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200 mL of 0.1 M NaOH solution is mixed with 200 mL of 0.025 M H2SO4 solution. Calculate |delta_H| (in joules) of mixing, given that HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l); delta_H = -57 kJ/mol.

1. (A) 570 J
2. (B) 285 J
3. (C) 1140 J
4. (D) 142.5 J

Correct answer: (A) 570 J

Solution

n(NaOH) = 0.2 L * 0.1 mol/L = 0.02 mol OH-. H2SO4 -> 2H+ + SO4²-: n(H+) = 2 * 0.2 * 0.025 = 0.01 mol. H+ is limiting. Moles of H2O formed = 0.01 mol. delta_H = 0.01 mol * (-57 kJ/mol) = -0.57 kJ = -570 J. |delta_H| = 570 J.

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