Question

`"H"_3"PO"_4`, `"H"_2"PO"_4^(-)`, `"HPO"_4^(2-)`: which species acts as an acid?

Answer 1

See explanation.

Explanation:

Out of the three compounds you listed, phosphoric acid, `"H"_3"PO"_4`, is always an acid. On the other hand, the other two compounds can act both as an acid and as a base.

So, phosphoric acid will give off one of its three acidic protons to form `"H"_2"PO"_4^(-)`. This polyatomic ion is called dihydrogen phosphate because it contains two acidic protons and one phosphate anion, `"PO"_4^(3-)`.

DIhydrogen phosphate can give off one of its two acidic protons to form `"HPO"_4^(2-)`. This polyatomic ion is called hydrogen phosphate because it contains one acidic proton and one phosphate anion.

Finally, hydrogen phosphate can give off its acidic proton to form the phosphate anion,

Now, dihydrogen phosphate and hydrogen phosphate can act both as an acid and a base because they can either release or accept a proton.

For example, you can have

`"H"_ 3"PO"_ (4(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 2"PO"_ (4(aq))^(-) + "H"_ 3"O"_ ((aq))^(+)`

and

`"H"_ 2"PO"_ (4(aq))^(-) + "H"_ 3"O"_ ((aq))^(+) rightleftharpoons "H"_ 3"PO"_ (4(aq)) + "H"_ 2"O"_ ((l))`

In the reverse reaction, dihydrogen phosphate accepts a proton to reform phosphoric acid, and so it acts as a base.

Similarly, you can have

`"H"_ 2"PO"_ (4(aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HPO"_ (4(aq))^(2-) + "H"_ 3"O"_ ((aq))^(+)`

and

`"HPO"_ (4(aq))^(2-) + "H"_ 3"O"_ ((aq))^(+) rightleftharpoons "H"_ 2"PO"_ (4(aq))^(-) + "H"_ 2"O"_ ((l))`

In this case, dihydrogen phosphate acts as an acid in the forward reaction because it donates a proton to form hydrogen phosphate. In the reverse reaction, hydrogen phosphate acts as a base because it accepts a proton to reform dihydrogen phosphate.

For hydrogen phosphate, you can have

`"HPO"_ (4(aq))^(2-) + "H"_ 2"O"_ ((l)) rightleftharpoons "PO"_ (4(aq))^(3-) + "H"_ 3"O"_ ((aq))^(+)`

Here hydrogen phosphate acts as an acid because it donates a proton to form the phosphate anion.