Respuesta :
Explanation:
For the given values of [tex]K_{a}[/tex] we will have the values of [tex]pK[/tex] as follows.
As, [tex]pK_{a} = -log K_{a}[/tex]
Therefore,
[tex]pK_{a1}[/tex] = 2.15, [tex]pK_{a2}[/tex] = 7.20
[tex]pK_{a3}[/tex] = 12.38
Now, at pH 6.50
[tex]H_{3}PO_{4} \rightarrow H_{2}PO^{-}_{4} + H^{+}[/tex]; [tex]K_{a1}[/tex]
At pH = 2.15; [tex]H_{3}PO_{4} = H_{2}PO^{-}_{4}[/tex]
[tex]H_{2}PO^{-}_{4} \rightarrow HPO^{2-}_{4} + H^{+}[/tex]; [tex]K_{a2}[/tex]
At pH 7.20; [tex]H_{2}PO^{-}_{4} = HPO^{2-}_{4}[/tex]
[tex]HPO^{2-}_{4} \rightarrow PO^{3-}_{4} + H^{+}[/tex]; [tex]K_{a3}[/tex]
Hence, we can conclude that most abundant species is [tex]H_{2}PO^{-}_{4}[/tex] and the second most abundant species is [tex]HPO^{2-}_{4}[/tex].
