Estimate the pH of 0.10 M Na₂HPO₄(aq) given pK_a1 = 2.12,
PK_a2 = 7.21,And pK_a3 = 12.68 for phosphoric acid.

Question

Estimate the pH of 0.10 M Na₂HPO₄(aq) given pK_a1 = 2.12, PK_a2 = 7.21,And pK_a3 = 12.68 for phosphoric acid.

Quizplus · Accepted Answer

The first step is to write the dissociation reactions for phosphoric acid:

H3PO4 ⇌ H+ + H2PO4-     pKa1 = 2.12 
H2PO4-  ⇌ H+ + HPO4^2- pKa2 = 7.21 
HPO4^2- ⇌ H+ + PO4^3-     pKa3 = 12.68

The equation for the net ionic reaction that occurs when Na3PO4 is dissolved in water is:

Na3PO4 + 3H2O ⇌ 3Na+ + H3PO4 + OH-

Since Na3PO4 is a strong electrolyte, it dissociates completely in water to form Na+ and PO43- ions. The PO43- ions undergo hydrolysis to produce OH- ions, which increase the pH of the solution.

The initial concentration of H3PO4 can be found using the stoichiometry of the reaction:

0.10 M Na3PO4 = 0.10 M H3PO4

Using the pKa values, we can determine the extent of dissociation of each H+ ion in phosphoric acid:

Ka1 = 10^-pKa1 = 7.94 x 10^-3 
Ka2 = 10^-pKa2 = 1.74 x 10^-8 
Ka3 = 10^-pKa3 = 3.98 x 10^-13

The first dissociation (Ka1) is relatively strong, so we can assume that most of the H3PO4 has lost its first H+. Using the expression for Ka1, we can calculate the concentration of H+ ions from the dissociation of the first H+ ion:

Ka1 = [H+][H2PO4-] / [H3PO4]

Assuming that the concentration of H+ ions is equal to the concentration of H2PO4-, and substituting the known values:

7.94 x 10^-3 = x^2 / (0.10 - x)

Solving for x gives:

[H+] = [H2PO4-] = 0.089 M

The second dissociation (Ka2) is weaker than the first, so we can assume that only a small fraction of the H2PO4- ions have lost their second H+. Using the expression for Ka2, we can calculate the concentration of H+ ions from the dissociation of the second H+ ion:

Ka2 = [H+][HPO42-] / [H2PO4-]

Assuming that the concentration of H+ ions is equal to the concentration of HPO42-, and substituting the known values:

1.74 x 10^-8 = x^2 / (0.10 - 0.089)

Solving for x gives:

[H+] = [HPO42-] = 1.15 x 10^-6 M

The third dissociation (Ka3) is much weaker than the first two, so we can assume that essentially all of the HPO42- ions remain intact.

Therefore, the pH of the solution is determined by the concentration of H+ ions from the first two dissociations:

pH = -log[H+] = -log(0.089 + 1.15 x 10^-6) ≈ 9.94

The best choice is B, 9.94, because it is the closest to the calculated pH value.

Estimate the PH of 0