The conductivity of 0.1 N NaOH solution is 0.022 S`cm^(-1)`.When equal volume of 0.1 N HCl solution is added, the conductivity of resultant solution is decreased to 0.0055 S `cm^(-1)` . The equivalent conductivity of NaCl solution is :

Options

A. 0.0055
B. 0.11
C. 110
D. None of these

Answer

To solve the problem, we need to find the equivalent conductivity of NaCl solution based on the given data about NaOH and HCl solutions. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Data
- The conductivity of 0.1 N NaOH solution is \( \kappa_{NaOH} = 0.022 \, S \, cm^{-1} \).
- When equal volumes of 0.1 N HCl are added, the conductivity of the resultant solution becomes \( \kappa_{resultant} = 0.0055 \, S \, cm^{-1} \).

### Step 2: Calculate the Normality of the Resultant Solution
When equal volumes of 0.1 N NaOH and 0.1 N HCl are mixed, the total volume doubles. Therefore, the normality of the resultant solution can be calculated as follows:

\[ \text{Normality of resultant solution} = \frac{0.1 \, N}{2} = 0.05 \, N \]

### Step 3: Use the Formula for Equivalent Conductivity
The equivalent conductivity (\( \Lambda_{eq} \)) is calculated using the formula:

\[ \Lambda_{eq} = \frac{\kappa \times 1000}{N} \]

Where:
- \( \kappa \) is the conductivity of the solution,
- \( N \) is the normality of the solution.

### Step 4: Substitute the Values into the Formula
Now, substitute the values we have into the formula:

\[ \Lambda_{eq} = \frac{0.0055 \, S \, cm^{-1} \times 1000}{0.05 \, N} \]

### Step 5: Perform the Calculation
Calculating the above expression:

\[ \Lambda_{eq} = \frac{0.0055 \times 1000}{0.05} = \frac{5.5}{0.05} = 110 \, S \, cm^2 \, eq^{-1} \]

### Conclusion
The equivalent conductivity of NaCl solution is \( 110 \, S \, cm^2 \, eq^{-1} \).

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