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# Thermometric Titration watch

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1. I am doing the lab to calculate the enthalpy of neutralization and calculate the concentration of base
I know very well about sources of error and theories, but I am not sure how to calculate the enthalpy of neutralization.

Can you tell me where I go wrong? or how to do it? Thank you!

Lab: Titrating sodium hydroxide with hydrogen chloride, measuring temperature change during this process.

2 mol/dm^3 HCl
50cm^3 NaOH
( I added 5cm^3 of HCl each time to wait for temperature to stabilize, and measured temperature change

For concentration of base, i used c=n/v

and C(a)V(a)=C(b)V(b) (acid, and base)

then calculated the volume of acid.

now when using Q=mcT

I THINK THAT MASS, M IS THE TOTAL OF BASE AND ACID..

BUT NOW Q IS NOT PER 1 MOLE.

HOW DO I GET THE NUMBER OF MOLES? IT IS THE NUMBER OF MOLES OF ACID AND BASE? OR JUST BASE?/ACID?

THANK YOU very much!
2. bump.... i am really sorry, but can any of you help?
3. enthalpy of neutralisation is per mole of water formed...
4. m is the mass of the water fro both the acidic and basic solutions

As to how many moles you use - you know that HCl + NaOH ---> H2O + NaCl so essentially the value you want is for the formation of water H+ + OH- ---> H2O. You should find that [acid] = [base] and so you can use the moles of acid or base but NOT both
5. THank you!
Can you please tell me where the values are off? or my methods are wrong? I am sorry!

N(a)=N(b)
C(a)V(a)=C(b)V(b)
2.0mol/dm3 X 0.0232dm3 = C(b) X (0,050dm3)
C(b) = 0.93mol/dm3

Note that the volumes were converted from cm3 to dm3.
For V(a), it was from the extrapolation of the graph (to take into account cooling)

Temperature change fro the graph = 11.6 oC
therefore, assuming water density as well. (can calcualte mass seeing volume)

Q=mcT
=(50+23.2)*4.18*11.6=3549 J
this is for this many number of moles (N(a) or N(b) = 0.0464moles

so 3549/0.0464=76487J=76kJ

while the accepted value is around 56 kJ.

is there some wrong calculation? or is it jsut data?

6. (Original post by EierVonSatan)
m is the mass of the water fro both the acidic and basic solutions

As to how many moles you use - you know that HCl + NaOH ---> H2O + NaCl so essentially the value you want is for the formation of water H+ + OH- ---> H2O. You should find that [acid] = [base] and so you can use the moles of acid or base but NOT both
hmm can you see my above post and please be kind enough once more to answer it? Thank you!
7. (Original post by Nuclear)
hmm can you see my above post and please be kind enough once more to answer it? Thank you!
I can't see anything you've done wrong - the value isn't that far away when you think about errors
8. (Original post by Nuclear)
hmm can you see my above post and please be kind enough once more to answer it? Thank you!
If your final value is too high then:

...your energy is too high - this comes from mcdeltaT
...or your moles are too low

only possibilities:

1. your mass is incorrectly weighed/calculated
2. your temperature measurement is too high.
3. the molarities of the acid and/or base are incorrect

One likely error is the temperature, as you had to carry out a back extrapolation. This is notoriously difficult to give good results.

Another serious source of error in thermometric titration is actually gauging the endpoint. This gives the moles of water and so is critical. It could be that you have underestimated.

As EVS says, your result isn't so wide of the mark...

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Updated: March 20, 2009

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