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Topic: 2nd equivalence point  (Read 14242 times)

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Offline Rutherford

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2nd equivalence point
« on: June 17, 2012, 11:22:19 AM »
What is the pH of a H2C2O4 solution (c=0.2M) at the 1st and the 2nd equivalence point if it was titrated with 0.2M NaOH? Ka1=5.4*10-2, Ka2=5.4*10-5.

At the 1st point I have to calculate the pH of a HC2O4- solution. pH=(pKa1+pKa2)/2=2.78, but how to calculate the pH for the second point? After the first step the volume of the solution will increase twice so c(HCO4-)=0.1M, here I stopped.

Offline Borek

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Re: 2nd equivalence point
« Reply #1 on: June 17, 2012, 11:50:36 AM »
You have a salt of a weak acid.
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Offline Rutherford

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Re: 2nd equivalence point
« Reply #2 on: June 17, 2012, 12:00:38 PM »
The volume change is the problem for me.
I thought maybe to add NaOH whose V is two times less that the V of the solution because its concentration is 2 times bigger (0.2M). The volume increaces 1.5 times so the c(C2O4)2-=0.067M, but don't know how to calculate c(H+), because I got some H+ ions from the first step and some from the second one.

Offline Borek

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Re: 2nd equivalence point
« Reply #3 on: June 17, 2012, 12:33:20 PM »
At the second end point you have a solution of disodium oxalate.

And the volume can be easily calculated from the reaction stoichiometry.
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Offline Rutherford

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Re: 2nd equivalence point
« Reply #4 on: June 17, 2012, 01:10:30 PM »
Maybe it's easier to write it this way:
H2C2O4+2NaOH :rarrow: Na2C2O4+2H2O
0.2V1     2*0.2V2
V2=0.5V1
The obtained volume is 1.5V1. c(C2O42-)=0.2V1/1.5V1=0.133M now it becomes protonized:
C2O42-+H2O :rarrow: HC2O4-+OH-
Kw/Ka2=x2/0.133
-logx=pOH, pH=14+logx=14+log(sqr0.133Kw/Ka2)=8.70 the right answer is 8.64. I suppose that this is the correct way. Thank you for the tips.

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