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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Rutherford on June 17, 2012, 11:22:19 AM

Title: 2nd equivalence point
Post by: Rutherford 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? K_a1=5.4*10^-2, K_a2=5.4*10^-5.

At the 1st point I have to calculate the pH of a HC₂O₄^- solution. pH=(pK_a1+pK_a2)/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.

Title: Re: 2nd equivalence point
Post by: Borek on June 17, 2012, 11:50:36 AM

You have a salt of a weak acid.

Title: Re: 2nd equivalence point
Post by: Rutherford 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.

Title: Re: 2nd equivalence point
Post by: Borek 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.

Title: Re: 2nd equivalence point
Post by: Rutherford 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(C₂O₄^2-)=0.2V₁/1.5V₁=0.133M now it becomes protonized:
C₂O₄^2-+H₂O :rarrow: HC₂O₄^-+OH^-
K_w/K_a2=x²/0.133
-logx=pOH, pH=14+logx=14+log(sqr0.133K_w/K_a2)=8.70 the right answer is 8.64. I suppose that this is the correct way. Thank you for the tips.