Explain the dissociation of a triprotic acid?

1 Answer
Truong-Son N.
Jun 5, 2015

Let's take phosphoric acid as an example of a triprotic acid.

#1) "H"_3"PO"_4(aq) + "H"_2"O"(l) rightleftharpoons cancel("H"_2"PO"_4^(-)(aq)) + "H"_3"O"^+(aq)#
#2) cancel("H"_2"PO"_4^(-)(aq)) + "H"_2"O"(l) rightleftharpoons cancel("HPO"_4^(2-)(aq)) + "H"_3"O"^+(aq)#
#3) cancel("HPO"_4^(2-)(aq)) + "H"_2"O"(l) rightleftharpoons "PO"_4^(3-)(aq) + "H"_3"O"^+(aq)#

As you can see, since it has three protons (triprotic), it dissociates three times, donating three protons to water (or three hydrogens to water).

Water's oxygen donates a lone pair of electrons to bond with a proton from #"H"_3"PO"_4#. Then that repeats with the resultant #"H"_2"PO"_4^(-)#, and then the resultant #"HPO"_4^(2-)#, to ultimately get #"PO"_4^(3-)#.

Overall we have:

#"H"_3"PO"_4(aq) + 3"H"_2"O"(l) rightleftharpoons "PO"_4^(3-)(aq) + 3"H"_3"O"^(+)(aq)#

or

#"H"_3"PO"_4(aq) rightleftharpoons "PO"_4^(3-)(aq) + 3"H"^(+)(aq)#

in Arrhenius notation.

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