# Can the solubility of a compound be used to estimate the degree of separation of a solute from a solution?

I'm confused exactly on how to use the concept of solubility to get the amount of separation from a certain solute in the solution.

This arises from attempting to solve a problem regarding this matter. The problem described is as follows:

$$400$$ grams of anhydrous sodium sulphate ($$Na_{2}SO_{4}$$) is dissolved in a liter of hot water. The solution is then let to cool carefully until reaching $$20^{\circ}C$$ to remain supersaturated with respect to the formed decahydrate $$Na_{2}SO_{4}\cdot 10H_{2}O$$. Then a small crystal of the latter salt is added to the solution, separating the excess of $$Na_{2}SO_{4}$$ dissolved, remaining a saturated solution. It is known that the saturated solution is equivalent to $$19.4$$ grams of $$Na_{2}SO_{4}$$ by $$100\,mL$$ of water. What amount of the decahydrate would had been separated?

The given alternatives are:

$$\begin{array}{ll} 1.&249\,g\,\textrm{to}\,259\,g\\ 2.&318\,g\,\textrm{to}\,327\,g\\ 3.&689\,g\,\textrm{to}\,698\,g\\ 4.&721\,g\,\textrm{to}\,730\,g\\ 5.&890\,g\,\textrm{to}\,899\,g\\ \end{array}$$

What I assumed is that:

$$400\,g\,Na_{2}SO_{4}\times\frac{142+180\,g\,Na_2SO_{4}\cdot 10 H_{2}O}{142\,g\,Na_{2}SO_{4}}-100\,mL\frac{19.4\,g}{100\,mL}\approx 807\,g$$

Therefore that would be the grams of sodium sulphate decahydrate but I'm not sure if that would be the ammount. I'm confused why do the alternatives features a range?. How does it appear that?. Can somebody explain how exactly to get to that given range?.

• I don't understand this part -- "Then a small sack of the latter salt is added to the solution, separating the excess of Na2SO4 by 100mL of water." – MaxW Dec 28 '19 at 0:37
• @MaxW I am also a bit confused about the intended meaning of the author. Since it is not indicated the ammount of sodium sulphate which was added, I don't know the reason for the inclusion of that information. – Chris Steinbeck Bell Dec 31 '19 at 3:36
• Then a small sack of the latter salt is added to the solution, so some $\ce{Na2SO4\cdot 10H2O}$ is added to the solution to nucleate precipitation separating the excess of $\ce{Na2SO4}$ of course $\ce{Na2SO4}$ was originally added to the solution but its is of course the decahydrate that is precipitating by 100mL of water. this is the part that absolutely baffles me... – MaxW Dec 31 '19 at 6:47
• @MaxW Does ignoring this portion (about adding the decahydrate) of the problem would help into solving it?. Its been days and I still can't get a clear answer. Needless to say that it seems that wikipedia data for the solubility of that decahydrate is not correct. Perhaps do you have that value or help me with this problem?. – Chris Steinbeck Bell Jan 2 '20 at 1:37
• @MaxW It turns out that you were right. This question was incorrectly typed thus sounded too strange. I went to the original source and found there was a missing piece of information and corrected the typo which caused confusion. As it stands now can this problem be solved?. – Chris Steinbeck Bell Jan 2 '20 at 7:39

## 2 Answers

$$400 g$$ $$\ce{Na2SO4}$$ contains $$400/124$$ = $$3.225$$ mole of $$\ce{Na2SO4}$$. As one mole $$\ce{Na2SO4}$$ will produce $$1$$ mole decahydrate $$\ce{Na2SO4}· \ce{10H_2O}$$ , one deduces that $$3.225$$ mole $$\ce{Na2SO4}· \ce{10H_2O}$$ weighs $$3.225 · 322 g = 1038g$$. But for doing this salt, $$1038 - 400 = 638$$ g water is necessary. Remains in solution $$1000 - 638 = 362$$ g water

As $$100$$ mL water may dissolve $$44$$ g decahydrate, $$362$$ g water will dissolve $$160$$ g decahydrate. The excess will be deposited in the container, namely $$1038 g - 160 g = 878$$ g

