CaSO4 precipitate in mixture of Na2SO4 and CaCl2 solutions

I have been stuck on a chemistry problem for a long time now and if anyone here can help me I would be eternally grateful.

40. Du tillsätter $$\pu{100 ml}$$ $$\pu{0,01 M}$$ $$\ce{Na2SO4}$$-lösning i $$\pu{100 ml}$$ $$\pu{0,02 M}$$ $$\ce{CaCl2}$$-lösning. Det bildas en $$\ce{CaSO4}$$-fällning. Hur mycket av $$\pu{0,01 M}$$ $$\ce{Na2SO4}$$-lösning måste ännu tillsättas för att den bildade $$\ce{CaSO4}$$-fällningen just och just ska upplösas helt? $$K_\mathrm{s}(\ce{CaSO4}) = \pu{2,5e-5 M}$$. $$(t = \pu{25 °C})$$ (2p)

a. $$\pu{0,43 l}$$
b. $$\pu{0,48 l}$$
c. $$\pu{0,53 l}$$
d. $$\pu{0,58 l}$$

Translation: We add 100 ml of 0.01 M $$\ce{Na2SO4}$$-solution to 100 ml of 0.02 M $$\ce{CaCl2}$$-solution. A precipitate of $$\ce{CaSO4}$$ is formed. What is the (minimal) volume of 0.01 M $$\ce{Na2SO4}$$-solution that needs to be added to the mix for the $$\ce{CaSO4}$$ precipitate to be just dissolved completely?

The problem then lists the solubility constant, $$K_\mathrm{s}$$, for $$\ce{CaSO4}$$ as $$2.5\cdot 10^{-5}$$.

The only way I can think of solving this myself is by looking at the added $$\ce{Na2SO4}$$-solution as just water and using the $$K_\mathrm{s}$$ value for $$\ce{CaSO4}$$ to see what concentration of $$\ce{CaSO4}$$ is possible in just water. Then calculate the amount of moles of precipitate that was actually formed to see how much water I would need to add to the mixture:

$$K_\mathrm{s}(\ce{CaSO4}) = \pu{2.5e-5 M^2} = [\ce{Ca^2+}][\ce{SO4^2-}]$$

$$\therefore [\ce{Ca^2+}] = [\ce{SO4^2-}] = \sqrt{\pu{2.5e-5 M^2}} = \pu{5e-3 M}$$

$$C = \frac{n}{V} \to n = CV$$

Amount of precipitate formed:

$$n(\ce{CaSO4}) = \pu{0.02 M} \cdot \pu{0.1 dm^3} = \pu{0.002 mol}$$

Required total volume of water to dissolve $$\pu{0.002 mol}$$ of $$\ce{CaSO4}$$:

$$V = \frac{n}{C} = \frac{\pu{0.002 mol}}{\pu{5e-3 M}} = \pu{0.4 dm^3}$$

This is wrong, but I can't figure out the right way to go about this. The answer surely has to do with how well $$\ce{CaSO4}$$ is dissolved in a $$\ce{Na2SO4}$$-solution as opposed to just water.

As you've already noted, there are 3 substances of interest here: $$\ce{Ca^{2+}}, \ce{SO4^{2-}}, \ce{H2O}$$.

Let $$V$$ be the additional volume of sodium sulfate solution. Then:

The total volume is $$V + 0.2\ \mathrm{L}$$.

The total amount of sulfate is $$(V + 0.1\ \mathrm{L}) (0.01\ \mathrm{M})$$.

The total amount of calcium is $$(0.1\ \mathrm{L}) (0.02\ \mathrm{M})$$.

We have then the following relationship:

$$\ce{[Ca^{2+}][SO4^{2-}]}=\frac{(0.1\ \mathrm{L}) (0.02\ \mathrm{M})}{V + 0.2\ \mathrm{L}}\frac{(V + 0.1\ \mathrm{L}) (0.01\ \mathrm{M})}{V + 0.2\ \mathrm{L}}\leq K_{\mathrm{sp}} = 2.5\times 10^{-5}$$

Now, it's just algebra to solve for the threshold value of $$V$$, keeping in mind that there are two solutions for $$V$$ because this is a quadratic equation. In this case, the correct solution is the positive root:

$$V \approx 0.48\ \mathrm{L}$$

The mistake you made is that the calcium and sulfate ion concentrations are definitely not equal, since the additional volume of solution contains only sulfate and not calcium.

