Question

The equivalent conductances of two strong electrolytes at infinite dilution in $H_2O$ (where ions move freely through a solution) at $25^°C$ are given below:
$\wedge^{\circ}_{CH_3COONa}=91.0\,S\,cm^{2} / equiv$
$\wedge^{\circ}_{HCl}=426.2cm^2 / equiv$
What additional information/quantity one needs to calculate $\wedge^{\circ}$ of an aqueous solution of acetic acid?

• A. $ \wedge^{\circ}$ of $NaCl$
• B. $ \wedge^{\circ}$ of $CH_3COOK$
• C. The limiting equivalent conductance of $H^{+}\left( \wedge^{\circ }_{H^{+}}\right) $
• D. $ \wedge^{\circ}$ of chloroacetic acid $(ClCH_2COOH)$

Answer 1

Answer: (A)

From Kohlrausch’s law
$ \Lambda^{\circ }_{CH_3COOH}=\Lambda^{\circ}_{CH_3COONa}+ \Lambda^{\circ }_{HCl}-\Lambda_{NaCl}^{\circ }$
Hence, (1) is the correct answer.