• The information it is a bit condensed. Can you split the steps?.Why do you subtract the grams of the sodium sulphate decahydrate to the weight of sodium sulphate to obtain the water?.The part where I'm more stuck is is why do you subtract $1000-638$? what's the meaning of $362\,g$ of water? what does it represent?. For the second step where you say $100\,mL$ may dissolve $44\,g$ of decahydrate and then $362\,g$ water will dissolve $160\,g$ decahydrate how did you obtained this number?.Is it from doing this?, $362\,g\textrm{water}\times\frac{44\,g}{100\,g\textrm{water}}=159.28\,g$ – Chris Steinbeck Bell Dec 28 '19 at 19:19
• I'm not very sure about value of the solubility of sodium decahydrate in water at $20^{\circ}C$. Can you check this data please?. Why in the given alternatives does it appear two values for the amount of separation for that salt?. What was the intended meaning?. None of the answers seem to match with what you obtained but only the fifth option does seem to get very close. Can you help me with this part?. – Chris Steinbeck Bell Dec 28 '19 at 19:23
• Hi. Sorry, but can you attend my questions in the comments? I'm still doubtful about those. – Chris Steinbeck Bell Dec 31 '19 at 3:36
• The original question was incorrectly stated. I went to the original source and found that mistake and corrected it. If you look at your arithmetic the division is not right. Can you fix that according to the updated information?. – Chris Steinbeck Bell Jan 2 '20 at 7:41

Assume:

• Water has a constant density of 1.000 g/ml from $$20 ^\circ\pu{C}$$ to $$100 ^\circ\pu{C}$$.

Given:

• $$400\pu{g}\ \ce{Na2SO4}$$ contains $$400/142.04 = 2.8161$$ mole of $$\ce{Na2SO4}$$.
• $$1000\ \pu{ml}\ \ce{H2O} = 1000\ \pu{g}\ \ce{H2O} = 1000/18.015 = 55.509\ \pu{moles}\ \ce{H2O}$$
• $$19.4\ \pu{grams}$$ of $$\ce{Na2SO4}$$ per $$100\pu{mL}$$ of water $$= 19.4/142.04$$ or $$0.13658\ \pu{moles}$$ of $$\ce{Na2SO4}$$ per $$100/18.015$$ or $$5.5509$$ moles of water. Thus in solution there are $$0.13658/5.5509=0.024605\ \pu{moles}$$ of $$\ce{Na2SO4}$$ per mole of water.

Let:

• $$x =$$ number of moles of $$\ce{Na2SO4\cdot 10H2O}$$ formed
• $$2.8161 - x =$$ number of moles of $$\ce{Na2SO4}$$ still in solution.
• $$10x =$$ number of moles of water in in $$\ce{Na2SO4\cdot 10H2O}$$
• $$55.509 - 10x =$$ number of moles of water still in solution

Thus:

$$2.8161 - x =(55.509 - 10x)\times0.024605$$
$$2.8161 - x = 1.3658 - 0.24605x$$
$$1.4503 = 0.75395x$$
$$x = 1.9236$$

Grams $$\ce{Na2SO4\cdot 10H2O} = 1.9236\times322.20 = 620\ \pu{grams}$$

The assumption about the density of water isn't valid of course, but it is within about 4%. 620*1.04 = 651 which is still not within one of the mass ranges given.

• The issue which still confuses me is why on earth ranges are given as an answer? Suspiciously all the alternatives are separated by a margin of 9 grams. Does it exist a justification for this? You used the density of water to get the mass in one liter and moles, but im stuck on the justification of the steps. What exactly is happening in lets say a beaker? You dissolve the salt and some part is transformed into the decahydrate form? But when you add a crystal what does it change there? – Chris Steinbeck Bell Jan 3 '20 at 12:31
• More decahydrate is formed? Does adding this crystal acts as a magnet to get all the possible decahydrate from the supersaturated solution? Can you illustrate this process to me because im confused on why you did those steps. The confusion i also have is the thing about supersaturated solution. In this case the precipitate is the decahydrate form or is it the anhydrous form? What's in the solution is the decahydrate form or is it the anhydrous form? Can you help me with this part please? – Chris Steinbeck Bell Jan 3 '20 at 12:34
• @ChrisSteinbeckBell - The cooled solution is super saturated with $\ce{Na2SO4}$. The added crystal of $\ce{Na2SO4\cdot 10H2O}$ is added to nucleate precipitation of $\ce{Na2SO4\cdot 10H2O}$ from the solution. After the precipitation is complete the solution is saturated with $\ce{Na2SO4}$ – MaxW Jan 3 '20 at 19:37
• @ChrisSteinbeckBell - I don't know why the choices are given as ranges. – MaxW Jan 3 '20 at 19:41
• @ChrisSteinbeckBell - I used the density of water but I could have worked the problem using volumes. However since no temperature was given for the hot water there would have been an implicit assumption that the water's volume didn't change. Stating that the density is constant makes clear where an poor assumption must be made. – MaxW Jan 3 '20 at 19:45