When 400 mL of 0.2 M H₂SO₄ solution is mixed with 600 mL of 0.1 M NaOH solution, the increase in temperature of the final solution is __________ × 10⁻² K. (Round off to the nearest integer).
[Use : H⁺ (aq) + OH⁻ (aq) → H₂O : Δ γ H = -57.1 kJ mol⁻¹]
Specific heat of H₂O = 4.18 J K⁻¹ g⁻¹
density of H₂O = 1.0 g cm⁻³
Assume no change in volume of solution on mixing.

Solution

Chemistry ma’am · 4 steps

Find the limiting ion first. Only that many moles of water form, and that sets the heat released.

1. Count the ions. Each H₂SO₄ gives two H⁺, so n(H⁺) = 2 × 0.2 × 0.4 = 0.16 mol. From the base, n(OH⁻) = 0.1 × 0.6 = 0.06 mol.
2. Limiting ion. OH⁻ is the smaller amount, so only 0.06 mol of water forms. Heat released = 0.06 × 57.1 = 3.426 kJ = 3426 J.
3. Mass of solution. The total volume is 1000 mL. With density 1.0 g cm⁻³, the mass is 1000 g.

   Δ T = q/(m c) = 3426/(1000 × 4.18) = 0.8196 K

4. So. Δ T = 81.96 × 10⁻² K, which rounds to 82.

The common slip: 219. Using all 0.16 mol of H⁺ instead of the limiting 0.06 mol of OH⁻ gives 219, not 82.

The idea to keep. In neutralisation heat problems, convert to moles of H⁺ and OH⁻ first. The smaller one decides the water formed and the heat released. The whole solution mass absorbs that heat.

Answer from the NTA final answer key. Solution by PrepWiser.