Why Does 0.10 M Dihydrogen Phosphate Dissociate Rather than Hydrolyse in a Neutral Solution of Water? Dihydrogen Phosphate has a Ka of 6.2x10^-8, yet it seems to form solutions with an acidic pH when it is in water.  Why is that?  Shouldn't it dissociate back to phosphoric acid and form a basic solution, since its Ka is less than 1.0x10^-7?  

It seems as though Ka does play a role in determining whether an amphoretic substance will act like a base or acid, since 0.10 M Hydrogen Phosphate has a Ka of 4.8x10^-13 and creates a basic solution.  However, I feel like I am missing another mechanism that determines the behavior of an amphoretic substance in a neutral solution.
The whole truth is that any time that you add any phosphate ion into an aqueous solution, then you will have all four phosphate species $\ce{H2PO4-, HPO4^{-2} , PO4^{-3} ,H3PO4}$  in solution, that ($\ce{H2PO4-}$)   is amphoteric that behaves both:

as an acidreleasing $\ce{H+}$ in aqueous solution and( forming $\ce{HPO4^{-2}}$ ions which can produce very few of $\ce{PO4^{-3}}$ ions)

$$\text{
Ionization:}\ce{H2PO4- &lt;=&gt; HPO4^{-2} +\color{red}{H^+} \quad\left(K_\mathrm{a}=\color{red}{6.2\times10^{-8}}\right)}$$
OR

as a base(can undergo hydrolysis forming ($\ce{H3PO4}$) acid 
$$\text{Hydrolysis:}\ce{H2PO4-  +  H2O &lt;=&gt; H3PO4 + \color{blue}{OH-} \quad\left(K_\mathrm{b}=\color{blue}{1.3\times10^{-12}}\right)}$$

So ,$\color{red}{K\mathrm{a}}&gt;\color{blue}{K\mathrm{b}}$ ,thus the first reaction is the dominant,so ,$\pu{pH}&lt;7 $ and the solution is$~\color{red}{\text{acidic}}$.