`^^_(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. `0.50`
B. `0.25`
C. `0.125`
D. `0.75`

Explanation

To find the degree of dissociation of HA, we will follow these steps:

### Step 1: Understand the given data
We have the following standard molar conductivities:
- \( \Lambda^0_{NaCl} = 126.4 \, \text{S cm}^2 \text{mol}^{-1} \)
- \( \Lambda^0_{HCl} = 425.9 \, \text{S cm}^2 \text{mol}^{-1} \)
- \( \Lambda^0_{NaA} = 100.5 \, \text{S cm}^2 \text{mol}^{-1} \)

The conductivity of \( 0.001 \, M \, HA \) is given as \( \kappa = 5 \times 10^{-5} \, \text{S cm}^{-1} \).

### Step 2: Calculate the molar conductivity of HA
The molar conductivity \( \Lambda_M \) can be calculated using the formula:
\[ \Lambda_M = \frac{\kappa \times 1000}{C} \]
where \( C \) is the concentration in mol/L. Here, \( C = 0.001 \, M \).

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

### Step 3: Calculate the total standard molar conductivity of HA
To find the standard molar conductivity of HA, we need to consider the contributions from HCl and NaA while excluding NaCl. The relationship is given by:
\[ \Lambda^0_{HA} = \Lambda^0_{HCl} + \Lambda^0_{NaA} - \Lambda^0_{NaCl} \]

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

### Step 4: Calculate the degree of dissociation (α) of HA
The degree of dissociation \( \alpha \) can be calculated using the formula:
\[ \alpha = \frac{\Lambda_M}{\Lambda^0_{HA}} \]

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

### Conclusion
The degree of dissociation of HA is \( 0.125 \).

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