50.0 mL of a solution of HCl is combined with 100.0 mL of 1.15 M NaOH in a calorimeter. The reaction mixture is initially at 22.4°C and the final temperature after reaction is 31.2°C. What is the molarity of the HCl solution? You may assume that there is an excess of base (so all of the HCl has reacted), that the specific heat of the reaction mixture is 0.96 cal/g°C, and that the density of the reaction mixture is 1.02 mg/mL. The heat of neutralization of HCl and NaOH is 13.6 kcal/mol.

How do I set this problem up?

HCl + NaOH ==> NaCl + H2O

q = heat generated = mass solution x specific heat solution x (Tfinal-Tinitial)
mass solution is 150 mL x 1.02 g/mL(NOTE: I think this is a typo and should be 1.02 g/mL and not 1.02 mg/mL)
That's q for how many mols?
That's 13.6 kcal/mol x #mols = q
Solve for # mols.
Then M = mols/L. You know mols and you know L (0.05), solve for M.

To find the molarity of the HCl solution, we need to calculate the heat produced during the reaction using the equation:

q = mass × specific heat × ΔT

Where:
q = heat produced (in calories)
mass = mass of the reaction mixture (in grams)
specific heat = specific heat of the reaction mixture (in cal/g°C)
ΔT = change in temperature (in °C)

First, let's calculate the mass of the reaction mixture. We'll use the given density of the reaction mixture:

density = mass/volume

mass = density × volume

mass = 1.02 mg/mL × (50.0 mL + 100.0 mL)

mass = 1.02 mg/mL × 150.0 mL

Since we're working with mL, we'll convert mg to g:

mass = 1.02 g/mL × 150.0 mL

mass = 153.0 g

Now, let's calculate the heat produced:

q = 153.0 g × 0.96 cal/g°C × (31.2°C - 22.4°C)

q = 153.0 g × 0.96 cal/g°C × 8.8°C

q ≈ 1335.86 cal

We can convert calories to kilocalories:

q ≈ 1335.86 cal ÷ 1000 cal/kcal

q ≈ 1.34 kcal

Next, knowing that the reaction is HCl + NaOH → NaCl + H2O, we can use the heat of neutralization of HCl and NaOH to relate the heat produced to the moles of HCl reacted.

The heat of neutralization is the heat released when one mole of a substance reacts with another:

13.6 kcal/mol = q ÷ moles of HCl

moles of HCl = q ÷ 13.6 kcal/mol

moles of HCl = 1.34 kcal ÷ 13.6 kcal/mol

moles of HCl ≈ 0.0985 mol

Finally, we can calculate the molarity of the HCl solution using the equation:

Molarity = moles of solute / volume of solution

Molarity = 0.0985 mol / 50.0 mL

Since we're working with mL, we'll convert to liters:

Molarity = 0.0985 mol / 0.0500 L

Molarity ≈ 1.97 M

Therefore, the molarity of the HCl solution is approximately 1.97 M.

To find the molarity of the HCl solution, we need to use the principles of calorimetry and the heat of neutralization. Calorimetry involves measuring the heat exchanged between the chemical reaction and its surroundings.

First, we need to determine the amount of heat released or absorbed during the reaction. This can be calculated using the formula:

q = m * c * ΔT

Where:
q is the amount of heat (cal),
m is the mass of the solution (g),
c is the specific heat of the solution (cal/g°C), and
ΔT is the change in temperature (°C).

We need to convert the volumes to mass using the density of the solution:

mass_HCl = volume_HCl * density_solution
mass_NaOH = volume_NaOH * density_solution

Next, we can find the heat released or absorbed during the reaction:

q_rxn = -q_solution

Given that the heat of neutralization is 13.6 kcal/mol, we can calculate the moles of NaOH used in the reaction:

moles_NaOH = Molarity_NaOH * volume_NaOH

Since the reaction is 1:1 between HCl and NaOH, the moles of HCl used will be the same as the moles of NaOH used.

Using the stoichiometry of the reaction, we can calculate the moles of HCl initially present:

moles_HCl = moles_NaOH

Finally, we can determine the molarity of the HCl solution:

Molarity_HCl = moles_HCl / volume_HCl

Substituting the known values into the equations above will allow us to find the molarity of the HCl solution.