When all of the $$\ce{CaSO4}$$ is dissolved completely, all the calcium ions originate from the $$\pu{100 mL} \; \ce{CaCl2}$$ solution, so it is easy to determine the amount:

$$n_\ce{Ca^2+} = \pu{100 mL \times 0.02 mol/L = 2 mmol}$$

To get the total sulfate amount, we have to consider what is present initially and what we add when we add a volume of $$V_\text{add}$$ of $$\ce{Na2SO4}$$.

$$n_\ce{SO4^2-} = \left(\pu{100 mL} + V_\text{add}\right) \times \pu{0.01 mol/L}$$

From this, we can figure out the concentration:

$$c_\ce{Ca^2+} = \frac{\pu{2 mmol}}{ V_\text{add} +\pu{200 mL}}$$

$$c_\ce{SO4^2-} = \frac{\pu{(100 mL} + V_\text{add}) \times \pu{0.01 mol/L}}{ V_\text{add} +\pu{200 mL}}$$

The product has to be equal to $$K_s$$:

$$K_s = \frac{\pu{2 mmol}}{ V_\text{add} +\pu{200 mL}} \times \frac{\pu{(100 mL} + V_\text{add}) \times \pu{0.01 mol/L}}{ V_\text{add} +\pu{200 mL}}$$

This gives a quadratic equation for $$V_\text{add}$$, giving you the numeric result upon solving it. The answer will be approximate because it assumes that volumes are additive and that activities are independent of spectator ion concentrations.

• I like the fact that you added to two caveats that make the answer approximate. We don't explicitly state the assumptions frequently enough here.
– MaxW
May 13 '19 at 15:34
• This is an exercise where I see a mismatch between the level of description of the chemistry and the level of math required to solve it. You might learn more chemistry if the question had been "What volume 10 mM $\ce{Na2SO4}$would it take to dissolve 2 mmol $\ce{CaSO4}$? Compare that to the volume of pure water it would take to dissolve the same amount of calcium sulfate." May 13 '19 at 15:45
• Nice observation, I agree. The chemistry problem is buried under so much math that any appreciation of the chemistry is lost. Students really need the problems to reinforce the chemistry so they can develop better chemical intuition.
– MaxW
May 13 '19 at 15:55
• Totally agree with @MaxW on the assumptions. Nice response. Agreed on the other comment as well. This would we a nicer problem as: (1) set up the equilibrium expression in terms of the added volume $V$ and (2) state any assumptions.
– Zhe
May 13 '19 at 17:12

Goggle translate from Swedish yields a slightly different translation.

You add $$\pu{100 ml}$$ of $$\pu{0.01 M} \; \ce{Na2SO4}$$ solution in $$\pu{100 ml}$$ of $$\pu{0.02 M} \; \ce{CaCl2}$$ solution. A $$\ce{CaSO4}$$ precipitate forms. How much of $$\pu{0.01 M}\; \ce{Na2SO4}$$ solution has yet to be added to completely dissolve the $$\ce{CaSO4}$$ precipitate formed? $$K_\mathrm{s}\ \ce{(CaSO4)} = \pu{2.5e−5 M}. (t =\pu{25 ^{\circ}C})$$

The key point being that the problem is looking for the "extra" amount of the $$\ce{Na2SO4}$$ solution to be added, not including the initial $$\pu{100 ml}$$.

To start let's convert the volumes to liters. So $$\pu{0.100 L}$$ of $$\pu{0.01 M} \; \ce{Na2SO4}$$ solution is added to $$\pu{0.100 L}$$ of $$\pu{0.02 M} \; \ce{CaCl2}$$ solution.

Let's let $$V_e$$ be the extra volume of the $$\pu{0.01 M} \; \ce{Na2SO4}$$ solution, which needs to be added.

We know that for the product of the final concentrations, $$\ce{[Ca^{2+}]}_f$$ and $$\ce{[SO4^{2-}]}_f$$, is equal to the $$K_\mathrm{sp}$$.

$$K_\mathrm{sp} = \ce{[Ca^{2+}]_f \cdot [SO4^{2-}]_f }$$

Now the amount of $$\ce{Ca^{2+}}$$ is $$\pu{0.1 mol/L} \times \pu{0.02 L} = \pu{0.002 mol}$$. There is $$\pu{0.2 L}$$ of solution initially to which $$V_e \; \pu{L}$$ of additional $$\pu{0.01 M} \; \ce{Na2SO4}$$ solution is added. So the final volume is $$\left(0.2 + V_e\right)$$ liters. Thus,

$$\ce{[Ca^{2+}]}_f = \dfrac{0.002}{0.2+V_e}$$

The amount of $$\ce{SO4^{2-}}$$ is $$(0.1+V_e)\times0.01$$ moles. There is $$0.2$$ liters of solution initially to which $$V_e$$ liters of additional $$\pu{0.01 M} \; \ce{Na2SO4}$$ solution is added. So the final volume is again $$\left(0.2 + V_e\right)$$ liters. Thus:

$$\ce{[SO4^{2-}]}_f = \dfrac{(0.1+V_e)\times0.01}{0.2+V_e}$$

Now making the substitutions:

$$K_\mathrm{sp} = \dfrac{0.002}{0.2+V_e} \times \dfrac{(0.1+V_e)\times0.01}{0.2+V_e}$$

Solving the quadratic equation yields $$V_e \approx \pu{0.483 L}$$