Reaction of one mole of (PO4)3- with one mole of H2CrO4

I am looking for the endproducts of a reaction of one mole of $\ce{PO4^3-}$ with one mole of $\ce{H2CrO4}$ in one litre of water.

First, the following reaction can happen:

$$\ce{PO4^3- + H2CrO4 <=> HPO4^2- + HCrO4-}$$

Then the following reactions may happen:

\begin{align} \ce{PO4^3- + HCrO4- &<=> HPO4^2- + CrO4^2-} \\ \ce{HPO4^2- + HCrO4- &<=> H2PO4^- + CrO4^2-} \\ \ce{HPO4^2- + H2CrO4 &<=> H2PO4^- + HCrO4^-} \\ \ce{H2PO4^- + H2CrO4 &<=> H3PO4 + HCrO4^-} \end{align}

Is there a way to determine how many moles there will be of each end product?

Note: this is not homework, rather just a thing I'm interested in. However, I have barely an idea where to start.

• Well, you seem to have solid math background, so it should be easier for you to restate this problem in a purely mathematical way. Just look up the pKa values for both acids, and you'll end up with a system of a few equations. Admittedly, it is non-linear, so don't expect to find an analytical solution. – Ivan Neretin Oct 20 '15 at 12:30
• Another thing is, you need to specify the original concentrations, not the amounts. This is because the acid dissociation constants are given in terms of concentrations. (Strictly speaking, it's activities, but whatever) – orthocresol Oct 20 '15 at 12:59
• @IvanNeretin Can you write these equations out, please? I found the pKa values: $\ce{H2CrO4}$: 0.74, $\ce{HCrO4-}$ : 6.49, $\ce{HPO4^2-}$ : 12.32 , $\ce{H2PO4^-}$ : 7.21. Of $\ce{PO4^3-}$ and $\ce{CrO4^2-}$ I can only find the pKb value, 1.68 and 7.51 respectively. – wythagoras Oct 20 '15 at 13:32
• $K_a =10^{-pK_a} = \mathrm{\frac{[A^-][H^+]}{[HA]}}$ for everything that contains H ($\ce{H3PO4,\;H2PO4-}$, etc.) As for $pK_b$, you don't need those at all (it is basically the same thing, only the other way around). – Ivan Neretin Oct 20 '15 at 14:02

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

$$\ce{H3PO4 + H2O <=> H2PO4^- + H3O^+}$$

$$\ce{H2PO4^- + H2O <=> HPO4^2- + H3O^+}$$

$$\ce{HPO4^2- + H2O <=> PO4^3- + H3O^+}$$

Then there is the chromic acid and its conjugate bases

$$\ce{H2CrO4 + H2O <=> HCrO4^- + H3O^+}$$

$$\ce{HCrO4^- + H2O <=> CrO4^2- + H3O^+}$$

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.