When 40.0 mL of 0.200 M HCI at 21.5°C is added to 40.0 mL of 0.200 M NaOH also at 21.5°C in a coffee-cup calorimeter, the temperature of the resulting solution rises to 22.8°C. Assume that the volumes are additive, the specific heat of the solution is 4.180 J/g°C, and that the density of the solution is 1.00 g/mL. Calculate the enthalpy change, AH, for the reaction: HCI (aq) + NaOH (aq) NaCI (aq) + H20 (I)

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Chapter1: Chemical Foundations
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The question is given like this, it is not incomplete. There is no heat capacity of the calorimeter given. Can it be solved without that? It is high school chemistry so maybe it can be solved in an easier way? Many thanks.

7. When 40.0 ml of 0.200 M HCI at 21.5°C is added to 40.0 ml of 0.200 M NaOH also at 21.5°C in a
coffee-cup calorimeter, the temperature of the resulting solution rises to 22.8°C. Assume that
the volumes are additive, the specific heat of the solution is 4.180 J/g°C, and that the density of
the solution is 1.00 g/mL. Calculate the enthalpy change, AH, for the reaction:
HCI (aq) + NaOH (aq) → NaCl (aq) + H20 (I)
Transcribed Image Text:7. When 40.0 ml of 0.200 M HCI at 21.5°C is added to 40.0 ml of 0.200 M NaOH also at 21.5°C in a coffee-cup calorimeter, the temperature of the resulting solution rises to 22.8°C. Assume that the volumes are additive, the specific heat of the solution is 4.180 J/g°C, and that the density of the solution is 1.00 g/mL. Calculate the enthalpy change, AH, for the reaction: HCI (aq) + NaOH (aq) → NaCl (aq) + H20 (I)
Expert Solution
Step 1

The reaction taking place in calorimeter :

HCl(aq)+NaOH(aq)NaCl(aq)+H2O(I)

Volume of HCl=40.ml

volume of NaOH=40 ml

total volume in calorimeter=Volume of HCl+Volume of NaOH=(40+40)ml=80 ml

density of the solution=1.0 g/ml

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