[meme]

Hi!
Wow, I love the new look of the forums - awesome! Anyway, I have some questions on calorimetry which I've tried, but I'm not sure I'm doing it right. I'd appreciate any help.

We did a lab on the heat of neutralization for an acid-base reaction. Anyway, these were my results:

• volume of acid (HCl) - 50mL
• temperature of acid - 20.9 C
• volume of NaOH - 50mL
• temperature of NaOH - 20.2 C
• exact molar concentration of NaOH - 1.0 mol/L
• maximum temperature from graph (we had to make a graph of our results) - 27.5 C

Calculations:

• Average initial temperature of acid + base: (20.2 + 20.9)/2 = 20.6
• Temperature change: 27.5 - 20.6 = 6.9
• Volume of final mixture: 50 + 50 = 100mL
• Mass of final mixture (assume density of solution is 1g/mL): (1)(100) = 100g
• specific heat of mixture(given): 4.18J/g C

Here's the part I'm not sure of:

• Heat evolved: (4.18)(100g)(6.9) = 2884.2 J
• Amount of OH^- reacted:
    HCl + NaOH --> NaCl + H₂O
    100g NaCl + H₂O  x  1 mol     x  1 mol NaOH  x  1 mol OH^-
                                    76.46g  x  1 mol            x  1 mol NaOH
     = 1.31 mol OH^- ??
  
• Amount of H₂O formed:
    1.31 mol NaOH  x  1 mol H₂O
                                1 mol NaOH
        = 1.31 mol H₂O ??
  
• Heat evolved per mole of H₂O, (heat of neutralization)
    heat of neutralization = - Cp(H₂O) x combined masses (acid + base) x change in temperature
    = - (4.18)(100g)(6.9) = -2884.2 J

I know there's something fishy going on there, please *delete me*


Other Questions:

1. A thermometer labeled as being miscalibrated at the factory – the freezing point of water reaches 1.2 ºC and the boiling point of water reads 101.2ºC at standard pressure. But, because you are in a hurry and the thermometer is the last one available, you decide to go ahead and use it in the experiment, recording the temperatures as they appear on the thermometer. How ill the use of this thermometer affect the determination of the specific heat of the metal?
So does it just affect the temperature by 1.2ºC? or does it affect it by something else? Then again, it probably wouldn't affect it at all, because change in temperature would still be the same, right?

2. The calorimeter, although a good insulator, absorbs some heat when the system is above room temperature. The consequence of this heat loss from the system is accounted for in the temperature extrapolation f the data. Describe an experiment that could be used to calibrate the calorimeter by measuring this heat loss.
I have NO idea about this one... hints, please?

Thanks again for all the help. I love you guys!

[Mitch]

You did sime hocus-pocus with those calculations. Did you use solid NaOH or pure HCl for this experiment? I'm assuming you didn't count in the dilution factor. I'm still looking over your numbers to see where it might of gone wrong.



1.You got it. Good answer

2. You basically just add boiling water to the calorimeter and see how the temperature goes down with time. And just calculate it's specific heat.

[Mitch]

do you know the molarity of the HCl or the NaOH?

[Donaldson Tan]

Hmm..

Heat evolved is correct

Amount of OH^- reacted:
I have no idea how u arrive at 100g NaCl. However, total volume of reaction mixture is 100ml and its density is assumed at 1g/ml. By Law of Mass Combination, there's no way NaCl present can be of 100g  

Heat of Neutralisation is expressed in J.mol^-1 with reference to water. U must divide the heat evolved by the amount of water formed by neutralisation.

[meme]

Wow.. you guys answered really fast! Anyway, here are the answers to your questions about my questions (hehe)

Mitch -

The HCl and the NaOH were diluted.
HCl - 1.1 M
NaOH - 1.0 M

Geodome:

Sorry about the typo, I actually checked everything a million times to take off any mistakes.. Guess I messed up. I had meant to put 100 g of NaOH + HCl (sorry again!) So are the numbers right, though? Is it 1.31 mol both H₂O and OH^-?

And about the heat of neutralization.. so I don't have to use the equation I wrote up there? So, basically, the answer would be heat evolved/mol of H₂O:
 2884.2 J/ 1.31 = 2201.7 J/mol = 2.20 kJ/mol
Is that right?

[Mitch]

Okay this might help out a lot then. You didn't use 100grams of NaOH, but you uses a 100 grams of a solution of NaOH. Use M=moles/volume, and you get that you used .1 moles of NaOH. Use that to solve for grams of NaOH used and use that grams to find out how many Joules were evolved. Let me know what you get, and if it makes more sense.

[Donaldson Tan]

I still dont understand why u use 100g NaOH. your reaction involves 50ml of NaOH solution and 50ml of HCl solution, both of known concentration. You can work from there directly

From the concentration of NaOH and HCl, it's obvious that HCl is in excess since they react according to a 1:1 molar ratio  

[meme]

no, I guess I didn't word it right

I meant to say that it's a 100 g of the NaOH PLUS the HCl. Together they make 100g. I got this because to make the solution we used 50mL of HCl and 50mL of NaOH, which equals 100mL of solution. Then, our lab book tells us that the density of the solution is 1.00 g/mL... so I just used m = dv, and got (1)(100) = 100g of solution. Sorry again!

[Mitch]

So you only used 50 mls of the NaOH solution. Then you used .05moles of NaOH. start calculating from there and let me know what you get.

[Donaldson Tan]

Amount of NaOH reacted
= Amount of NaOH used
= 50/1000 X 1.0
= 0.050mol

Amount of HCl used
= 50/1000 X 1.1
= 0.055mol [HCl in excess, given 1:1 molar ratio for rxn]

Amount of H₂O formed from Neutralisation
= 0.050mol [given 1:1 molar ratio with NaOH reacted]

[meme]

Sorry for replying back late, I just got out of class.

Okay.. so I see how you got .05 mol NaOH. Our book already tells us that the OH^- is the limiting reactant, so I really didn't need to figure that out.. okay.. so since it is a 1:1 ratio, it would be .05 mols of H₂O...

I'm guessing since everything is in a 1:1 ratio, the amount of OH^- reacted is also the same, right?

Okay.. so now for the heat of neutralization: Geodome said to divide the heat evolved by the amount of water. Soooooo... it's 2884.2/.05
My answer for the heat evolved per mole of water (heat of neutralization) is 57684 J/mol = 57.7 kJ/mol.... right?

[Mitch]

Why don't you try doing it my way?

[meme]

lol, okay...

Use M=moles/volume, and you get that you used .05 moles of NaOH. Use that to solve for grams of NaOH used and use that grams to find out how many Joules were evolved. Let me know what you get, and if it makes more sense.

.05 mol NaOH x (38.99g/1mol) = 1.95g NaOH
heat evolved: (4.18)(1.95g)(6.9C) = 56.2J

so, now what? divide it by mols of water?

[Mitch]

Don't divide, your done. That's the amout of joules released.

[meme]

great! okay.. so how do I figure out the heat of neutralization though?