Why did the HCl and NaOH go to therapy? They had neutralization issues!
Alright, let's crunch some numbers and find the molar heat of neutralization.
First, we need to calculate the mass of the final solution. Since the density of the solution is 1.05 g/mL, and the total volume of the solution is 100 mL, we have:
Mass of solution = Volume of solution × Density
Mass of solution = 100 mL × 1.05 g/mL
Mass of solution = 105 g
Next, let's find the heat absorbed by the solution using the formula:
q = m × C × ΔT
Where:
q = heat absorbed (in J)
m = mass of the solution (in g)
C = specific heat of the solution (in J/g°C)
ΔT = change in temperature (in °C)
Since the specific heat of the solution is 4.184 J/g°C, and the change in temperature is 29.8°C - 24.1°C = 5.7°C, we have:
q = 105 g × 4.184 J/g°C × 5.7°C
q = 2547.07 J
Now, let's find the heat absorbed by the calorimeter and the reaction:
q_total = q_solution + q_calorimeter
Since the heat capacity of the calorimeter is 32.5 J/°C, we have:
q_total = 2547.07 J + 32.5 J/°C × 5.7°C
q_total = 2547.07 J + 185.25 J
q_total = 2732.32 J
The molar heat of neutralization is the heat evolved per mole of HCl reacting. Since the number of moles of HCl and NaOH are equal (due to stoichiometry), we use the moles of HCl in the calculation.
To find the moles of HCl, we use the formula:
moles = Molarity × Volume (in L)
Since the initial volume of HCl is 0.050 L (50 mL = 0.050 L) and the molarity is 1.2 M, we have:
moles HCl = 1.2 M × 0.050 L
moles HCl = 0.06 mol
Finally, to find the molar heat of neutralization (ΔH_rxn) in J/mol, we divide the total heat absorbed by the moles of HCl:
ΔH_rxn = q_total / moles HCl
ΔH_rxn = 2732.32 J / 0.06 mol
ΔH_rxn ≈ 45,538.66 J/mol
So, the molar heat of neutralization is approximately 45,538.66 J/mol.