Conductivity of an aqueous solution of `0.1M HX` (a weak mono-protic acid) is `5 xx 10^(-4) Sm^(-1)`
Given: `^^_(m)^(oo) [H^(+)] = 0.04 Sm^(2) mol^(-1): ^^_(m)^(oo) [X^(-)] = 0.01 Sm^(2)mol^(-1)`
Find `pK_(a)[HX]`

Explanation

To find the \( pK_a \) of the weak monoprotic acid \( HX \) given the conductivity of its solution and the molar conductivities at infinite dilution, we can follow these steps:

### Step 1: Calculate the Molar Conductivity at Given Concentration (\( \lambda_{mc} \))
We know that the molar conductivity at a given concentration can be calculated using the formula:

\[ \lambda_{mc} = \frac{\text{Conductivity} \times 1000}{\text{Concentration}} \]

Given:
- Conductivity \( K = 5 \times 10^{-4} \, \text{S/m} \)
- Concentration \( C = 0.1 \, \text{mol/m}^3 \)

Convert concentration to \( \text{mol/m}^3 \) (which is already given as \( 0.1 \)).

Now, substituting the values:

\[ \lambda_{mc} = \frac{5 \times 10^{-4} \times 1000}{0.1} = \frac{5 \times 10^{-1}}{0.1} = 5 \times 10^{-6} \, \text{S m}^2 \text{mol}^{-1} \]

### Step 2: Calculate the Molar Conductivity at Infinite Dilution (\( \lambda_{m}^{\infty} \))
The molar conductivity at infinite dilution for the acid \( HX \) can be expressed as:

\[ \lambda_{m}^{\infty} = \lambda_{m}^{\infty} [H^+] + \lambda_{m}^{\infty} [X^-] \]

Given:
- \( \lambda_{m}^{\infty} [H^+] = 0.04 \, \text{S m}^2 \text{mol}^{-1} \)
- \( \lambda_{m}^{\infty} [X^-] = 0.01 \, \text{S m}^2 \text{mol}^{-1} \)

Now, substituting the values:

\[ \lambda_{m}^{\infty} = 0.04 + 0.01 = 0.05 \, \text{S m}^2 \text{mol}^{-1} \]

### Step 3: Calculate the Degree of Ionization (\( \alpha \))
The degree of ionization \( \alpha \) can be calculated using the formula:

\[ \alpha = \frac{\lambda_{mc}}{\lambda_{m}^{\infty}} \]

Substituting the values:

\[ \alpha = \frac{5 \times 10^{-6}}{0.05} = 10^{-4} \]

### Step 4: Calculate the Acid Dissociation Constant (\( K_a \))
For a weak monoprotic acid, the dissociation constant \( K_a \) can be calculated using:

\[ K_a = C \alpha^2 \]

Where \( C = 0.1 \, \text{mol/m}^3 \) and \( \alpha = 10^{-4} \).

Substituting the values:

\[ K_a = 0.1 \times (10^{-4})^2 = 0.1 \times 10^{-8} = 10^{-9} \]

### Step 5: Calculate \( pK_a \)
The \( pK_a \) is calculated using the formula:

\[ pK_a = -\log K_a \]

Substituting the value of \( K_a \):

\[ pK_a = -\log(10^{-9}) = 9 \]

### Final Answer
Thus, the value of \( pK_a \) for the weak monoprotic acid \( HX \) is:

\[ \boxed{9} \]