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 $\ce{Pb^{2+}}$?