If the molar conductivities at infinite dilution for CH_3COOK, K_2SO_…

The correct answer is Option (1) → \((x - y + 2z)/2\)

Kohlrausch’s law of independent migration of ions states that the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and anion, each counted as many times as it appears in the formula. A weak electrolyte such as acetic acid cannot be measured at infinite dilution directly, so its value is built up from strong electrolytes that contain its ions.

The given data are:

    \( \displaystyle \Lambda^\circ(\ce{CH3COOK}) = \lambda^\circ(\ce{CH3COO-}) + \lambda^\circ(\ce{K+}) = z \)

    \( \displaystyle \Lambda^\circ(\ce{K2SO4}) = 2\lambda^\circ(\ce{K+}) + \lambda^\circ(\ce{SO4^2-}) = y \)

    \( \displaystyle \Lambda^\circ(\ce{H2SO4}) = 2\lambda^\circ(\ce{H+}) + \lambda^\circ(\ce{SO4^2-}) = x \)

The target is \( \lambda^\circ(\ce{CH3COO-}) + \lambda^\circ(\ce{H+}) \). Taking \( \ce{CH3COOK} \), adding half of \( \ce{H2SO4} \) and subtracting half of \( \ce{K2SO4} \) brings in one \( \ce{H+} \) and removes the \( \ce{K+} \) and the sulphate:

    \( \displaystyle \Lambda^\circ(\ce{CH3COOH}) = z + \frac{x}{2} - \frac{y}{2} \)

    \( \displaystyle = \lambda^\circ(\ce{CH3COO-}) + \lambda^\circ(\ce{K+}) + \lambda^\circ(\ce{H+}) + \frac{1}{2}\lambda^\circ(\ce{SO4^2-}) - \lambda^\circ(\ce{K+}) - \frac{1}{2}\lambda^\circ(\ce{SO4^2-}) \)

    \( \displaystyle = \lambda^\circ(\ce{CH3COO-}) + \lambda^\circ(\ce{H+}) \)

So

    \( \displaystyle \Lambda^\circ(\ce{CH3COOH}) = \frac{x - y + 2z}{2} \)

Shortcut: One acetate ion is needed, and it comes only from \( \ce{CH3COOK} \), so z must appear with coefficient 1, that is, as 2z/2. This rules out Options (3) and (4); between (1) and (2), half of \( \ce{K2SO4} \) removes the one \( \ce{K+} \), so y appears as y/2, not 2y/2.