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When a student mixes 50 mL of 1.0 M HCI and 50 mL of 1.0 M NaOH in a coffee cup calorimeter, the temperature of the resultant solution increases from 21.0°C to 27.5°C. Assuming that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is 100 mL, its density 1.0 g mL-1 and that its specific heat is 4.18 J/g, calculate : (a) the heat change during mixing. (b) the enthalpy change for the reaction :HCl(aq ) + NaOH(aq) → NaCl(aq) + H2O |
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Answer» mM of HCI = 50 x 1.0 = 50; mM of NaOH = 50 x 1.0 = 50 Heat changes during mixing = m.s.△T 50 mM of HCI and 50 mM of NaOH = (100 x 1) x 4.18 x 6.5 = 2717 J = 2.72 kJ Further, on mixing 50 mM of HCI and 50 mM of NaOH produces heat = 2.72 kJ 1000 mM of HCI (or 1000 Meq.) and 1000 mM of NaOH produces (acid and bases are monobasic and monoacidic, respectively).
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