Aqueous Chemistry: Ranking Ion Concentration I am trying to rank the concentration of $\ce{PO4^3-}$, $\ce{NO3^-}$, and $\ce{Na+}$ that remain in a reaction. The reaction occurs when 100mL of 1.0M $\ce{Na3PO4}$ solution is mixed with 100mL of 1.0M $\ce{AgNO3}$ soln. I have written the equation as:

$\ce{Na3PO4(aq) + 3AgNO3(aq) -&gt; 3NaNO3(aq) + Ag3PO4(s)}$

From this equation I find that there are 0.3mol of $\ce{Na+}$ by 0.1L * 1.0M * 3mol Na, and 
0.1mol of $\ce{PO4^3-}$ by: 0.1L * 1.0M * 1mol $\ce{PO4^3-}$ ,
and finally 0.3mol of $\ce{NO3-}$ by 0.1L * 1.0M * 1mol $\ce{NO3-}$.

By these calculations, the concentration of the nitrate ion and sodium ion should by equal. However, the correct answer is that the most concentrated ion is sodium followed by nitrate followed by phosphate. Could someone please explain where I'm going wrong.

Thanks!
Let's look at the overall reaction:

$\ce{Na3PO4(aq) + 3AgNO3(aq) -&gt; 3NaNO3(aq) + Ag3PO4(s)}$

As you can see $\ce{Na^+}$ and  $\ce{NO_3^-}$ are spectators ions, they don't participate in the main reaction. Their concentrations, taking into account the dilution, are: 

$\ce{[Na^+]= \frac {3 \times 1.0 \times 100}{200}}= 1.5 \ce{M}$.

${[\ce{NO_3^-}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.

The net ionic equation is:

$\ce{PO4^{3-}(aq) + 3Ag^+(aq) -&gt;  Ag3PO4(s)}$

The initial concentration of phosphate and silver ions taking into account the dilution are:

${[\ce{PO_4^3-}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.

${[\ce{Ag^+}]= \frac { 1.0 \times 100}{200}}= 0.5 \ce{M}$.

According to the stoichiometry of the reaction: 3 moles of silver ions react with one mole of phosphate ion. This means that silver ion is a limiting reactant. It will completely react to form the precipitate of silver phosphate. On the other hand, the concentration of phosphate at the end of the reaction will be:
${[\ce{PO_4^3-}]= 0.5-\frac{0.5}{3}=\frac{2\times 0.5}{3}=0.33 \ce{M}}$.

In order to determine exactly the concentration of silver ion, we have to use the solubility product of silver phosphate:
${[\ce{Ag^+}]= \frac{K_{sp}}{[\ce{PO_4^3-}]^3}=\frac{8.89\times10^{-17}}{(0.5/3)^3}}=0.04\times10^{-17}\ce{M}$