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:

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

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.


  • $\begingroup$ What's the concentration of Na+ in 1 M trisodium phosphate, and what's the concentration of (NO3)- in 1M silver nitrate? $\endgroup$ Nov 24, 2014 at 2:44
  • $\begingroup$ The concentration of Na+ in 1M trisodium phosphate should be 3.0M. I found this by finding the number of moles of trisodium phosphate in 100mL of 1.0M solution: 0.1L and then multiplying that by 1M to get 0.1mol trisodium phosphate and then multiplying by 3mol to account to get the number of Na ions. Then I divided by 0.1L to get 3.0M. With similar steps I found the [NO3-] to be 1.0M. I'm not sure why this doesn't work when I try to find concentration after the reaction is complete. Thanks for your reply. $\endgroup$
    – Blakeasd
    Nov 24, 2014 at 2:59
  • $\begingroup$ You seem to have overlooked the fact that the volume of the solution has doubled after combining them both. The concentration of phosphate ions could be calculated from the solubility product of silver phosphate, but should be very low. $\endgroup$ Nov 24, 2014 at 19:26

1 Answer 1


Let's look at the overall reaction:

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

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:

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

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: ${[\ce{Ag^+}]= \frac{K_{sp}}{[\ce{PO_4^3-}]^3}=\frac{8.89\times10^{-17}}{(0.5/3)^3}}=0.04\times10^{-17}\ce{M}$


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