1
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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: $(0.0125-0.00453)/0.125=0.06376 \ce{M}$

$\endgroup$
  • $\begingroup$ Thank you, this makes so much more sense now! One question: does that mean limiting reactants only apply to solids, liquids, and gases in the products? If so, is this because aqueous substances are dissociated in the reaction? $\endgroup$ – Kootling Oct 24 '14 at 12:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.