# Calorimetry homework help

1. Jan 26, 2008

### Mitchtwitchita

can anybody here help me out with this problem?

A quantity of 4.00 x 10^2 mL of 0.600 M HNO3 is mixed with 4.00 x 10^2 mL of 0.300 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of both solutions is the same at 18.46 degrees C. What is the final temperature of the solution? (Use the result in Example 6.8 for your calculation).

Results for previous example: qrxn = -2.81 KJ.

qrxn + qsoln = 0
qrxn = -qsoln
therefore, qsoln = 2.81

qsoln = ms(deltaT)
2.81 = (400 g + 400 g)(4.184 KJ/g * degrees C)(x - 18.46 degrees C)
x = [2.81/(800 g)(4.184 KJ/g * degrees C)] + 18.46 degrees C
=18.46 degrees C?

Can anybody please show me where I'm going wrong?

2. Jan 26, 2008

### rocomath

How did you get 400g?

Please be more clear with your units. Which information refers to Nitric acid or Barium Hydroxide?

3. Jan 26, 2008

### Mitchtwitchita

I thought the 400g was in relation to the 400 mL of nitric acid and that the 400 g was in relation to the barium hydroxide.

4. Jan 26, 2008

### rocomath

No that's the volume, you want the mass, in grams. So solve for grams of each and we'll go from there.

5. Jan 26, 2008

### Mitchtwitchita

I think HNO3 = 0.400 L x 0.600 mol/1L
=0.24 mol x 63.02 g/1 mol
=15.12 g HNO3

and Ba(OH)3 = 0.400 L x 0.300 mol/1 L
=0.12 mol x 188.3 g/1 mol
=22.60 g Ba(OH)3

15.12g + 22.60g = 37.72g solution?

6. Jan 26, 2008

### rocomath

$$-q_{rxn}=q_{soln}$$

$$2.81kJ=m_{a}c_{a}\Delta{T}_{a}+m_{b}c_{b}\Delta{T}_{b}$$

$$\Delta{T}_a=\Delta{T}_b=\Delta{T}$$

$$2.81kJ=\Delta{T}(m_{a}c_{a}+m_{b}c_{b})$$

a = Nitric acid
b = Barium hydroxide

You can't just add their weights because they have different specific heats. Are you able to find their specific heats?

Last edited: Jan 26, 2008
7. Jan 26, 2008

### Mitchtwitchita

Yeah, the assumption is that the specific heats are the same as water.

8. Jan 26, 2008

### rocomath

Oh ok. I was wondering why you were just straight up adding them from your original post.

Anyways, yeah the equation simplifies nicely. Just do what you did originally using the actual mass.

$$2.81\times 10^3 J=m_{soln}c\Delta T$$

Last edited: Jan 26, 2008
9. Jan 26, 2008

### Mitchtwitchita

2810 J = (37.7 g)(4.184 J/g * degrees C)(x - 18.46 degrees C)
x = [2810 J/(37.7 g)(4.184 J/g * degrees C)] + 18.46 degrees C
=36.27 degrees C

However, this still doesn't match the answer at the back of the book (22.49 degrees C). Is there something wrong with my math?

10. Jan 26, 2008

### rocomath

I don't know what else to tell you, I got the same thing. Sorry, you'll have to wait for someone else to clean up my mess :p I'll try one more time to see if I can figure it out.

11. Jan 26, 2008

### rocomath

Oh, I think I got it. Let me finish working it out b4 I type it up.

12. Jan 26, 2008

### Mitchtwitchita

Thanks a lot rocophysics! you've been a great help. I can't see any error that I could be making now. Maybe the book has a typo which tends to happen.

13. Jan 26, 2008

### Mitchtwitchita

Oh my god, the previous question also says that the heat of neutralization is -56.2 KJ/mol. But that still doesn't help me out because I don't know how to apply it or if it even pertains to this question.

14. Jan 26, 2008

### rocomath

15. Jan 26, 2008

### Mitchtwitchita

A quantity of 4.00 x 10^2 mL of 0.600 M HNO3 is mixed with 4.00 x 10^2 mL of 0.300 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of both solutions is the same at 18.46 degrees C. What is the final temperature of the solution? (Use the result in Example 6.8 for your calculation).

Results for previous example: qrxn = -2.81 KJ.

qrxn + qsoln = 0
qrxn = -qsoln
therefore, qsoln = 2.81

HNO3 = 0.400 L x 0.600 mol/1L
=0.24 mol x 63.02 g/1 mol
=15.12 g HNO3

and Ba(OH)3 = 0.400 L x 0.300 mol/1 L
=0.12 mol x 188.3 g/1 mol
=22.60 g Ba(OH)3

15.12g + 22.60g = 37.72g solution?

2810 J = (37.7 g)(4.184 J/g * degrees C)(x - 18.46 degrees C)
x = [2810 J/(37.7 g)(4.184 J/g * degrees C)] + 18.46 degrees C
=36.27 degrees C

The final answer is 22.49 degrees C
and the heat of neutralization is -56.2 KJ/mol