How to calculate the amount of sodium sulfate needed to lower the silver ion concentration of a saturated silver sulfate solution?

Question

How many moles of $\ce{Na2SO4}$ we have to add into a saturated $\pu{0.5 L}$ solution of $\ce{Ag2SO4}$ so that the concentration of $\ce{Ag}$ is $\pu{4.0 \times 10^-3 M}$? ($K_\mathrm{sp} = 1.4 \times 10^{-5}$)

• I found out first $\ce{[SO4^2-]}$ in a normal saturated solution: \begin{align} K_\mathrm{sp} &= [\ce{Ag+}]^2[\ce{SO4^2-}],\\ 1.4 \times 10^{-5} &= 4s^3,\\ s &= 0.015 \end{align} So we have $\pu{0.015 M}$ of $\ce{SO4^2-}$ in this solution.
• Then, I find out $\ce{[SO4^2-]}$ needed for a saturated solution of $\ce{Ag2SO4}$ but with $[\ce{Ag}] = \pu{4.0 \times 10^-3 M}$: \begin{align} 1.4 \times 10^{-5} &= (4.0 \times 10^{-3})^2 \times [\ce{SO4^2-}],\\ [\ce{SO4^2-}] &= 0.875 \end{align} So we need $\pu{0.875 M}$ of $\ce{SO4^2-}$ in such saturated solution.
• So the difference of $\ce{SO4^2-}$ is $\pu{0.875 M} - \pu{0.015 M} = \pu{0.86 M}$. So we need to add $\pu{0.86 M}$ of $\ce{SO4^2-}$, that is $\pu{0.43 mol}$ of $\ce{SO4^2-}$ ($\pu{0.86 M} \times \pu{0.5 L} = \pu{0.43 mol}$).
• Eventually, we obtained we need to add $\pu{0.43 mol}$ of $\ce{Na2SO4}$.

The above are my calculations, can anyone tell me if my steps can be justified?

Answer 1

Your calculations are true to a certain extend. However, you neglected the amount of $\ce{SO4^{2-}}$ ions precipitated as $\ce{Ag2SO4}$. And also, since we are calculating in the means of concentration terms, we have to assume that addition of $\ce{Na2SO4}$ solid did not change the original $\pu{0.5 L}$-volume.

You calculated correctly the concentration of $\ce{[SO4^2-]}$ ions ($s$) in the original saturated solution, which is equal to the solubility of $\ce{Ag2SO4}$ ($s$):

$$K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{SO4^2-}]= (2s)^2 \times s= 4s^3= 1.4 \times 10^{-5} $$ $$\therefore \; s = \left(\frac{1.4 \times 10^{-5}}{4}\right)^\frac{1}{3} = 1.52 \times 10^{-2}$$

You also calculated correctly the concentration of $\ce{[SO4^2-]}$ ions in the solution after certain amount of $\ce{Na2SO4}$ solid was added (suppose it is $m$ mols) to make the concentration of $\ce{Ag+}$ ions is equal to $\pu{4.0 \times 10^{-3} M}$:

$$K_\mathrm{sp} = [\ce{Ag+}]^2[\ce{SO4^2-}]= (4.0 \times 10^{-3})^2 \times [\ce{SO4^2-}] = 1.4 \times 10^{-5}$$ $$\therefore \; [\ce{SO4^2-}] = \left(\frac{1.4 \times 10^{-5}}{16.0 \times 10^{-6}}\right) = \left(\frac{1.4}{1.6}\right) = \pu{0.875 M}$$

Thus, you have $\pu{0.875 M}$ of $\ce{SO4^2-}$ ions in this solution. However, note that some of $\ce{SO4^{2-}}$ ions would be precipitated as solid $\ce{Ag2SO4}$ when you keep adding $\ce{Na2SO4}$ solid (vide supra) in order to maintain the equilibrium (Le Chatelier's principle). This amount is equal to the half of $[\ce{Ag+}]$ ions precipitated with it:

$$\therefore \; [\ce{SO4^2-}]_{ppt} = \frac{1}{2}\left(2 \times 1.52 \times 10^{-2} - 4.0 \times 10^{-3} \right)= \pu{1.32 \times 10^{-2} M}$$

Suppose you add $m$ amount (in moles) of $\ce{Na2SO4}$ to the original solution. which already had $\pu{1.52 \times 10^{-2} M}$ of $\ce{SO4^{2-}}$ ions. Thus, $[\ce{SO4^{2-}}]$ remain in the solution:

$$[\ce{SO4^{2-}}]_{remain} = \pu{0.875 M} = m + 1.52 \times 10^{-2} - 1.32 \times 10^{-2} $$

$$\therefore \; m = 0.875 - 0.20 \times 10^{-2} \approx \pu{0.873 M}$$

Thus, you need to add $\pu{0.873 mol/L}\times \pu{0.5 L} \approx \pu{0.437 mol}$ of $\ce{Na2SO4}$ to the original solution.