( k = 0.78 \, \text{cm}^{-1} \)
- Limiting molar conductivity, \( \Lambda_0 = 390 \, \text{S cm}^{2} \text{mol}^{-1} \)

### Step 1: Calculate the conductivity (\( \kappa \))
The conductivity (\( \kappa \)) can be calculated using the formula:
\[ \kappa = \frac{k}{R} \]
Substituting the values:
\[ \kappa = \frac{0.78 \, \text{cm}^{-1}}{2.0 \times 10^{3} \, \Omega} = 3.9 \times 10^{-4} \, \text{S cm}^{-1} \]

### Step 2: Calculate the molar conductivity (\( \Lambda \))
The molar conductivity (\( \Lambda \)) is given by:
\[ \Lambda = \frac{1000 \times \kappa}{C} \]
Substituting the values:
\[ \Lambda = \frac{1000 \times 3.9 \times 10^{-4} \, \text{S cm}^{-1}}{0.10 \, \text{mol L}^{-1}} = 3.9 \, \text{S cm}^{2} \text{mol}^{-1} \]

### Step 3: Calculate the degree of dissociation (\( \alpha \))
The degree of dissociation (\( \alpha \)) can be calculated using the formula:
\[ \alpha = \frac{\Lambda}{\Lambda_0} \]
Substituting the values:
\[ \alpha = \frac{3.9 \, \text{S cm}^{2} \text{mol}^{-1}}{390 \, \text{S cm}^{2} \text{mol}^{-1}} = 0.01 \]

### Step 4: Calculate the concentration of \( H^+ \) ions
The concentration of \( H^+ \) ions produced from the dissociation of the weak acid can be calculated as:
\[ [H^+] = C \cdot \alpha = 0.10 \, \text{mol L}^{-1} \times 0.01 = 1.0 \times 10^{-3} \, \text{mol L}^{-1} \]

### Step 5: Calculate the pH of the solution
The pH can be calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the value:
\[ \text{pH} = -\log(1.0 \times 10^{-3}) = 3 \]

### Step 6: Calculate the \( K_a \) and \( pK_a \)
The dissociation constant (\( K_a \)) can be calculated as:
\[ K_a = \frac{[H^+]^2}{C(1 - \alpha)} = \frac{(1.0 \times 10^{-3})^2}{0.10(1 - 0.01)} = \frac{1.0 \times 10^{-6}}{0.099} \approx 1.01 \times 10^{-5} \]
Then, calculate \( pK_a \):
\[ pK_a = -\log(K_a) \approx -\log(1.01 \times 10^{-5}) \approx 5 \]

### Summary of Results:
1. pH of the acid solution = 3 (Correct)
2. \( pK_a \) of the acid = 5 (Correct)
3. Degree of dissociation of the acid = 0.01 (Correct)

### Conclusion:
All the statements given are correct.