The pH of a 0.01 M weak acid HX (Ka = 4 × 10–10) is found to be 5. Now the acid — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

# Step-by-Step Solution **Step 1: Set up the equilibrium expression** For weak acid HX: Ka = [H⁺][X⁻]/[HX] **Step 2: Determine [H⁺] from the given pH** pH = 5, so [H⁺] = 10⁻⁵ M **Step 3: Apply the ICE table at equilibrium** - Initial [HX] = 0.01 M - Change: -x where x = [H⁺] = 10⁻⁵ M - Equilibrium: [HX] ≈ 0.01 - 10⁻⁵ ≈ 0.01 M (negligible change) - At equilibrium: [H⁺] = [X⁻] = 10⁻⁵ M **Step 4: Verify Ka with the given value** Ka = (10⁻⁵)(10⁻⁵)/(0.01) = 10⁻¹⁰/10⁻² = 10⁻⁸ This does NOT match the given Ka = 4 × 10⁻¹⁰. **Step 5: Conclusion** The measured pH of 5 is **inconsistent** with a 0.01 M solution of HX (Ka = 4 × 10⁻¹⁰). The actual pH should be approximately 6, not 5. Therefore, the answer is: **The given pH value is inconsistent with the stated Ka and concentration values** (or the acid has been diluted/additional H⁺ source is present).