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 25.00°C. Assuming
that the reaction goes to completion, and that the resulting solution has a mass of 100.00 g and a heat
capacity the same as water, calculate the final temperature of the solution.
Ca2+(aq) + 2 F-(aq) → CaF2(s)
ΔH° = –11.5 kJ

When trying to solve this I get an answer of 26.1 degrees C.

I am basically doing T2 = 11500(.02)/(100x2.09) - 25 = 26.1. However I know this to be the wrong answer. Can someone help me see my error?

Answer 1

Answer – We are given, 50.0 mL of 0.400 M Ca(NO3)2 is added to 50.0 mL of 0.800 M NaF , ti = 25.0^oC , mass = 100 g

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

First we need to calculate the limiting reactant

Moles of Ca(NO₃)₂ = 0.400 M * 0.050 L

                                 = 0.02 moles

Moles of NaF = 0.800 M * 0.050 L

                        = 0.040 moles

Moles of CaF₂(s) from Ca(NO₃)₂

1 moles of Ca(NO₃)₂ = 1 moles of CaF₂(s)

So, 0.02 moles = ?

= 0.020 moles of CaF₂(s)

Moles of CaF₂(s) from NaF

2 moles of NaF = 1 moles of CaF₂(s)

So, 0.04 moles NaF= ?

= 0.020 moles of CaF₂(s)

So both are limiting reactant and moles of CaF₂(s) = 0.020 moles

So, form 1 mole of CaF₂(s) = -11.5 kJ

So, 0.020 moles of CaF₂(s) = ?

= -0.23 kJ

So, q = 0.23 kJ = 230 J

We know, q = m*C*∆t

230 J = 100 g * 4.184 J/g^oC *(tf -25.0)

230 J = 418.4tf – 10460

So, 418.4 tf = 230+10460

tf = 25.5^oC