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

Explanation

To solve the problem step by step, we will calculate the value of \( pK_a \) for the weak acid \( HA \) given the molar conductivity, pH, and other provided values.

### Step 1: Calculate the Molar Conductivity at Infinite Dilution (\( \lambda_m^{\infty} \)) for \( HA \)

According to Kohlrausch's Law, the molar conductivity at infinite dilution for \( HA \) can be calculated using the following equation:

\[ \lambda_m^{\infty}(HA) = \lambda_m^{\infty}(HCl) + \lambda_m^{\infty}(NaA) - \lambda_m^{\infty}(NaCl) \]

Substituting the given values:

\[ \lambda_m^{\infty}(HA) = 425 \, \text{S cm}^2 \text{mol}^{-1} + 100 \, \text{S cm}^2 \text{mol}^{-1} - 125 \, \text{S cm}^2 \text{mol}^{-1} \]

Calculating this gives:

\[ \lambda_m^{\infty}(HA) = 425 + 100 - 125 = 400 \, \text{S cm}^2 \text{mol}^{-1} \]

### Step 2: Calculate the Concentration of \( H^+ \) Ions

Given that the pH of the solution is 4, we can find the concentration of \( H^+ \) ions using the formula:

\[ [H^+] = 10^{-pH} \]

Substituting the value of pH:

\[ [H^+] = 10^{-4} \, \text{mol/L} \]

### Step 3: Calculate the Degree of Dissociation (\( \alpha \))

The degree of dissociation \( \alpha \) can be calculated using the formula:

\[ \alpha = \frac{\lambda_m}{\lambda_m^{\infty}(HA)} \]

Substituting the values:

\[ \alpha = \frac{200 \, \text{S cm}^2 \text{mol}^{-1}}{400 \, \text{S cm}^2 \text{mol}^{-1}} = 0.5 \]

### Step 4: Calculate the Ionization Constant (\( K_a \))

The ionization constant \( K_a \) can be calculated using the formula:

\[ K_a = \frac{C \alpha^2}{1 - \alpha} \]

Where \( C \) is the concentration of the acid, which is equal to \( [H^+] = 10^{-4} \, \text{mol/L} \).

Substituting the values:

\[ K_a = \frac{(10^{-4}) (0.5)^2}{1 - 0.5} \]

Calculating this gives:

\[ K_a = \frac{10^{-4} \cdot 0.25}{0.5} = \frac{10^{-4} \cdot 0.25}{0.5} = 0.5 \times 10^{-4} = 2.5 \times 10^{-5} \]

### Step 5: Calculate \( pK_a \)

Finally, we can calculate \( pK_a \) using the formula:

\[ pK_a = -\log(K_a) \]

Substituting the value of \( K_a \):

\[ pK_a = -\log(2.5 \times 10^{-5}) \]

Using a calculator:

\[ pK_a \approx 4.6 \]

### Final Answer

Thus, the value of \( pK_a(HA) \) at \( 25^\circ C \) is approximately **4.6**.