Reaction of one mole of (PO4)3- with one mole of H2CrO4
To answer this question, you should stop thinking about the acids and bases reacting with each other, but rather they react with water exclusively.
Firstly, the phosphoric acid and its conjugate bases will have the following reactions with water

$$\ce{H3PO4 + H2O &lt;=&gt; H2PO4^- + H3O^+}$$

$$\ce{H2PO4^- + H2O &lt;=&gt; HPO4^2- + H3O^+}$$

$$\ce{HPO4^2- + H2O &lt;=&gt; PO4^3- + H3O^+}$$

Then there is the chromic acid and its conjugate bases

$$\ce{H2CrO4 + H2O &lt;=&gt; HCrO4^- + H3O^+}$$

$$\ce{HCrO4^- + H2O &lt;=&gt; CrO4^2- + H3O^+}$$

So now you have five equlibrium reactions.  You have eight unknowns, $[\ce{H3PO4}]$, $[\ce{H2PO4^-}]$, $[\ce{HPO4^2-}]$, $[\ce{PO4^3-}]$, $[\ce{H2CrO4}]$, $[\ce{HCrO4^-}]$, $[\ce{CrO4^2-}]$, and $[\ce{H3O^+}]$. 

The Ka expressions from the above reactions give you five equations.  Then you have the initial concentrations of phosphate and chromic acids, which will give you two more, then the fact that the number of protons is conserved will give you the eighth and final equation.