The conductivity of a weak acid HA of concentration 0.001 mol `L^(-1)` is `2.0 xx 10^(-5)S cm^(-1)`. If `Lambda_(m)^(0)(H A) = 190 S cm^(2) mol^(-1)`, the ionization constant `(K_(a))` of HA is equal to ________ ` xx 10^(-6)`.
(Round off to the Nearest Integer)

Answer

To solve the problem, we need to find the ionization constant \( K_a \) of the weak acid \( HA \) given its conductivity and molar conductance at infinite dilution.

### Step-by-Step Solution:

1. **Given Data**:
- Concentration of the weak acid \( [HA] = 0.001 \, \text{mol L}^{-1} \)
- Conductivity \( \kappa = 2.0 \times 10^{-5} \, \text{S cm}^{-1} \)
- Molar conductance at infinite dilution \( \Lambda_m^0(HA) = 190 \, \text{S cm}^2 \text{mol}^{-1} \)

2. **Calculate Molar Conductance at Given Concentration**:
Molar conductance \( \Lambda \) can be calculated using the formula:
\[ \Lambda = \frac{\kappa \times 1000}{C} \]
where \( C \) is the concentration in mol L\(^{-1}\).

Substituting the values:
\[ \Lambda = \frac{2.0 \times 10^{-5} \, \text{S cm}^{-1} \times 1000}{0.001} = \frac{2.0 \times 10^{-2}}{0.001} = 20 \, \text{S cm}^2 \text{mol}^{-1} \]

3. **Calculate Degree of Ionization (\( \alpha \))**:
The degree of ionization \( \alpha \) can be calculated using:
\[ \alpha = \frac{\Lambda}{\Lambda_m^0} \]
Substituting the values:
\[ \alpha = \frac{20}{190} \approx 0.10526 \]

4. **Calculate the Ionization Constant \( K_a \)**:
The ionization constant \( K_a \) for the weak acid can be calculated using the formula:
\[ K_a = \frac{C \cdot \alpha^2}{1 - \alpha} \]
Since \( \alpha \) is small, we can approximate \( 1 - \alpha \approx 1 \):
\[ K_a \approx C \cdot \alpha^2 \]
Substituting the values:
\[ K_a \approx 0.001 \cdot (0.10526)^2 \approx 0.001 \cdot 0.0111 \approx 1.108 \times 10^{-5} \]

5. **Express \( K_a \) in Required Format**:
To express \( K_a \) in the form of \( x \times 10^{-6} \):
\[ K_a \approx 11.08 \times 10^{-6} \]
Rounding off to the nearest integer gives:
\[ K_a \approx 11 \times 10^{-6} \]

### Final Answer:
The ionization constant \( K_a \) of the weak acid \( HA \) is approximately \( 11 \times 10^{-6} \).

---