Question

The molar conductivity ($\Lambda_m^0$) at infinite dilution for HCl, NaCl and CH$_3$COONa at 25\(^{\circ}\)C are 426, 126 and 91 S cm\(^2\) respectively. The \(\Lambda_m^0\) for acetic acid at the same temperature will be

• A. 391 S cm\(^2\)
• B. 209 S cm\(^2\)
• C. 461 S cm\(^2\)
• D. 643 S cm\(^2\)

Hint:

For calculating the limiting molar conductivity of a weak electrolyte using Kohlrausch's Law, typically you add the \(\Lambda_m^0\) of two strong electrolytes whose ions, when combined and one common ion subtracted, yield the ions of the weak electrolyte. A common pattern is: $\Lambda_m^0(\text{Weak Acid}) = \Lambda_m^0(\text{Salt of Weak Acid}) + \Lambda_m^0(\text{Strong Acid}) - \Lambda_m^0(\text{Salt of Strong Acid})$.

Answer 1

The Correct Option is A

Step 1: Assign the limiting ionic conductivities as unknowns: let \(\lambda^0(H^+) = h\), \(\lambda^0(Cl^-) = c\), \(\lambda^0(Na^+) = s\), \(\lambda^0(CH_3COO^-) = a\). From the three given salts: \(h + c = 426\) (for \(HCl\)), \(s + c = 126\) (for \(NaCl\)), and \(a + s = 91\) (for \(CH_3COONa\)).

Step 2: The quantity needed is \(\Lambda_m^0(CH_3COOH) = h + a\). Adding the first and third equations gives \(h + c + a + s = 426 + 91 = 517\).

Step 3: Subtracting the second equation \((s + c = 126)\) from this sum removes the unwanted \(s\) and \(c\) terms: \(h + a = 517 - 126 = 391\).
\[ \boxed{\Lambda_m^0(CH_3COOH) = 391 \text{ S cm}^2} \]