Calculate the amount of heat evolved when 

(i) 500 cm³ of 0.1 M hydrochloric acid is mixed with 200 cm³ of 0.2 M sodium hydroxide solution 

(ii) 200 cm³ of 0.2 M sulphuric acid is mixed with 400 cm³ of 0.5 M potassium hydroxide solution. 

Assuming that the specific heat of water is 4.18 J K^–1 g^–1 and ignoring the heat absorbed by the container, thermometer, stirrer etc., what would be the rise in temperature in each of the above cases? 

(i) 500 cm³ of 0.1 M HCl = 0.1/1000 × 500 mole of HCl = 0.05 mole of H⁺ ions  200 cm³ of 0.2 M NaOH

= 0.2/1000 × 200 mole of NaOH = 0.04 mole of NaOH = 0.04 mole of OH⁻ ions

Thus, 0.04 mole of H⁺ ions will combine with 0.04 mole of OH⁻ ions to form 0.04 mole of H₂O and 0.01 mole of H⁺ ions will remain unreached.

Heat evolved when 1 mole of H⁺ ions combine with 1 mole of OH⁻ ions = 57.1 kJ.

∴  Heat evolved when 0.04 mole of H⁺ ions combine with 0.04 mole of OH⁻ ions = 57.1 × 0.04 = 2.284 kJ

(ii) 200 cm³ of 0.2 M H₂SO₄ =0.2/1000× 200 mole of H₂SO₄ = 0.04 mole of H₂SO₄ = 0.08 mole of H⁺ ions

400 cm³ of 0.5 M KOH =0.5/1000× 400 mole of KOH = 0.2 mole of KOH = 0.2 mole of OH⁻ ions

Thus, 0.08 mole of H⁺ ions will neutralize 0.08 mole of OH⁻ ions. (out of 0.2 mole of OH⁻ ions) to form 0.08 mole of H₂O.

Hence, heat evolved = 57.1 × 0.08 = 4.568 kJ

In case (i), heat product = 2.284 kJ = 2284 J

Total mass of the solution = 500 + 200 = 700 g

Specific heat = 4.18 J K⁻¹ g⁻¹

Q = m × C × Δt

∴   Δt =Q/(m × C)=2284/(700 × 4.18)= 0.78°

In case (ii), heat produced = 4.568 kJ = 4568 J

Total mass of the solution = 200 + 400 = 600 g

∴   Δt =Q/(m × C)=4568/(600 × 4.18)= 1.82°