200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is -57.1 kJ. The increase in temperature in °C of the system on mixing is x × 10⁻². The value of x is ___________. (Nearest integer)
[Given : Specific heat of water = 4.18 J g⁻¹ K⁻¹, Density of water = 1.00 g cm⁻³]
[Assume no volume change on mixing)

Solution

Chemistry ma’am · 4 steps

Heat comes only from the acid and base that actually react, so find the limiting one first.

1. Count the moles. HCl: 0.2 L × 0.2 M = 0.04 mol. NaOH: 0.3 L × 0.1 M = 0.03 mol. The reaction is 1 : 1, so NaOH is limiting and only 0.03 mol of water forms.
2. Heat released. q = 0.03 × 57.1 kJ = 1.713 kJ = 1713 J. The extra 0.01 mol of HCl has nothing to react with, so it releases no heat.

   Δ T = q/mc = 1713/(500 × 4.18) = 0.8196 °C

3. Mass of solution. The total volume is 200 + 300 = 500 mL. With density 1.00 g cm⁻³, that gives m = 500 g, the value used above.
4. So. Δ T = 0.8196 °C = 81.96 × 10⁻² °C, so x rounds to 82.

The common slip: 109. Using the 0.04 mol of HCl instead of the limiting 0.03 mol of NaOH gives Δ T = 1.09 °C, that is 109, not 82.

The idea to keep. In neutralization problems, heat released equals the moles of water formed times the molar heat. That count comes from the limiting reagent. Use the total mass of the mixed solution in q = mcΔ T.

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