$$ \ce{H3PO4 <--> H+ + H2PO4-} \hskip{2em} K_{a_1}=7.2 \times 10^{-3} $$ $$ \ce{H2PO4- <--> H+ + HPO4^{2-}} \hskip{1.5em} K_{a_2}=6.3 \times 10^{-8} $$ $$ \ce{HPO4^{2-} <--> H+ + PO4^{3-}} \hskip{2em} K_{a_3}=4.5 \times 10^{-13} $$

A solution is prepared by mixing 50 mL of 1 \(M \ce{ NaH2PO4}\) with 50 mL of 1 \(M  \ce{Na2HPO4}\). On the basis of the information above, which of the following species is present in the solution at the lowest concentration?

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\(\ce{ NaH2PO4}\) and \(\ce{ Na2HPO4}\) will dissociate completely into ions. The major species present before any reactions occur are:

$$ \ce{Na+} \text{ and } \ce{ H2PO4^-} \text{ and } \ce{ H2PO4^{2-}} $$

We can consider the production of all ions:

\(\ce{Na+}\) ions will be abundant from the initial dissociation of salts.

\(\ce{HPO4^{2-}}\) will be present from the forward reaction of the second reaction and from the initial dissociation.

\(\ce{H2PO4^-}\) will be present from the initial dissociation. It can also be produced from the reverse reaction of the second reaction.

\(\ce{PO4^{3-}}\) will be present from the forward reaction of the third reaction.

The \(K_{a_3}\) value of the third reaction is very small, meaning that very little product is made. Therefore, \(\ce{PO4^{3-}}\) will be present at the lowest concentration.