Question

When 50.0 mL of 0.400 M Ca(NO3)2 is added to 50.0 mL of 0.800 M NaF, CaF2 precipitates, as shown in the net ionic equation below. The initial temperature of both solutions is 30.00°C. Assuming that the reaction goes to completion, and that the resulting solution has a mass of 100.00 g and a specific heat of 4.18 J/(g ∙ °C), calculate the final temperature of the solution.

            Ca2+(aq) + 2 F-(aq) → CaF2(s)   ΔH° = -11.5 kJ

30.55°C

If there is additional information about the heat capacity of the calorimeter (58.3 J/°C) in the above questions, how would you approach your answer?

Answer 1

Given reaction is

Initial temperature= 30 degree Celsius

Moles of Calcium ions available= Moles of Calcium nitrate=Molarity x Volume=50mL x 0.4M= 20mmol.

Moles of Flouride ion = Moles of NaF= 50mL x 0.8M=40 mmol

According to stoichiometry when we compare both the reactants, Both get completely consumed in the reaction. So

Moles of CaF₂ formed= Moles of Ca²⁺= 20mmol.

Total energy liberated =

Now Mass of the solution formed= 100g

Specific heat capacity= 4.18J/

From Q= mC

Change in Temperature, = =^oC

So final temperature is = 30.55^oC

If there is additional heat capacity of Calorimeter, then it has to be taken into consideration.

Heat given out by reaction= Heat absorbed by calorimeter+ Heat absorbed by Solution

230J= ( Heat capacity of Calorimeter+ mxSpecific heat capacity of solution)

=230/(58.3+100*4.18)=0.483^oC

Final Temperature would be =30.48^oC