Answer:
HPO₄⁻² predominates at pH 9.3
Explanation:
These are the equilibriums of the phosphoric acid, a tryprotic acid where 3 protons (H⁺) are realesed.
H₃PO₄ + H₂O ⇄ H₂PO₄⁻ + H₃O⁺ pKa 2.14
H₂PO₄⁻ + H₂O ⇄ HPO₄⁻² + H₃O⁺ pKa 6.86
HPO₄⁻² + H₂O ⇄ PO₄⁻³ + H₃O⁺ pKa 12.38
The H₂PO₄⁻ works as amphoterous, it can be a base and acid, according to these equilibriums.
H₂PO₄⁻ + H₂O ⇄ HPO₄⁻² + H₃O⁺
H₂PO₄⁻ + H₂O ⇄ H₃PO₄ + OH⁻
pH 9.3 is located between 6.86 and 12.38 where we have this buffer system HPO₄⁻² / PO₄⁻³, where the HPO₄⁻² is another amphoterous:
HPO₄⁻ + H₂O ⇄ H₂PO₄⁻ + OH⁻
HPO₄⁻² + H₂O ⇄ PO₄⁻³ + H₃O⁺
The media from the two pKa, indicates the pH where the protonated form is in the same quantity as the unpronated form, so:
(6.86 + 12.38) /2 = 9.62
Above this pH, [PO₄⁻³] > [HPO₄⁻²].
In conclussion, at pH 9.3, [HPO₄⁻²] > [PO₄⁻³]
