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Topic: Titration Problem  (Read 6849 times)

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Offline nozo

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Titration Problem
« on: March 31, 2006, 10:58:09 PM »
Help, I'm not sure if I'm doing this right...

"The volume of a sample of pure HCl gas was 189 ml @ 25 deg celsius and 108 mmHg. It was completely dissolved in about 60 ml water and titrated with an NaOH soln; 15.7 ml of NaOH soln were required to neutralize the HCl. Calculate the molarity of the NaOH soln."

Here's what I did:

NaOH + HCl --> H2O + NaCl

PV = nRT
(.142 atm)(.189 L) / (.082057)(298.15 K) = .001097 mol HCl

.001097 mol HCl * (1 mol NaOH / 1 mol HCl) = .001097 mol NaOH

Molarity (M) = .001097 / .0157 L = .0698 M for NaOH

Am I on the right track, or way off?

Tia!

Offline Albert

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Re:Titration Problem
« Reply #1 on: April 01, 2006, 03:39:05 AM »
Looks correct.

Offline Borek

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Re:Titration Problem
« Reply #2 on: April 01, 2006, 03:42:10 AM »
OK

Classic question for EBAS, in fact I even didn't bothered myself with checking your approach - I just calculated what the final result should be :)

Oops, Albert was faster :(
« Last Edit: April 01, 2006, 03:50:58 AM by Borek »
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Offline nozo

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Re:Titration Problem
« Reply #3 on: April 01, 2006, 08:14:05 AM »
Thanks guys! :)

Jeez, I wish I had the answer keys, so I don't have to ask everytime :P -I always doubt myself when doing problems...

Anyway, back to doing more...   :P

SoFTiCE

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Re:Titration Problem
« Reply #4 on: April 01, 2006, 08:33:37 AM »
R=8.314 J/(K*mol)!!!!
You have mistake!

Offline Borek

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Re:Titration Problem
« Reply #5 on: April 01, 2006, 08:59:16 AM »
R=8.314 J/(K*mol)!!!!
You have mistake!

No, the value used is correct. It all depends on the units used:

0.08205783 L*atm/(K*mol)
8.314510 kPa*dm3/(K*mol)
8.314510 J/(K*mol)
8,314472 L*kPa/(K*mol)
62,3637 L*mmHg/(K*mol)
83,14472 L*mbar/(K*mol)
1.987216 cal/(K*mol)

8.314510 and 8,314472 are basically the same, they are from different sources and I don't remember now why do they differ.
« Last Edit: April 01, 2006, 10:48:32 AM by Borek »
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SoFTiCE

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Re:Titration Problem
« Reply #6 on: April 01, 2006, 10:29:06 AM »
Yes, I'm sorry. ;D

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