Question

Explain Kohlrausch's law. If \(\Lambda^\circ_m\) for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm² mol^−1, calculate \(\Lambda^\circ\) for HAc. (4)

Hint:

By Kohlrausch's law of independent ion migration, \(\Lambda^\circ(HAc) = \Lambda^\circ(HCl) + \Lambda^\circ(NaAc) - \Lambda^\circ(NaCl)\).

Answer 1

Step 1 (Statement of Kohlrausch's law): Kohlrausch's law of independent migration of ions states that at infinite dilution (when the electrolyte is completely dissociated), the limiting molar conductivity of an electrolyte is the sum of the individual limiting molar conductivities of its cation and anion.
\[ \Lambda^\circ_m = \nu_+ \,\lambda^\circ_+ + \nu_- \,\lambda^\circ_- \]
where \(\lambda^\circ_+\) and \(\lambda^\circ_-\) are the limiting molar conductivities of the cation and anion, and \(\nu_+, \nu_-\) are the number of cations and anions per formula unit.
Step 2 (Use): This law lets us calculate \(\Lambda^\circ_m\) of a weak electrolyte (like acetic acid, HAc) which cannot be found by extrapolation, by combining the \(\Lambda^\circ_m\) values of suitable strong electrolytes.
Step 3 (Set up the combination): Write HAc in terms of the given salts:
\[ \Lambda^\circ(HAc) = \Lambda^\circ(HCl) + \Lambda^\circ(NaAc) - \Lambda^\circ(NaCl) \]
Check the ions: HCl gives \(H^+ + Cl^-\); NaAc gives \(Na^+ + Ac^-\); subtracting NaCl removes \(Na^+ + Cl^-\), leaving \(H^+ + Ac^-\) = HAc. Correct.
Step 4 (Substitute the values):
\[ \Lambda^\circ(HAc) = 425.9 + 91.0 - 126.4 \]
Step 5 (Arithmetic):
\[ 425.9 + 91.0 = 516.9 \]
\[ 516.9 - 126.4 = 390.5 \]
\[\boxed{\Lambda^\circ(HAc) = 390.5 \; S\,cm^2\,mol^{-1}}\]