Problem: Why can't we determine Λ°m of $CH_{3}COOH$ by graph, and how does Kohlrausch's law solve it?

Step 1 — Understand the problem with weak electrolytes: Weak electrolytes (like $CH_{3}COOH$) do NOT fully dissociate in solution. At concentration C, only a small fraction α dissociates: $CH_{3}COOH$ ⇌ $CH_{3}COO^{-}$ + $H^{+}$. As C decreases (dilution), α increases, meaning more ions are present per mole dissolved. The increase in ion count with dilution causes Λm to rise steeply.

Step 2 — Why extrapolation fails: For strong electrolytes (fully ionized), the Λm vs. √C graph is a straight line (Λm = Λ°m − A√C), so extrapolating to C = 0 is reliable. But for weak electrolytes, the Λm vs. √C graph curves sharply upward near C = 0. The curve doesn't follow a predictable straight-line pattern that can be reliably extrapolated. Attempting to extrapolate introduces massive error.

Step 3 — Kohlrausch's key insight: At infinite dilution, ALL electrolytes (both strong and weak) are completely dissociated. The limiting molar conductivity Λ°m depends only on the individual ion conductivities — which are independent properties of each ion type, regardless of whether the ion came from a "strong" or "weak" source.

Step 4 — The algebraic trick: We know strong electrolyte Λ°m values from reliable graph extrapolation:

• Λ°m($CH_{3}COONa$) = contribution from $CH_{3}COO^{-}$ + $Na^{+}$
• Λ°m(HCl) = contribution from $H^{+}$ + $Cl^{-}$
• Λ°m(NaCl) = contribution from $Na^{+}$ + $Cl^{-}$

Adding ($CH_{3}COONa$) + (HCl) and subtracting (NaCl): = ($CH_{3}COO^{-}$ + $Na^{+}$) + ($H^{+}$ + $Cl^{-}$) − ($Na^{+}$ + $Cl^{-}$) = $CH_{3}COO^{-}$ + $H^{+}$ = Λ°m($CH_{3}COOH$)

Step 5 — Conclusion: This is an indirect determination using the additive property of limiting ionic conductivities. It works because Kohlrausch's law (Λ°m = sum of individual ionic conductivities) is universal — at infinite dilution, each ion contributes independently to conductance.

Final result: Λ°m($CH_{3}COOH$) = Λ°m($CH_{3}COONa$) + Λ°m(HCl) − Λ°m(NaCl)