The decreasing order of electrical conductivity of the following aqueous solutions is:
(A) 0.1 M Formic acid,
(B) 0.1 M Acetic acid,
(C) 0.1 M Benzoic acid.

Electrical conductivity in aqueous solutions of weak acids depends entirely on the concentration of mobile ions produced. At identical concentrations ($$0.1 \text{ M}$$), a stronger acid dissociates more completely, releasing more ions into the solution and exhibiting a higher electrical conductivity. Therefore, the conductivity order perfectly follows the acid dissociation constant ($$K_a$$) order.

Detailed Evaluation of Each Option:

• Formic Acid (A): $$\text{HCOOH}$$

  Formic acid has no electron-donating alkyl groups attached to the carboxyl group. It has the highest dissociation constant ($$K_a \approx 1.8 \times 10^{-4}$$) among the three, making it the strongest electrolyte and the most conductive solution.




• Benzoic Acid (C): $$\text{C}_6\text{H}_5\text{COOH}$$

  The phenyl ring behaves as a weaker electron-donating group via resonance/inductive pathways compared to a normal aliphatic chain when considering overall carboxylic acid strength ($$K_a \approx 6.3 \times 10^{-5}$$). It is weaker than formic acid but significantly stronger than acetic acid.




• Acetic Acid (B): $$\text{CH}_3\text{COOH}$$

  The methyl group ($$\text{--CH}_3$$) exerts a strong electron-donating inductive effect (+I), which destabilizes the carboxylate anion conjugate base. It has the lowest dissociation constant ($$K_a \approx 1.75 \times 10^{-5}$$), making it the weakest acid and the least conductive solution.

Conclusion:

Comparing the ion concentrations produced at equilibrium gives the decreasing order of electrical conductivity as:

$$\mathbf{(A) > (C) > (B)}$$

Answer: Option B — (A) > (C) > (B)