You mix 50.0 ml of a weak monoprotic acid with 50.0 ml of NaOH solution in a cof

Question

You mix 50.0 ml of a weak monoprotic acid with 50.0 ml of NaOH solution in a coffee cup calorimeter. Both solutions (and the calorimeter) were initially at 22.0 C. The final temperature of the neutralization reaction was determined to be 33.5 C. The calorimeter constant was knonw to be 20.6 J/C. Specific heat of H2O= 4.182 j/g.C

a) what is the total amount of heat evolved in this reaction?

b) if 135 mmol of the monoprotic acid were neutralized in this reaction, what is the molar heat of neutralization for this reaction?

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Explanation / Answer

The 100 g (100 mL) of solution gained 11.5 deg C (33.5-22).

100 g x 4.184 J/g deg c x 11.5 deg c = 4812 J

Joules yo heat the calorimeter -

20.5 J/deg C x 11.5 deg c = 236 J

a. heat evolved in reaction = 5048 J (4812 + 236)

135 mmol = 0.135 mol

0.135 mole H2O produced

b. 5048 J / 0.135 mol = 37393 J/mol (37.9 kJ/mol)

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