Question

In the lab you mix 50.0 mL of 0.250 M Ca(NO3)2 with 50.0 mL of 0.500 M NaF in a coffee cup calorimeter to form a CaF2 precipitate. The initial temperature of each solution is 23 degrees celsius. Assuming that the final solution has a total mass of 100.0 g and a specific heat of 4.18 J/g degrees celsius, calculate the final temperature you expect for the solution. Assume no heat is lost to the calorimeter.

Ca2+ (aq) + 2F-(aq) --> CaF2 (s)
∆H° = -115 kJ/mol

Answer 1

In the lab you mix 50.0 mL of 0.250 M Ca(NO3)2 with 50.0 mL of 0.500 M NaF in a coffee cup calorimeter to form a CaF2 precipitate. The initial temperature of each solution is 23 degrees celsius. Assuming that the final solution has a total mass of 100.0 g and a specific heat of 4.18 J/g degrees celsius, calculate the final temperature you expect for the solution. Assume no heat is lost to the calorimeter.

$$\mathrm{Ca}^{2+}(\mathrm{aq}) + 2\mathrm{F-}(\mathrm{aq}) \rightarrow \mathrm{CaF_2}(s)$$ $$\Delta H^{\circ} = -115\ \mathrm{kJ/mol}$$

Solution:

First you need to determine which reactant is the limiting one. That is, which reactant will run out first. It will determine how much heat is given off.

moles $\mathrm{Ca}^{2+} = \mathrm{M}\ \mathrm{Ca}^{2+} \times \mathrm{L}\ \mathrm{Ca}^{2+} = (0.400)(0.0500) = 0.0200$ moles $\mathrm{Ca}^{2+}$

moles $\mathrm{F-} = \mathrm{M}\ \mathrm{F-} \times \mathrm{L}\ \mathrm{F-} = (0.800)(0.0500) = 0.0400$ moles $\mathrm{F-}$

The balanced equation tells us that it takes 2 moles of F- to react with 1 mole of $\mathrm{Ca}^{2+}$, and that's exactly what we have: 0.0400 moles F- / 0.0200 moles $\mathrm{Ca}^{2+} = \frac{2}{1}$. So both reactants will run out at the same time.

The equation also tells us that 1 mole of $\mathrm{Ca}^{2+}$ (or 2 moles of F-) will produce -11.5 kJ of heat. So how much heat will 0.0200 moles of $\mathrm{Ca}^{2+}$ produce?

0.0200 moles $\mathrm{Ca}^{2+} \times (-11.5\ \mathrm{kJ}\ \mathrm{heat} / 1\ \mathrm{mole}\ \mathrm{Ca}^{2+}) = 0.230\ \mathrm{kJ}\ \mathrm{heat} = 230\ \mathrm{J}\ \mathrm{heat}$

This amount of heat was absorbed by the water, causing the water temperature to increase.

Heat gained by water = (mass H2O)(specific heat H2O)(Tf - Ti)

$$230\ \mathrm{J} = (100\ \mathrm{g}\ \mathrm{H_2O})(4.18\ \mathrm{J/g}\ \mathrm{C})(T_f - 23.0)$$ $$230 = 418\ \mathrm{Tf} - 9614$$ $$9844 = 418\ \mathrm{Tf}$$ $$T_f = 23.55\ \mathrm{C}.$$

Answer: Tf = 23.55 C.

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