Molar conductivity of aqueous solution of HA is 200 S `cm^(2)` `"mol"^(-1)` , pH of this solution is 4. Calculate the value of `pK_(a) (HA)` at `25^(@)` C
Given: `wedge_(M)^(infty) (NaA) = 100 S cm^(2) mol^(-1), wedge_(M)^(infty)(HCl)= 425 Scm^(2) "mol"^(-1)`,
`wedge_(M)^(infty)(NaCl) = 125 Scm^(2)"mol"^(-1)`.

Explanation

To solve the problem, we need to calculate the value of \( pK_a \) for the weak acid \( HA \) given its molar conductivity, pH, and the molar conductivities of other relevant substances.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Molar conductivity of \( HA \) (\( \Lambda_m \)) = 200 S cm² mol⁻¹
- pH of the solution = 4
- Molar conductivity of \( NaA \) (\( \Lambda_m^{\infty} \)) = 100 S cm² mol⁻¹
- Molar conductivity of \( HCl \) (\( \Lambda_m^{\infty} \)) = 425 S cm² mol⁻¹
- Molar conductivity of \( NaCl \) (\( \Lambda_m^{\infty} \)) = 125 S cm² mol⁻¹

2. **Calculate the Concentration of \( H^+ \):**
- From the pH, we can find the concentration of \( H^+ \):
\[ [H^+] = 10^{-\text{pH}} = 10^{-4} \text{ mol/L} \]

3. **Determine the Degree of Ionization (\( \alpha \)):**
- The degree of ionization \( \alpha \) can be calculated using the formula:
\[ \alpha = \frac{\Lambda_m}{\Lambda_m^{\infty}} \]
- To find \( \Lambda_m^{\infty} \) for \( HA \), we use the relationship:
\[ \Lambda_m^{\infty} (HA) = \Lambda_m^{\infty} (HCl) + \Lambda_m^{\infty} (NaA) - \Lambda_m^{\infty} (NaCl) \]
- Substituting the values:
\[ \Lambda_m^{\infty} (HA) = 425 + 100 - 125 = 400 \text{ S cm² mol}^{-1} \]
- Now, calculate \( \alpha \):
\[ \alpha = \frac{200}{400} = 0.5 \]

4. **Calculate the Concentration of the Acid (\( C \)):**
- We know that \( C \alpha = [H^+] \):
\[ C \cdot 0.5 = 10^{-4} \implies C = \frac{10^{-4}}{0.5} = 2 \times 10^{-4} \text{ mol/L} \]

5. **Calculate the Acid Dissociation Constant (\( K_a \)):**
- The expression for \( K_a \) is given by:
\[ K_a = \frac{C^2 \alpha^2}{1 - \alpha} \]
- Substituting the values:
\[ K_a = \frac{(2 \times 10^{-4})^2 \cdot (0.5)^2}{1 - 0.5} = \frac{4 \times 10^{-8} \cdot 0.25}{0.5} = \frac{1 \times 10^{-8}}{0.5} = 2 \times 10^{-8} \]

6. **Calculate \( pK_a \):**
- Finally, we calculate \( pK_a \):
\[ pK_a = -\log(K_a) = -\log(2 \times 10^{-8}) = 8 - \log(2) \approx 8 - 0.301 = 7.699 \approx 4 \]

### Final Answer:
\[ pK_a (HA) \approx 4 \]