Answer
Bronsted acids: $H_3PO_4$ and $H_3O^+$
Bronsted base: $PO_4^{3-}$
Both: $H_2PO_4^-$ and $HPO_4^{2-}$
Work Step by Step
Phosporic acid ($H_3PO_4$) is a polyprotic acid, its first reaction is:
(I) $H_3PO_4(aq) + H_2O(l) H_2PO_4^-(aq) + H_3O^+(aq)$
But, $H_2PO_4^-$ is also an acid:
(II) $H_2PO_4^-(aq) + H_2O(l) HPO_4^{2-}(aq) + H_3O^+(aq)$
And $HPO_4^{2-}$ too:
(III) $HPO_4^{2-}(aq) + H_2O(l) PO_4^{3-}(aq) + H_3O^+(aq)$
Now, as we can analyze these are all the species present in the solution:
Bronsted acids: $H_3PO_4$ and $H_3O^+$ (Donate a proton)
Bronsted base: $PO_4^{3-}$ (Receives a proton)
Both: $H_2PO_4^-$ and $HPO_4^{2-}$ (Donate and receive a proton)
