# How does a limiting reactant relate to the concentration of ions in solution?

Given the question

A sample of $\pu{1.50 g}$ of lead(II) nitrate is mixed with $\pu{125 mL}$ of $\pu{0.100 M}$ sodium sulfate solution. What is the limiting reactant in the reaction? Calculate the concentrations of all ions that remain in the solution after reaction.

I first found the chemical equation for this problem to be:

$$\ce{Pb(NO3)2 (aq) + Na2SO4 (aq) -> PbSO4 (s) + 2NaNO3 (aq)}$$

I found the limiting reactant to be $\ce{Pb(NO3)2}$ because there are fewer moles of that substance than the $\ce{Na2SO4}$, with there being $0.00453 \,\mathrm{mol}\, \ce{Pb(NO3)2}$.

I then used the limiting reactant to find the concentration of each ion. I multiplied 0.00453 mol by the number of moles of each ion in the equation, and put that over 0.125 L to get the molarity. However, the correct answers are

• no $\ce{Pb^{2+}}$ ions are left in solution
• $[\ce{SO4^{2-}}] = 0.0638 \,\mathrm{M}$
• $[\ce{Na^+}] = 0.200 \,\mathrm{M}$
• $[\ce{NO3^-}] = 0.0725 \,\mathrm{M}$

I only got the concentration of $\ce{NO3^-}$ correct. What have I done wrong? I think my mistake has something to do with using the limiting reactant in the equation. Also, I understand why there is no $\ce{Pb^{2+}}$ in the solution, because it is in solid form, but doesn't that mean the $\ce{SO4^{2-}}$ should also produce no ions since it is bonded to the $\ce{Pb^{2+}}$?

Of course, the concentration of $\ce{Na^+}$ is $0.2\ce{M}$, because it is a spectator ion. It doesn't participate in the main reaction between $\ce{Pb^2+}$ and $\ce{SO4^2-}$. The concentration of $\ce{Na^+}$ is $0.1 \times 2=0.2 \ce{M}$ as there are two ions of sodium in sodium sulfate.
The number of mols of ion nitrate, as a spectator ion, is $2\times1.5/331.2= 0.00906 \ce{mol}$. The concentration of ion nitrate is $0.00906/0.125=0.0725 \ce{M}$.
Let's go back to the main reaction between $\ce{Pb^2+}$ and $\ce{SO4^2-}$: The limiting ion is $\ce{Pb^2+}$ as the number of moles of lead nitrate $1.5/331.2=0.00453 \ce{mol}$ is smaller than the number of moles of sodium sulfate $0.0125 \ce{mol}$. So, after the reaction, there are no $\ce{Pb^2+}$ ions left in solution.
As for the remaining $\ce{SO4^2-}$; its concentration is: $(0.0125-0.00453)/0.125=0.06376 \ce{M}$