Answer:
[tex]H_3PO_4(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+H_2PO_4^{-}(aq); \ Ka_1[/tex]
[tex]H_2PO_4^{-}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+HPO_4^{-2}(aq); \ Ka_2[/tex]
[tex]HPO_4^{-2}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+PO_4^{-3}(aq); \ Ka_3[/tex]
Explanation:
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In this case, since the phosphoric acid is a triprotic acid, we infer it has three stepwise ionization reactions in which one hydrogen ion is released at each step, considering they are undergone due to the presence of water, thus, we proceed as follows:
[tex]H_3PO_4(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+H_2PO_4^{-}(aq); \ Ka_1[/tex]
[tex]H_2PO_4^{-}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+HPO_4^{-2}(aq); \ Ka_2[/tex]
[tex]HPO_4^{-2}(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+PO_4^{-3}(aq); \ Ka_3[/tex]
Moreover, notice each step has a different acid dissociation constant, which are quantified in the following order:
Ka1 > Ka2 > Ka3
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