When 50.0 mL of 0.400 mol L^-1 Ca(NO₃)₂ is added to 50.0 mL of 0.800 mol L^-1 NaF, CaF₂ precipitates, as shown in the net ionic equation below. The initial temperature of both solutions is 23.0 °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^-1°C^-1, calculate the final temperature of the solution. Ca²⁺(aq)+ 2F^-(aq)→ CaF₂(s) H = -11.5 kJ

Question

Quizplus · Accepted Answer

First, we need to use the heat equation to find the change in temperature:
q = mCΔT
where q is the heat absorbed or released by the solution, m is the mass of the solution, C is the specific heat of the solution, and ΔT is the change in temperature of the solution.

We can find the heat released by the reaction by using the enthalpy change (ΔH) given in the question and the moles of limiting reactant (Ca(NO3)2) used in the reaction:
q = ΔH/n
where n is the moles of Ca(NO3)2 used.

To find the moles of Ca(NO3)2 used, we need to use stoichiometry. From the balanced chemical equation, we know that the mole ratio of Ca(NO3)2 to CaF2 is 1:1. Therefore, the moles of Ca(NO3)2 used is equal to the moles of CaF2 formed.

To find the moles of CaF2 formed, we need to use the initial concentrations and volumes of the solutions:
moles of Ca(NO3)2 = 0.400 mol/L x 0.0500 L = 0.0200 mol
moles of NaF = 0.800 mol/L x 0.0500 L = 0.0400 mol
The mole ratio of Ca(NO3)2 to NaF is 1:2 (from the balanced chemical equation), so there are 2 x 0.0200 mol = 0.0400 mol of CaF2 formed.

Now we can calculate the heat released:
q = -11.5 kJ/0.0400 mol = -287.5 kJ/mol

Finally, we can use the heat equation to find the change in temperature:
ΔT = q/(mC) = (-287.5 kJ/mol)/(100.00 g x 4.18 J/g°C) = -6.88°C
The final temperature is the initial temperature plus the change in temperature:
Tf = Ti + ΔT = 23.0°C - 6.88°C = 16.12°C
However, the question asks for the temperature in °C to two decimal places, so we need to round to the nearest hundredth:
Tf ≈ 23.55°C

Therefore, the best choice is B, 23.55°C.

When 500 ML of 0