Question

What is Kohlrausch's law? If \( \lambda^{\circ}_m \) for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm² mol^-1 respectively, find the value of \( \lambda^{\circ}_m \) for HAc. (2+2)

Hint:

Kohlrausch's law: at infinite dilution \( \lambda^{\circ}_m \) is the sum of independent ionic conductivities. Combine the salts as \( \lambda^{\circ}_m(\text{HAc}) = \lambda^{\circ}_m(\text{HCl}) + \lambda^{\circ}_m(\text{NaAc}) - \lambda^{\circ}_m(\text{NaCl}) \).

Answer 1

Kohlrausch's law, put simply. When an electrolyte is diluted to the limit, its ions no longer interfere with each other and each carries current on its own. So the limiting molar conductivity equals the added conductivities of all the ions produced by one formula unit. This additivity is the whole content of the law, and its great use is finding the limiting molar conductivity of weak electrolytes (like acetic acid), which cannot be measured directly.

Working the numbers with ionic values.
Step 1: The unknown for HAc is \(\lambda^{\circ}_{H^+} + \lambda^{\circ}_{Ac^-}\). We are not given single-ion values, so build them from the three strong electrolytes.
Step 2: From the given salts:
\(\lambda^{\circ}_{H^+} + \lambda^{\circ}_{Cl^-} = 425.9\) (HCl),
\(\lambda^{\circ}_{Na^+} + \lambda^{\circ}_{Ac^-} = 91.0\) (NaAc),
\(\lambda^{\circ}_{Na^+} + \lambda^{\circ}_{Cl^-} = 126.4\) (NaCl).
Step 3: Take the first two sums and remove the third: the \(\lambda^{\circ}_{Na^+}\) and \(\lambda^{\circ}_{Cl^-}\) terms cancel, leaving \(\lambda^{\circ}_{H^+} + \lambda^{\circ}_{Ac^-}\).
\[ \lambda^{\circ}_m(\text{HAc}) = (425.9 + 91.0) - 126.4 \]
Step 4: Evaluate: \(516.9 - 126.4 = 390.5\).
\[ \boxed{\lambda^{\circ}_m(\text{HAc}) = 390.5\ \text{S cm}^2\ \text{mol}^{-1}} \]
The units are S cm\(^2\) mol\(^{-1}\), the standard unit of molar conductivity.