Calculating Ka Values of H3PO4 from Titration Data

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To calculate the Ka values of H3PO4 from titration data, the first and second equivalence points (EP) were identified at 4.76 mL with a pH of 4.65 and 9.41 mL with a pH of 9.23, respectively. The concentrations of H3PO4 and HCl were 0.106 M and 0.003 M. An attempt to use pKa = pH at half the second EP yielded a Ka value of 1.51*10^(-7), which is significantly different from the accepted pKa2 of 6.8*10^(-8). The discussion highlights confusion regarding the calculation of the first Ka value and the influence of remaining HCl at the first EP. Clarification on thermodynamic corrections and proper methodologies for determining Ka values is sought.
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Homework Statement


I need to determine the Ka values of H3PO4 having titrated a solution of H3PO4 and HCl with 0.1142 M NaOH. The concentrations of H3PO4 and HCl were found to be 0.106 M and 0.003 M respectively. The first EP was reached at 4.76 mL with a pH of 4.65 and the second EP was reached at 9.41 mL and a pH of 9.23.


Homework Equations



Kw=Ka*Kb
pH=pKa+log([A-]/[HA])
Ka=([H+][H2PO4-])/[H3PO4]

The Attempt at a Solution


Struggled with this for a while now and I feel like I am missing something obvious.

Attempted to use the fact that pKa = pH at 1/2 the EP for the second EP received an answer of 1.51*10^(-7), which is about an order of magnitude off of the accepted pKa2 of 6.8*10^(-8). Not sure how this could be - all of the HCl has been neutralized and you are just converting H2PO4- to HPO4 2-, however this is all based off data I collected. I'm not even sure how to determine the first Ka value or even where to start. I don't believe you can say that pKa=pH at 1/2 the first EP because you still have the HCl in solution. Not sure how else you could go about solving it. Any help would be great appreciated.
 
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Have you taken into account thermodynamic corrections?
 

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