Question

What is the pKa of `CH_3COOH` and `H_3PO_4`?

Answer 1

`\mathbf("CH"_3"COOH")`, ACETIC ACID

Acetic acid is a monoprotic acid, so it only has one `"pKa"`, and it is `4.74`.

If you are more familiar with its `K_a`, which is commonly given as `1.8xx10^(-5)`, then its `"pKa"` is `-log(K_a) = 4.74`.

`\mathbf("H"_3"PO"_4)`, PHOSPHORIC ACID

You can see that phosphoric acid is not monoprotic.

It is polyprotic. It has three protons. So it has three `"pKa"`s.

• `"pKa"_1 = 2.16`
• `"pKa"_2 = 7.21`
• `"pKa"_3 = 12.32`

Each of its `K_a`'s corresponds to a sequential dissociation.

`K_(a1) = (["H"_2"PO"_4^(-)]["H"^(+)])/(["H"_3"PO"_4])`:

`"H"_3"PO"_4(aq) rightleftharpoons "H"_2"PO"_4^(-)(aq) + "H"^(+)(aq)`

`K_(a2) = (["HPO"_4^(2-)]["H"^(+)])/(["H"_2"PO"_4^(-)])`:

`"H"_2"PO"_4^(-)(aq) rightleftharpoons "HPO"_4^(2-)(aq) + "H"^(+)(aq)`

`K_(a3) = (["PO"_4^(3-)]["H"^(+)])/(["HPO"_4^(2-)])`:

`"HPO"_4^(2-)(aq) rightleftharpoons "PO"_4^(3-)(aq) + "H"^(+)(aq)`

Combining all three reactions, we get:

`"H"_3"PO"_4(aq) rightleftharpoons cancel("H"_2"PO"_4^(-)(aq)) + "H"^(+)(aq)`
`cancel("H"_2"PO"_4^(-)(aq)) rightleftharpoons cancel("HPO"_4^(2-)(aq)) + "H"^(+)(aq)`
`cancel("HPO"_4^(2-)(aq)) rightleftharpoons "PO"_4^(3-)(aq) + "H"^(+)(aq)`
`"------------------------------------------------------"`
`color(blue)("H"_3"PO"_4(aq) rightleftharpoons "PO"_4^(3-)(aq) + 3"H"^(+)(aq))`

The full hypothetical one-step dissociation here has the `K_a` of:

`color(blue)(K_(a,"one-step") = K_(a1)K_(a2)K_(a3))`

`= (cancel(["H"_2"PO"_4^(-)])["H"^(+)])/(["H"_3"PO"_4])(cancel(["HPO"_4^(2-)])["H"^(+)])/(cancel(["H"_2"PO"_4^(-)]))(["PO"_4^(3-)]["H"^(+)])/(cancel(["HPO"_4^(2-)]))`

`= color(blue)((["PO"_4^(3-)]["H"^(+)]^3)/(["H"_3"PO"_4]))`

`= 10^(-2.16)*10^(-7.21)*10^(-12.32) = 10^(-21.69)`

`= color(blue)(2.04xx10^(-22))`

In other words, the one-step `K_a` is so small that phosphoric acid heavily favors getting deprotonated one proton at a time, rather than all three in one go.

Hence, its `"pKa"` is NOT given as `-log(10^(-21.69)) ~~ 21.69`. Instead, it is given as `"pKa"_1`, `"pKa"_2`, and `"pKa"_3`.