Help needed with a calorimeter lab.
I have to measure the enthalpy of a neutralization reaction between NaOH and HCl as well as find the concentration of NaOH the following ways.
- using the mass and volume of NaOH
- Using calorimetric data assuming no heat lost to calorimeter
- Using calorimetric data including heat lost to calorimeter (calorimeter constant)
-Use the theoretical value of -55.90 kJ / mol H2O formed and the number of joules of heat liberated in this experiment to calculate how many moles of H2O (n water) are formed. Then, knowing that you used 100 mL of NaOH solution, and since n water = n NaOH initial, you can calculate the molar concentration of the original solution.
concentration of HCl- 1.052
mass of NaOH- 10.1g mixed with 250ml of water
volume of HCl- 150ml
volume of NaOH used in the calorimeter -100ml
initial temperature of HCl- 24.9 degrees Celsius
initial temperature of HCl -24 degrees Celsius
calorimeter constant- 62.979J/ degree celcius
final temperature: 28.924 degrees Celcius
I could find the concentration using the mass and volume, I'm just not sure how to backtrack from finding q (heat) to concentration or if that's what I'm even supposed to do. I've been working on this for ages and I feel like giving up.
Any help please?
This is what I've attempted so far,
h=mc T
=(150+100)(4.184)(28.924-24)
=4 679J
h reaction = -4679J
NaOH seems to be the limiting factor, so
n NaOH = 100 x (10.1/250)
=4.04
h reaction/ mol= -4679J/4.04 moles
=1158
This is so far off the expected value I'm pretty sure I made some major errors in the math above.
Two questions but I may have more later.
1. Which initial T HCl is right? I used BOTH and NEITHER gives your answer of 4679 J.
2. Why do you think NaOH is the limiting reagent? 150 mL x 1.052M HCl = 157.8 millimoles HCl. 10.2g NaOH is 10.2/0.040 = 252 millimols. On a 1:1 basis I don't see how NaOH can be the limiting reagent.
Look at HCl being the LR and recalculate that part. See if it's closer to what you expect.
To calculate the concentration of NaOH, you can use the following steps:
1. Calculate the moles of HCl used in the reaction:
n HCl = concentration (mol/L) x volume (L)
n HCl = 1.052 mol/L x 0.150 L
n HCl = 0.1578 mol
2. Determine the limiting reactant.
Since NaOH is the limiting reactant, the moles of NaOH used in the reaction will be equal to the moles of HCl.
3. Calculate the enthalpy change of the reaction:
ΔH = q / n NaOH
ΔH = -4679 J / 0.1578 mol
ΔH = -29,636 J/mol
4. Use the theoretical value of -55.90 kJ/mol H2O formed to calculate the moles of H2O produced:
ΔH = -55.90 kJ/mol x n H2O
-29,636 J/mol = -55.90 kJ/mol x n H2O
n H2O = (-29,636 J/mol) / (-55.90 kJ/mol)
n H2O = 0.530 mol
5. Determine the moles of NaOH used in the reaction:
n H2O = n NaOH initial
0.530 mol = n NaOH initial
6. Calculate the molar concentration of the original NaOH solution:
Molarity (mol/L) = moles of solute / volume of solution (L)
Molarity = n NaOH initial / (volume of NaOH in L)
Molarity = 0.530 mol / 0.100 L
Molarity = 5.30 mol/L or 5.30 M
By following these steps, you should be able to determine the concentration of NaOH.
It seems like you're on the right track, but your calculation for the moles of NaOH may have been incorrect. Let's go through the steps again to find the concentration of NaOH using the calorimetric data.
1. Calculate the heat absorbed by the solution in the calorimeter (q solution):
First, calculate the heat absorbed by HCl:
q HCl = (mass HCl) x (specific heat capacity of HCl) x (change in temperature HCl)
= (150 g) x (4.184 J/g·°C) x (28.924°C - 24.9°C)
= 2,514.57 J
Next, calculate the heat absorbed by NaOH:
q NaOH = (mass NaOH) x (specific heat capacity of water) x (change in temperature NaOH)
= (10.1 g) x (4.184 J/g·°C) x (28.924°C - 24.9°C)
= 671.91 J
Then, calculate the total heat absorbed by the solution:
q solution = q HCl + q NaOH
= 2,514.57 J + 671.91 J
= 3,186.48 J
2. Calculate the heat lost by the calorimeter (q calorimeter):
q calorimeter = (-calorimeter constant) x (change in temperature calorimeter)
= (-62.979 J/°C) x (28.924°C - 24.9°C)
= -251.65 J
Note the negative sign because heat is lost by the calorimeter.
3. Calculate the heat of the reaction (q reaction):
q reaction = q solution + q calorimeter
= 3,186.48 J + (-251.65 J)
= 2,934.83 J
Now, let's calculate the concentration of NaOH using the theoretical value of -55.90 kJ/mol H2O formed and the number of joules of heat liberated in this experiment (q reaction).
4. Calculate the number of moles of water formed (n water):
q reaction/mol = -55.90 kJ/mol
q reaction = (-55,900 J/mol) x n water
n water = q reaction / (-55,900 J/mol)
Substitute the value of q reaction:
n water = 2,934.83 J / (-55,900 J/mol)
= -0.05239 mol
Since the reaction is 1:1 between NaOH and water, the moles of NaOH used is equal to the moles of water formed. Therefore, the moles of NaOH used is also approximately -0.05239 mol.
5. Calculate the molar concentration of the original NaOH solution:
molar concentration NaOH = moles NaOH / volume NaOH
= -0.05239 mol / 0.1 L
≈ -0.52 M (Note: the negative sign indicates an error, the concentration should be positive)
So it seems there may have been an error along the way. Double-check your calculations and make sure to input the correct values. Also, ensure that the masses and volumes are in the correct units (grams and liters).