The equivalent conductances of two strong electrolytes at infinite dilution in H2O (where ions move freely through a solution) at `25^(@)C`are given below:
`wedge_(CH_(3)COONa^(@) = 91.0Scm^(2)//equ iv` , `wedge_HCl^(@)= 426.25cm^(2)//equiv`
What additional information/quantity one needs to calculate `overset(o)(wedge)` of an aqueous solution of acetic acid?

Options

A. `overset(o)(wedge)` of NaC1
B. `3 overset(o)(wedge)` of `CH_(3)COOK`
C. The limitation equivalent conductance of `H^(+) (wedge_(H^+)^(@))`
D. `overset(o)wedge` of chlorocetic acid `(C1CH_(2)COOH)`

Explanation

To find the equivalent conductance of acetic acid (\( \overset{o}{\lambda} \) of CH₃COOH) at infinite dilution, we can use Kohlrausch's law, which states that the equivalent conductance of a weak electrolyte can be calculated using the equivalent conductances of its ions and the equivalent conductance of a strong electrolyte that dissociates completely.

### Step-by-Step Solution:

1. **Understand the Given Data**:
- The equivalent conductance of sodium acetate (\( \text{CH}_3\text{COONa} \)) at infinite dilution is \( \overset{o}{\lambda}_{\text{CH}_3\text{COONa}} = 91.0 \, \text{S cm}^2/\text{equiv} \).
- The equivalent conductance of hydrochloric acid (\( \text{HCl} \)) at infinite dilution is \( \overset{o}{\lambda}_{\text{HCl}} = 426.25 \, \text{S cm}^2/\text{equiv} \).

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