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 netionic equation below. The initial temperature of both solutions is 23.0°C. Assuming that the reaction goes tocompletion, and that the resulting solution has a mass of 100.00 g and a specific heat of 4.18 J/(g ∙°C), calculatethe final temperature of the solution. Ca2+(aq) + 2 F-(aq) → CaF2(s)∆H°=-11.5 kJ

23.55°C

Explanation:

Based on the equation:

Ca²⁺(aq) + 2F⁻(aq) → CaF₂(s) ∆H°=-11.5 kJ

When 1 mole of Ca²⁺ and 2 of F⁻ reacts, 11.5kJ are released.

Thus, we need to find moles of reaction to find the heat released and using:

C = SₓmₓΔT

We can find the final temperature as follows:

Moles of reaction:

0.0500L * (0.400mol / L) = 0.0200moles Ca²⁺ = Moles of reaction

Heat produced is:

0.0200 moles * (11.5kJ / mol) = 0.23kJ

Using:

C = SₓmₓΔT

Where C is heat = 230J

S is specific heat = 4.18J/g

m is mass of solution = 100.00g

And ΔT is change in temperature

230J = 4.18J/gₓ100.00gₓΔT

ΔT = 0.55°C

As initial temperature is 23.0°C

Final temperature = 23.0°C + 0.55°C =

the answer is 303.15 or 303