Equivalent conductivity at infinite dilution for sodium potassium oxalate, `(COO^(-))_(2)Na^(+)K^(+)`, will be (given, molar conductivities of oxalate, `K^(+) and Na^(+)` ions at infinite diluton are `148.2,50.1,73.5" S "cm^(2)mol^(-1)` respectively).

Options

A. `271.8" S "cm^(2)eq^(-1)`
B. `67.96" S "cm^(2)eq^(-1)`
C. `543.6" S "cm^(2)eq^(-1)`
D. `135.9" S "cm^(2)eq^(-1)`.

Explanation

To find the equivalent conductivity at infinite dilution for sodium potassium oxalate \((COO^-)_{2}Na^+K^+\), we will use the molar conductivities of the individual ions at infinite dilution. The formula for equivalent conductivity \(\Lambda_{eq}^0\) is given by:

\[ \Lambda_{eq}^0 = \Lambda_{m}^0(Na^+) + \Lambda_{m}^0(K^+) + 2 \Lambda_{m}^0(COO^{-}) \]

Where:
- \(\Lambda_{m}^0(Na^+)\) is the molar conductivity of the sodium ion at infinite dilution.
- \(\Lambda_{m}^0(K^+)\) is the molar conductivity of the potassium ion at infinite dilution.
- \(\Lambda_{m}^0(COO^{-})\) is the molar conductivity of the oxalate ion at infinite dilution.

Given values:
- \(\Lambda_{m}^0(Na^+) = 73.5 \, S \, cm^2 \, mol^{-1}\)
- \(\Lambda_{m}^0(K^+) = 50.1 \, S \, cm^2 \, mol^{-1}\)
- \(\Lambda_{m}^0(COO^{-}) = 148.2 \, S \, cm^2 \, mol^{-1}\)

Now, substituting these values into the formula:

\[ \Lambda_{eq}^0 = \Lambda_{m}^0(Na^+) + \Lambda_{m}^0(K^+) + 2 \Lambda_{m}^0(COO^{-}) \]

\[ \Lambda_{eq}^0 = 73.5 + 50.1 + 2 \times 148.2 \]

Calculating the oxalate contribution:

\[ 2 \times 148.2 = 296.4 \]

Now, adding all the contributions together:

\[ \Lambda_{eq}^0 = 73.5 + 50.1 + 296.4 = 420.0 \, S \, cm^2 \, mol^{-1} \]

Thus, the equivalent conductivity at infinite dilution for sodium potassium oxalate is:

\[ \Lambda_{eq}^0 = 420.0 \, S \, cm^2 \, mol^{-1} \]