`^^_(m)^(@)` for `NaCI,HCI` and `NaA` are `126.4,425.9` and `100.5Scm^2 m ol^(-1)`, respectively. If the conductivity of `0.001 MHA` is `5xx10^(-5)Scm^(-1)`, degree of dissociation of HA is :

Options

A. `0.25`
B. `0.125`
C. `0.50`
D. `0.75`

Explanation

To solve the problem, we need to find the degree of dissociation (α) of the weak acid HA using the given data. Here’s a step-by-step solution:

### Step 1: Calculate the Molar Conductivity (λm) of HA
The molar conductivity (λm) can be calculated using the formula:
\[ \lambda_m = \frac{\kappa \times 1000}{C} \]
where:
- \(\kappa\) is the conductivity of the solution,
- \(C\) is the concentration of the solution in mol/L.

Given:
- \(\kappa = 5 \times 10^{-5} \, \text{S cm}^{-1}\)
- \(C = 0.001 \, \text{mol/L}\)

Substituting the values:
\[ \lambda_m = \frac{5 \times 10^{-5} \times 1000}{0.001} = \frac{5 \times 10^{-2}}{0.001} = 50 \, \text{S cm}^2 \text{mol}^{-1} \]

### Step 2: Calculate the Standard Molar Conductivity (λ°m)
Using Kohlrausch's Law, we can calculate the standard molar conductivity (λ°m) for HA:
\[ \lambda^0_m = \lambda_{HCl} + \lambda_{Na^+} - \lambda_{NaCl} \]

Given:
- \(\lambda_{HCl} = 425.9 \, \text{S cm}^2 \text{mol}^{-1}\)
- \(\lambda_{Na^+} = 100.5 \, \text{S cm}^2 \text{mol}^{-1}\)
- \(\lambda_{NaCl} = 126.4 \, \text{S cm}^2 \text{mol}^{-1}\)

Substituting the values:
\[ \lambda^0_m = 425.9 + 100.5 - 126.4 = 400 \, \text{S cm}^2 \text{mol}^{-1} \]

### Step 3: Calculate the Degree of Dissociation (α)
The degree of dissociation (α) can be calculated using the formula:
\[ \alpha = \frac{\lambda_m}{\lambda^0_m} \]

Substituting the values:
\[ \alpha = \frac{50}{400} = \frac{1}{8} = 0.125 \]

### Final Answer
The degree of dissociation of HA is:
\[ \alpha = 0.125 \]