Equal volumes of 0.015 M `CH_(3)COOH` and 0.0015M NaOH are mixed together. What would be molar conductivity of mixture if conductity of `CH_(3)COONa` is `6.3xx10^(-4)S cm^(-1)`?

Options

A. `8.4 S cm^(2) "mol"^(-1)`
B. `84 S cm^(2)"mol"^(-1)`
C. `4.2S cm^(2)"mol"^(-1)`
D. `42 S cm^(2) "mol"^(-1)`

Answer

To solve the problem, we need to calculate the molar conductivity of the mixture formed by mixing equal volumes of 0.015 M CH₃COOH (acetic acid) and 0.0015 M NaOH (sodium hydroxide).

### Step-by-Step Solution:

1. **Calculate the initial moles of CH₃COOH and NaOH:**
- Let the volume of each solution be \( V \) liters.
- Moles of CH₃COOH = Concentration × Volume = \( 0.015 \, \text{mol/L} \times V \, \text{L} = 0.015V \, \text{mol} \)
- Moles of NaOH = Concentration × Volume = \( 0.0015 \, \text{mol/L} \times V \, \text{L} = 0.0015V \, \text{mol} \)

2. **Determine the reaction between CH₃COOH and NaOH:**
- The reaction is:
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]
- From the stoichiometry of the reaction, 1 mole of CH₃COOH reacts with 1 mole of NaOH to produce 1 mole of CH₃COONa.

3. **Calculate the moles of CH₃COONa formed:**
- Since NaOH is the limiting reagent (as it has fewer moles), the moles of CH₃COONa formed will be equal to the moles of NaOH:
\[ \text{Moles of CH}_3\text{COONa} = 0.0015V \, \text{mol} \]

4. **Calculate the total volume of the mixture:**
- The total volume after mixing is:
\[ V_{\text{total}} = V + V = 2V \]

5. **Calculate the concentration of CH₃COONa in the mixture:**
- The concentration of CH₃COONa in the total volume is:
\[ \text{Concentration of CH}_3\text{COONa} = \frac{0.0015V}{2V} = \frac{0.0015}{2} = 0.00075 \, \text{M} \]

6. **Calculate the molar conductivity of the mixture:**
- Molar conductivity (\( \Lambda_m \)) is given by the formula:
\[ \Lambda_m = \frac{\text{Conductivity} (\kappa)}{\text{Concentration} (C)} \]
- Given that the conductivity of CH₃COONa is \( 6.3 \times 10^{-4} \, \text{S/cm} \) and the concentration is \( 0.00075 \, \text{M} \):
\[ \Lambda_m = \frac{6.3 \times 10^{-4} \, \text{S/cm}}{0.00075 \, \text{mol/L}} = 0.84 \, \text{S cm}^2/\text{mol} \]

### Final Answer:
The molar conductivity of the mixture is \( 0.84 \, \text{S cm}^2/\text{mol} \).