Which of the following equimolar solutions has the highest conductivity?

Step 1: Analyze the question Conductivity of a solution depends on the concentration of free mobile ions. Since the solutions are equimolar (same concentration), the one that produces the most ions will conduct best.

Step 2: Concept

• Strong electrolytes ionize completely.
• Weak electrolytes ionize only partially, so they have few ions.
• The number of ions produced per formula unit matters.
• H⁺ ions have unusually high mobility, which boosts conductivity.

Step 3: Examine each option

• A. CH₃COOH is a weak acid, only slightly ionized, so it has very few ions and low conductivity.
• B. H₂CO₃ is a weak acid, very slightly ionized, so it has the lowest conductivity of all.
• C. H₂SO₄ is a strong acid, and each mole gives up to 2 H⁺ and 1 SO₄²⁻, about 3 moles of ions: H₂SO₄ → 2H⁺ + SO₄²⁻ In practice, the second ionization is not fully complete, but the ion concentration is still very high, and the H⁺ ions are very mobile.
• D. NaOH is a strong base, fully ionized: NaOH → Na⁺ + OH⁻. This gives 2 moles of ions per mole. OH⁻ is also highly mobile, but there are fewer ions than from H₂SO₄.

Step 4: Compare Strong electrolytes (C and D) beat weak ones (A and B). Between C and D, H₂SO₄ gives more ions per mole (about 3 vs 2) and contains the very mobile H⁺. So H₂SO₄ has the highest conductivity.

Answer: C. H₂SO₄(aq)

Exam tip: For "equimolar" conductivity questions, first eliminate the weak electrolytes (acetic acid, carbonic acid, ammonia solution). Then pick the strong electrolyte that gives the most ions per formula unit. Remember that H₂SO₄ is dibasic, so it gives more ions than HCl or NaOH.