Question

If the `"pH"` is `6.49` and the three values of `"pK"_a = 2.148, 7.199, 12.15` for phosphoric acid, what species dominates in solution?

Answer 1

Well, a `"pH"` below a given `"pKa"` indicates that the solution is more acidic than the species dissociation to which it corresponds.

`"pK"_(a1) harr "H"_3"PO"_4` dissociation

`"pK"_(a2) harr "H"_2"PO"_4^(-)` dissociation

`"pK"_(a3) harr "HPO"_4^(2-)` dissociation

As a result, we can say:

• `"pH" < "pK"_(a1)` indicates the acidic form of `"H"_3"PO"_4` will exist in greater quantity (than the conjugate base).
• `"pK"_(a1) < "pH" < "pK"_(a2)` indicates the basic form of `"H"_3"PO"_4` (the conjugate base of this), or the acidic form of `"HPO"_4^(2-)` (the conjugate acid of this), will exist in greater quantity.
• `"pK"_(a2) < "pH" < "pK"_(a3)` indicates the basic form of `"H"_2PO"_4^(-)` will exist in greater quantity than the acidic form, i.e. than `"H"_2"PO"_4^(-)`.
• `"pH" > "pK"_(a3)` indicates the most basic form, i.e. the `"PO"_4^(3-)` species will exist in greater quantity than all that came before.

(We can assume that the remaining species not mentioned in an acid/conjugate base equilibrium are dominated by the main ones mentioned.)

A `"pH"` between `"pK"_(a1)` and `"pK"_(a2)` means the solution is more basic than `"H"_3"PO"_4` and more acidic than `"HPO"_4^(2-)`. Thus, we expect that `bb("H"_2"PO"_4^(-))` dominates in solution.

And this can be shown mathematically from the Henderson-Hasselbalch equation, as we are still in the buffer region.

`"pH" = "pK"_(a1) + log((["H"_2"PO"_4^(-)])/(["H"_3"PO"_4]))`

`= "pK"_(a2) + log((["HPO"_4^(2-)])/(["H"_2"PO"_4^(-)]))`

• Since `"pH" > "pK"_(a1)`, the first logarithm is positive and `["H"_2"PO"_4^(-)] > ["H"_3"PO"_4]` (i.e. the argument is greater than `1`).
• Since `"pH" < "pK"_(a2)`, the second logarithm is negative and `["HPO"_4^(2-)] < ["H"_2"PO"_4^(-)]` (i.e. the argument is between `0` and `1`).

And we conclude that `"H"_2"PO"_4^(-)` is the dominant species in solution at this `"pH"`.