Why does hydrogen phosphate act as a base?

Let's look at question c:

a) Write a balanced equation for the reaction. $$\ce{2 NaOH + H3PO4 -> Na2HPO4 + 2 H2O (l)}$$

b) When some crystals of $\ce{Na2HPO4}$ were dissolved in water, the $\mathrm{pH}$ of the resulting solution was found to be $9.5$. Calculate the hydrogen ion concentration of this solution. $$[\ce{H+}] = 10^\mathrm{-pH}\, ;\qquad [\ce{H+}] = 10^{-9.5} = \pu{3.2e-10 mol L-1}$$

c) Write an equation for the reaction of the $\ce{HPO4^2-}$ ion with water to account for the measured $\mathrm{pH}$. $$\ce{HPO4^2- + H2O (l) <=> H2PO4- + OH-}$$

So, I'm a bit confused with what's happening. It seems that water is acting as an acid in this reaction, and donating protons to the ion. Can someone just elaborate on this whole process please because I'm not really sure what is happening. Why is it $\ce{HPO}$ and not $\ce{H2PO}$ or $\ce{H3PO}$ (as is in the original equation?) Is it a step wise process and only one step included?

• I am a little confused and may have answered a question that you did not ask. Are you actually asking why in part c you should consider $\ce{HPO4^2-}$ as oposed to $\ce{H2PO4-}$ or $\ce{H3PO4}$? Please also check you sum formulae of the ions, they are not correct as written now. Let me know how I can improve my answer. – Martin - マーチン Jul 17 '15 at 6:53

When hydrogen phosphate salts are dissolved in water there are two main equilibria formed. This is based on the fact, that hydrogen phosphate can act as a Brønsted–Lowry base, i.e. accept protons, or as an acid, i.e. donate protons. For water the same is true. In addition to this it can react with itself, which is known as the autoprotolysis of water: $$\ce{H2O + H2O <=> H3+O + {}^{-}OH}\tag1$$
To a lesser extent there is also the following equilibrium happening: $$\ce{HPO4^2- + 2H2O <=> H3PO4 + 2 {}^{-}OH}\tag4$$
From the acidity constant you know that $\ce{HPO4^2-}$ will react mostly as a base, i.e. $\mathrm{p}K_\mathrm{a}(\ce{Na2HPO4})=12.35$, so the equilibrium $(3)$ will be predominant.