The molar conductivities `Lambda_(NaOAc)^@` and `Lambda_(HCI)^@` at infinite dilution is watter at `25^@C` are `91.0` and `426.2 S cm^@ //`mol respectively. To calculate `Lambda_(HOAc,)^2` the additional value required is:

Options

A. `KCI`
B. `NaOH`
C. `NaCI`
D. `H_(2)O`

Answer

To solve the problem of calculating the molar conductivity of acetic acid (HOAc) at infinite dilution, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Molar Conductivities**:
We are given the molar conductivities at infinite dilution for sodium acetate (NaOAc) and hydrochloric acid (HCl).
- \(\Lambda^0_{NaOAc} = 91.0 \, S \, cm^2 \, mol^{-1}\)
- \(\Lambda^0_{HCl} = 426.2 \, S \, cm^2 \, mol^{-1}\)

2. **Using Kohlrausch's Law**:
According to Kohlrausch's law, the molar conductivity of an electrolyte at infinite dilution can be expressed as the sum of the molar conductivities of its individual ions.
- For NaOAc:
\[ \Lambda^0_{NaOAc} = \Lambda^0_{Na^+} + \Lambda^0_{OAc^-} \]
- For HCl:
\[ \Lambda^0_{HCl} = \Lambda^0_{H^+} + \Lambda^0_{Cl^-} \]

3. **Setting Up the Equation for HOAc**:
To find the molar conductivity of acetic acid (HOAc), we can express it as:
\[ \Lambda^0_{HOAc} = \Lambda^0_{H^+} + \Lambda^0_{OAc^-} \]

4. **Finding the Required Values**:
To calculate \(\Lambda^0_{HOAc}\), we need the values of \(\Lambda^0_{H^+}\) and \(\Lambda^0_{OAc^-}\). However, we do not need the values for \(\Lambda^0_{Na^+}\) and \(\Lambda^0_{Cl^-}\).

5. **Combining the Equations**:
We can combine the equations for NaOAc and HCl:
\[ \Lambda^0_{NaOAc} + \Lambda^0_{HCl} = (\Lambda^0_{Na^+} + \Lambda^0_{OAc^-}) + (\Lambda^0_{H^+} + \Lambda^0_{Cl^-}) \]
Rearranging gives:
\[ \Lambda^0_{HOAc} = \Lambda^0_{NaOAc} + \Lambda^0_{HCl} - \Lambda^0_{Na^+} - \Lambda^0_{Cl^-} \]

6. **Identifying the Additional Value Needed**:
To eliminate \(\Lambda^0_{Na^+}\) and \(\Lambda^0_{Cl^-}\), we need the value of \(\Lambda^0_{NaCl}\), which is the sum of the conductivities of the individual ions:
\[ \Lambda^0_{NaCl} = \Lambda^0_{Na^+} + \Lambda^0_{Cl^-} \]

7. **Conclusion**:
Therefore, the additional value required to calculate \(\Lambda^0_{HOAc}\) is \(\Lambda^0_{NaCl}\).

### Final Answer:
The additional value required is \(\Lambda^0_{NaCl}\).