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

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?

• Why divide 0.86 moles by 2? It is $\ce{SO4^{2-}}$ that is needed in the solution, not $\ce{Na+}$.
– MaxW
Commented Dec 15, 2019 at 18:53
• Oh, I just added a note to that, and I changed all mol/L to M so it might be more visible. It is 0.86 mol/L, so I multiply it by 0.5 L in order to obtain the mole needed which is 0.43 mol Commented Dec 15, 2019 at 18:56
• Dah... I didn't read the problem correctly. Dividing by 2 was correct.
– MaxW
Commented Dec 15, 2019 at 21:49
• Haha. No worries, it happens Commented Dec 15, 2019 at 22:03
• In the given data, $\ce{K_s_p}$ = $\ce{1.4 x 10^{-5}}$. But two lines later, you write another value for the same constant. You write that $\ce{K_s_p}$ = $\ce{4.0 x 10^{-3}}$. Why ? All the following calculations are wrong. Commented Dec 15, 2019 at 22:11

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.