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Topic: acid and base titration  (Read 1894 times)

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Offline hanuelee

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acid and base titration
« on: October 23, 2012, 08:05:09 PM »
Hello!
This is my first time posting. Nice to meet you :)

I did a acid and base titration lab in my class with NaOH, strong acid(HCl, with unknown molarity) and weak acid(HC2H3O2, with unknown molarity).

I used 10.0mL of HCl with 0.0951M of NaOH and the equivalence point was 11.45mL

I also used 10.0mL of HC2H3O2, with 0.0850M of NaOH(It's different from the one I used for HCl because our lab group made a mistake during the experiment and we didn't have enough time so we had to get someone else's data) and the equivalence point was 11.43mL

So this is what I did:

M = mol/L
0.0951M = mol/0.01145L
mol = 0.00109

M = mol/L
0.0850M = mol/0.01143L
mol = 0.00109

M1V1 = M2V2
M1(0.0100L) = (0.0951M)(0.01145L)
M1 = 0.109M of HCl

M1V1 = M2V2
M1(0.0100L) = (0.0850M)(0.01143L)
M1 = 0.00971M HC2H3O2

and my teacher told me the actual molarity of HCl and HC2H3O2 was 1.0M..

|(1.0M - 0.109M)/1.0M| * 100 = 89.1%

|(1.0M - 0.00971M)/1.0M| * 100 = 99.03%

I'm not so sure what I did wrong for my calculations to get around 90 percent error! Could you guys help me identify my mistakes in the calculations? Or could it just be that I accidentally copied down the wrong concentrations for HCl and HC2H3O2??

Thanks in advance!  ;D

Offline Hunter2

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Re: acid and base titration
« Reply #1 on: October 24, 2012, 01:02:47 AM »
It can not be 1 M. If you have a solution of 0.0951 M and for 10 ml you consume 11.45 ml you can see at the numbers that the molarity has to be similar.

1 ml 0.1 M NaOH consumes 1 ml 0.1 M HCl for example.



Offline Aegis6

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Re: acid and base titration
« Reply #2 on: October 24, 2012, 05:52:16 PM »
I ran through your math and then I did the problem for myself and the results that I got concurred with yours. If you test for your teacher's values, you get that the Molarity of NaOH was 0.87M, which is clearly not the case given that it was a provided, standardized value:

HCl = 1M

M=Mol/L

1=mol/.01L

0.01mol HCl = 0.01mol NaOH

M=0.01/0.01145

M=0.87

And it runs through essentially the same for the Acetic Acid component. So either the value for the Sodium Hydroxide is wrong, or your teacher gave you the wrong HCl molarity.

Offline hanuelee

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Re: acid and base titration
« Reply #3 on: October 24, 2012, 06:23:15 PM »
Thank you for the replies!!  :D

I didn't have a chance to ask my teacher but apparently everyone in my class got almost 90 percent error :o

I also had to use the balanced equation to get the molarity of HCl and HC2H3O2 instead of M1V1 = M2V2 because M1V1 = M2V2 is used for diluting~ :o I made a huge mistake there..

But thanks for the replies!

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