When 50.0 mL of 0.400 M Ca(NO3)2 is added to 50.0 mL of 0.800 M NaF, CaF2 precip

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)   H° = -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?

Explanation / Answer

Initial concentrations:
50.0 mL of 0.400 M Ca(NO3)2 is added to 50.0 mL of 0.800 M NaF
[Ca2+] = 50/1000*0.4 = 0.02 moles
[F-] = 50/1000*0.8 = 0.04 moles
CaF2 =0

Final concentrations:
Ca2+(aq) + 2 F-(aq) -> CaF2(s);   DH° = -11.5 kJ
[Ca2+] = 0.02 -0.02 =0moles
[F-] = 0.04 -2*0.02 = 0moles
CaF2 = 0.02 mole

Enthalpy of neutralization, DH° = -11.5 kJ/mole of Ca+2
Total heat released, q = -DH°*moles of Ca+2 = 11.5*0.02 = 0.23 kJ = 230 J

mass, m = 100g
specific heat, cp = 4.18 J/(g · °C)
Initial temperature, Ti = 30.0C
Final temperature = Tf
Heat released, q = m*cp*(Tf-Ti) = 230 J
Final temperature, Tf = 230/(m*cp)+Ti = 230/(100*4.18) + 30 = 30.55 C

Calculation with heat capacity of the calorimeter (58.3 J/°C)
cpcal = 58.3 J/°C
Heat released, q = cpcal*(Tf-Ti)+m*cp*(Tf-Ti) = 230 J
Final temperature, Tf = 230/(m*cp+cpcal)+Ti = 230/(100*4.18+58.3) + 30 = 30.48 C

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