The specific conductance `(k)`of a `0.1M NaOH` solution is `0.0221 S cm^(-1)`. On addition of an equal volume `(V)` of `0.10 M HCI` solution, the value of specific conductance `(k)` falls to `0.0056 S cm^(-1)`. On further addition of same volume `(V)` of `0.10 M HCI`, the value of k rises to `0.017 S cm^(-1)`. calculate equivalent conductivity `(^^)` for `HCI` in `S cm^(2) mol^(-1)`.

Explanation

To solve the problem, we need to calculate the equivalent conductivity (λ) for HCl based on the specific conductance values provided for the NaOH and HCl solutions.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- Specific conductance of 0.1 M NaOH solution (k₁) = 0.0221 S cm⁻¹
- Specific conductance after adding equal volume of 0.1 M HCl (k₂) = 0.0056 S cm⁻¹
- Specific conductance after adding another equal volume of 0.1 M HCl (k₃) = 0.017 S cm⁻¹

2. **Calculate the Concentration of Ions After First Addition**:
- When equal volumes of NaOH and HCl are mixed, the total volume doubles. Therefore, the concentration of NaOH after mixing becomes:
\[ [\text{NaOH}] = \frac{0.1 \, \text{M}}{2} = 0.05 \, \text{M} \]
- The concentration of HCl after mixing becomes:
\[ [\text{HCl}] = \frac{0.1 \, \text{M}}{2} = 0.05 \, \text{M} \]
- The reaction between NaOH and HCl produces NaCl:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- After the first addition, both NaOH and HCl will neutralize each other completely, resulting in:
\[ [\text{NaCl}] = 0.05 \, \text{M} \]

3. **Calculate the Specific Conductance of NaCl**:
- The specific conductance of the solution after the first addition (k₂) is due to the ions present in the solution. Since NaOH and HCl neutralize each other, the remaining specific conductance is due to NaCl.
- The specific conductance of NaCl can be calculated as:
\[ k_{\text{NaCl}} = k_3 - k_2 = 0.017 \, \text{S cm}^{-1} - 0.0056 \, \text{S cm}^{-1} = 0.0114 \, \text{S cm}^{-1} \]

4. **Calculate the Equivalent Conductivity (λ) for HCl**:
- The equivalent conductivity (λ) can be calculated using the formula:
\[ \lambda = \frac{k \times 1000}{C} \]
- Here, \( k \) is the specific conductance of NaCl and \( C \) is the concentration of NaCl after the first addition:
\[ C = 0.05 \, \text{M} \]
- Substituting the values:
\[ \lambda_{\text{NaCl}} = \frac{0.0114 \times 1000}{0.05} = 228 \, \text{S cm}^2 \text{mol}^{-1} \]

5. **Final Calculation for Equivalent Conductivity of HCl**:
- Since the equivalent conductivity for HCl is equal to the conductivity of NaCl formed, we can conclude:
\[ \lambda_{\text{HCl}} = \lambda_{\text{NaCl}} = 228 \, \text{S cm}^2 \text{mol}^{-1} \]

### Final Answer:
The equivalent conductivity (λ) for HCl is **228 S cm² mol⁻¹**.