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.

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

To determine whether half of the HPO4^2- (aq) can be deprotonated, we need to consider the acid dissociation equilibrium for phosphoric acid (H3PO4):

H3PO4 + H2O ⇌ HPO4^2- + H3O+

The Ka value provided (4.79 x 10^-13) is the acid dissociation constant, which measures the degree of dissociation of a weak acid.

In this case, the third equivalence point of the titration indicates complete neutralization of all HPO4^2- present in the solution. We can use this information to calculate the pH at this point and determine whether half of the HPO4^2- can be deprotonated.

First, let's write down the balanced chemical equation for the deprotonation of HPO4^2-:

HPO4^2- + H2O ⇌ PO4^3- + H3O+

Now, let's assign variables to unknown concentrations:
[HPO4^2-] = x
[PO4^3-] = y

The equilibrium expression for this reaction is:
Ka = [PO4^3-][H3O+] / [HPO4^2-]

We know that [HPO4^2-] = 0.5x because we are looking to see if half of the HPO4^2- can be deprotonated.

Since the third equivalence point was not observed, we can assume that the concentration of H3O+ is negligible at this point. Hence, we can neglect [H3O+] in the equilibrium expression.

Therefore, the equilibrium expression simplifies to:
Ka = y / (0.5x)

Now, we can substitute the values of Ka and solve for y:

(4.79 x 10^-13) = y / (0.5x)

Next, we need to consider the equilibrium expression for the overall dissociation of phosphoric acid:

H3PO4 + H2O ⇌ HPO4^2- + H3O+

The initial concentration of H3PO4 is equal to the concentration of sodium hydroxide used in the standardization experiment, which is 0.0125 M. At the third equivalence point, all of the H3PO4 will be neutralized, so its concentration will become zero.

Using the stoichiometry of the balanced chemical equation, we can relate the concentration of y (PO4^3-) to the initial concentration of H3PO4:

y = (0.0125 M)/2

Now, we substitute this value of y into the equilibrium expression:

(4.79 x 10^-13) = (0.0125 M)/2 / (0.5x)

Simplifying further:

(4.79 x 10^-13) = 0.0125 M / 2x

Cross-multiplying:

(4.79 x 10^-13) * (2x) = 0.0125 M

Solving for x:

x = (0.0125 M) / (2 * (4.79 x 10^-13))

x ≈ 1.31 x 10^-11 M

From this calculation, we find that the concentration of HPO4^2- at the third equivalence point is approximately 1.31 x 10^-11 M.

Therefore, half of the HPO4^2- (aq) can indeed be deprotonated at this equilibrium point, indicating that the pH is above the pKa value of HPO4^2- and the deprotonation process is favorable.

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