# 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?

Aug 17, 2017

Well, a $\text{pH}$ below a given $\text{pKa}$ indicates that the solution is more acidic than the species dissociation to which it corresponds.

${\text{pK"_(a1) harr "H"_3"PO}}_{4}$ dissociation

${\text{pK"_(a2) harr "H"_2"PO}}_{4}^{-}$ dissociation

${\text{pK"_(a3) harr "HPO}}_{4}^{2 -}$ dissociation

As a result, we can say:

• ${\text{pH" < "pK}}_{a 1}$ indicates the acidic form of ${\text{H"_3"PO}}_{4}$ will exist in greater quantity (than the conjugate base).
• ${\text{pK"_(a1) < "pH" < "pK}}_{a 2}$ indicates the basic form of ${\text{H"_3"PO}}_{4}$ (the conjugate base of this), or the acidic form of ${\text{HPO}}_{4}^{2 -}$ (the conjugate acid of this), will exist in greater quantity.
• ${\text{pK"_(a2) < "pH" < "pK}}_{a 3}$ indicates the basic form of ${\text{H"_2PO}}_{4}^{-}$ will exist in greater quantity than the acidic form, i.e. than ${\text{H"_2"PO}}_{4}^{-}$.
• ${\text{pH" > "pK}}_{a 3}$ indicates the most basic form, i.e. the ${\text{PO}}_{4}^{3 -}$ 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 $\text{pH}$ between ${\text{pK}}_{a 1}$ and ${\text{pK}}_{a 2}$ means the solution is more basic than ${\text{H"_3"PO}}_{4}$ and more acidic than ${\text{HPO}}_{4}^{2 -}$. Thus, we expect that $\boldsymbol{{\text{H"_2"PO}}_{4}^{-}}$ 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 ${\text{pH" > "pK}}_{a 1}$, the first logarithm is positive and $\left[{\text{H"_2"PO"_4^(-)] > ["H"_3"PO}}_{4}\right]$ (i.e. the argument is greater than $1$).
• Since ${\text{pH" < "pK}}_{a 2}$, the second logarithm is negative and $\left[{\text{HPO"_4^(2-)] < ["H"_2"PO}}_{4}^{-}\right]$ (i.e. the argument is between $0$ and $1$).

And we conclude that ${\text{H"_2"PO}}_{4}^{-}$ is the dominant species in solution at this $\text{pH}$.