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)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the limiting molar conductivity of HA
We can calculate the limiting molar conductivity of HA using the formula:
\[ \Lambda_m^{\infty}(HA) = \Lambda_m^{\infty}(HCl) + \Lambda_m^{\infty}(NaA) - \Lambda_m^{\infty}(NaCl) \]
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 degree of dissociation (α)
The degree of dissociation (α) can be calculated using the formula:
\[ \alpha = \frac{\Lambda_m}{\Lambda_m^{\infty}(HA)} \]
Where:
- \(\Lambda_m\) is the molar conductivity of the solution (given as 200 S cm² mol⁻¹).
- \(\Lambda_m^{\infty}(HA)\) is the limiting molar conductivity we just calculated (400 S cm² mol⁻¹).

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

### Step 3: Calculate the concentration of H⁺ ions from pH
Given that the pH of the solution is 4, we can find the concentration of H⁺ ions using:
\[ \text{pH} = -\log[H^+] \]
Thus,
\[ [H^+] = 10^{-4} \, \text{mol L}^{-1} \]

### Step 4: Write the expression for the dissociation constant (Kₐ)
The dissociation of HA can be represented as:
\[ HA \rightleftharpoons H^+ + A^- \]
The dissociation constant \(K_a\) can be expressed as:
\[ K_a = \frac{[H^+][A^-]}{[HA]} \]
At equilibrium, we have:
- \([H^+] = C \alpha\)
- \([A^-] = C \alpha\)
- \([HA] = C(1 - \alpha)\)

Substituting these into the expression for \(K_a\):
\[ K_a = \frac{(C \alpha)(C \alpha)}{C(1 - \alpha)} = \frac{C^2 \alpha^2}{C(1 - \alpha)} = \frac{C \alpha^2}{1 - \alpha} \]

### Step 5: Substitute the known values into the \(K_a\) expression
We know:
- \([H^+] = 10^{-4} \, \text{mol L}^{-1}\) which is equal to \(C \alpha\).
- \(\alpha = 0.5\).

Thus,
\[ C \alpha = 10^{-4} \implies C = \frac{10^{-4}}{0.5} = 2 \times 10^{-4} \, \text{mol L}^{-1} \]

Now substituting \(C\) and \(\alpha\) into the \(K_a\) expression:
\[ K_a = \frac{(2 \times 10^{-4})(0.5)^2}{1 - 0.5} = \frac{(2 \times 10^{-4})(0.25)}{0.5} = \frac{5 \times 10^{-5}}{0.5} = 10^{-4} \]

### Step 6: Calculate pKₐ
Finally, we calculate \(pK_a\):
\[ pK_a = -\log K_a = -\log(10^{-4}) = 4 \]

### Conclusion
The value of \(pK_a\) for HA at 25°C is **4**.