Molar conductivities `(Lambda_(m)^(@))` at infinite dilution of `NaCl, HCl` and `CH_(3)COONa` arc `126.4, 425.9` and `91.0 S cm^(2) mol^(-1)` respectively. `Lambda_(m)^(@)` for `CH_(3)COOH` will be

Options

A. `425.5 S cm^(2) mol^(-1)`
B. `180.5 S cm^(2) mol^(-1)`
C. `290.8 S cm^(2) mol^(-1)`
D. `390.5 S cm^(2) mol^(-1)`

Answer

To find the molar conductivity of acetic acid (CH₃COOH) at infinite dilution, we can use the relationship between the molar conductivities of the ions involved in the dissociation of acetic acid. The process involves using the known molar conductivities of sodium acetate (CH₃COONa), hydrochloric acid (HCl), and sodium chloride (NaCl).

### Step-by-Step Solution:

1. **Identify the Given Molar Conductivities**:
- Molar conductivity of NaCl (Λₘ(NaCl)) = 126.4 S cm² mol⁻¹
- Molar conductivity of HCl (Λₘ(HCl)) = 425.9 S cm² mol⁻¹
- Molar conductivity of sodium acetate (Λₘ(CH₃COONa)) = 91.0 S cm² mol⁻¹

2. **Write the Reaction**:
Acetic acid can be formed from sodium acetate and hydrochloric acid:
\[ \text{CH}_3\text{COONa} + \text{HCl} \rightarrow \text{CH}_3\text{COOH} + \text{NaCl} \]

3. **Apply the Concept of Molar Conductivity**:
The molar conductivity of acetic acid can be calculated using the formula:
\[ \Lambda_m(\text{CH}_3\text{COOH}) = \Lambda_m(\text{CH}_3\text{COONa}) + \Lambda_m(\text{HCl}) - \Lambda_m(\text{NaCl}) \]

4. **Substitute the Values**:
Plug in the values we have:
\[ \Lambda_m(\text{CH}_3\text{COOH}) = 91.0 + 425.9 - 126.4 \]

5. **Perform the Calculation**:
\[ \Lambda_m(\text{CH}_3\text{COOH}) = 91.0 + 425.9 - 126.4 = 390.5 \, \text{S cm}^2 \text{mol}^{-1} \]

6. **Final Result**:
The molar conductivity of acetic acid (CH₃COOH) at infinite dilution is approximately:
\[ \Lambda_m(\text{CH}_3\text{COOH}) \approx 390.5 \, \text{S cm}^2 \text{mol}^{-1} \]