Determine the pH at the equivalence point of a titration of phosphoric acid with

Question

Determine the pH at the equivalence point of a titration of phosphoric acid with NaOH.

H3PO4 + 2 NaOH Na2HPO4+ 2 H2O

Assume that the concentration of salt at the equivalence point is 0.03 M.

*So this is a polyprotic titration approach. I think I know how to do it but what confuses me is the 2 coefficent in front of the NaOH.

So basically we got 3 Ka's for this species. Ka1= 7.11e-3 Ka2=6.32e-8 Ka3=4.5e-13

I'm thinking I have to find the [-OH]. So I set it up like this HPO4-2+ H2O   H2PO4- + -OH

Solve for Kb2=Kw/Ka2

Kb=1.58e-7 1.58e-7=x2/(.03-x), x=6.88e-5 which = [-OH] so -log(6.88e-5)=pOH

ph=14-pOH, 9.84: However the answer on my hw gave 9.65. I am almost positive it has something to do with 2 equivalents of NaOH or that I am just setting up the whole problem wrong. Either way pls help, Thank You!

Explanation / Answer

The best way to calculate an equivalence point in a triprotic acid is via:

pH = 1/2*(pKax + pKay)

where x and y are the ionization involved in the equivalence point

for Na2HPO4; this is HPO4-2, meaning that we need pKas

between pka2 and pKa3

reason --> this is exactly between pKa2 and pKa3 points, reclal that

Vequivalence = 3 Vneutralization

then

pH = 1/2*(pKax + pKay)

pH = 1/2*(7.2 + 12.3)

pH = 9.75

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