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.




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.


 4

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.


 3

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 $V\_e \approx \pu{0.483 L}$ 


 3