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

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.