I really don't understand the problem. What are you to judge by? Of course you can deprotonate half of it if you add enough NaOH. If you add enough NaOH I should think you could deprotonate all of the HPO4^2-. But I don't know how you would tell when you were there since you can't see the eq pt with indicators.
Deprotonating half of the HPO4^= will be the pKa value or 12.3
In an acid-base experiment at 25.0°C, a sodium hydroxide solution was
standardized against KHP. The average value for the concentration of the sodium
hydroxide solution was calculated to be 0.0125 M. This solution was used to titrate
the phosphoric acid present in coke. The first two equivalence points were clearly
visible in the titration curve of the coke sample. However, the third equivalence point
could not be seen. Using calculations, demonstrate whether half of the HPO4^2- (aq) (Ka = 4.79 x 10-13) can be deprotonated.
2 answers
It is not possible to reach the half equivalence point of HPO42- using 0.0125 M of NaOH because the maximum pH possible with this concentration of base is 12.10 (the pH of the NaOH solution itself) and half of the third proton does not dissociate until the pH is equal to 12.32 (the third half equivalence point), which isn’t isn’t possible since that pH is greater than that of the solution.
pOH= -log(0.0125)= 1.903089987
pH+pOH=14 (at 25 degrees C)
14-1.903089987= 12.09691001
pH= 12.10
Pka=- log(Ka)
pKa= -log (4.79 x 10^-13) = 12.3
pOH= -log(0.0125)= 1.903089987
pH+pOH=14 (at 25 degrees C)
14-1.903089987= 12.09691001
pH= 12.10
Pka=- log(Ka)
pKa= -log (4.79 x 10^-13) = 12.3