A 50.0 mL solution of 1.2 M HCl at 24.1C is mixed with 50.0 mL of 1.3 M NaOH, also at 24.1C, in a coffee-cup calorimeter. After the reaction occurs, the temperature of the resulting mixture is 29.8C. The density of the final solution is 1.05 g/mL. Calculate the molar heat of neutralization. Assume the specific heat of the solution is 4.184 J/gC. The heat capacity of the calorimeter is 32.5 J/C

So you reacted 1.2M x 0.050L = 0.06 mols HCl with 0.06 mols NaOH in 100 mL solution.

mass solution = 100 mL x 1.05 g/mL = 105 grams.
delta T = dT = 29.8-24.1 = 5.7 degrees.
q = (mass soln x specific heat x dT) + (Ccal x dT)
q = (105 x 4.184 x 5.7) + (32.5 x 5.7) = ? J for 0.06 mols H2O formed.
Convert to J/1 mol and change that to kJ/mol.

A 50.0 mL solution of 1.2 M HCl at 24.1°C is mixed with 50.0 mL of 1.3 M NaOH, also at 24.1°C, in a coffee-cup calorimeter. After the reaction occurs, the temperature of the resulting mixture is 29.8°C. The density of the final solution is 1.05 g/mL. Calculate the molar heat of neutralization. Assume the specific heat of the solution is 4.184 J/g·°C. The heat capacity of the calorimeter is 32.5 J/°C.

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.

To calculate the molar heat of neutralization, we need to consider the heat exchanged during the reaction.

First, we need to find the total heat absorbed or released by the solution. We can use the formula:

q = m * c * ΔT

Where:
q = heat (in Joules)
m = mass of the solution (in grams)
c = specific heat capacity of the solution (in J/g°C)
ΔT = change in temperature (in °C)

To find the mass of the solution, we need to calculate the sum of the masses of the HCl and NaOH solutions. First, convert the volume of each solution to mass using the density of the final solution:

Mass of HCl solution = 50.0 mL * 1.05 g/mL = 52.5 g
Mass of NaOH solution = 50.0 mL * 1.05 g/mL = 52.5 g

Now, let's calculate the heat absorbed or released by the reaction using the first equation:

q = (mHCl + mNaOH) * c * ΔT
= (52.5 g + 52.5 g) * 4.184 J/g°C * (29.8°C - 24.1°C)

Next, we need to calculate the moles of HCl and NaOH used in the reaction. We can use the molarity (M) and volume (V) of each solution:

moles HCl = M HCl * V HCl
= 1.2 mol/L * 0.050 L
= 0.060 mol HCl

moles NaOH = M NaOH * V NaOH
= 1.3 mol/L * 0.050 L
= 0.065 mol NaOH

Since HCl and NaOH react in a 1:1 ratio, the moles of HCl used is equal to the moles of NaOH used.

Now, we can calculate the molar heat of neutralization:

Molar heat of neutralization = q / (moles HCl + moles NaOH)
= q / (0.060 mol + 0.065 mol)

2.69kJ/mol