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The way I was told is that a saturated solution contains the maximum amount of solute and a supersaturated solution contains more solute than a solution can hold at a particular temperature. But I was also told that if you add more solute than it can hold, that it will not dissolve and will settle at the bottom as a solid. So how is that supersaturated? Since it didn't dissolve, isn't it still just saturated? It was also mentioned that a supersaturated solution needs to be cooled, but if the temperature dropped, wouldn't the solute just come out of solution and settle at the bottom (like sugar and cold coffee) and remain saturated?

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    $\begingroup$ You are right about the impossibility to prepare a super saturated solution by dissolving more solute. But you can prepared an almost saturated solution and cool it fast. In particular cases and ckean environments separation of the solute can be slow enough to get in a supersarurated regime. $\endgroup$ – Alchimista Mar 23 '18 at 18:06
  • $\begingroup$ en.wikipedia.org/wiki/Supersaturation $\endgroup$ – aventurin Mar 23 '18 at 18:30
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A saturated solution has the maximal amount of solute at equilibrium.

If equilibrium is not required, solutions can have more solute. For example, when a bottle of soda is first opened, it is supersaturated in CO2, but it is not at equilibrium and CO2 effervesces.

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  • $\begingroup$ When would equilibrium not be required? Does it depend on the solute? $\endgroup$ – Sarah Mar 23 '18 at 18:44
  • $\begingroup$ Also, the reason I'm asking is we're doing a lab with sodium thiosulfate. We heat it up, adding all but one crystal, and then put it in a ice bath and add the last crystal and observe. Is it supersaturated when it is heated? and then the one last crystal causes it to fall out of solution and crystallize? And if so, why is it supersaturated? Is it because it isn't at equilibrium? $\endgroup$ – Sarah Mar 23 '18 at 18:56
  • $\begingroup$ Think about it this way. When the solution is heated and you add all but one crystal, you probably saturated the solution. The solution is then at equilibrium with solute dissolving at the same rate as it crystallizes. When you cool down the solution in general the solubility decreases. Therefore, for a while the dissolved solute is greater than the solubility of the solvent. However, eventually the system returns to equilibrium, with the consequence of having solute crystallize out of solution. $\endgroup$ – JavaScriptCoder Mar 23 '18 at 21:38
  • $\begingroup$ Feel free to take my understanding of it with a grain of salt though. I am in middle school and taught myself chemistry, so I might be wrong. $\endgroup$ – JavaScriptCoder Mar 23 '18 at 21:39
  • $\begingroup$ @Sarah It is not supersaturated until you cool it, which lowers the equilibrium solubility. $\endgroup$ – DavePhD Mar 23 '18 at 23:09
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A saturated solution has the maximal amount of solute at equilibrium.

If equilibrium is not required, solutions can have more solute. For example, when a bottle of soda is first opened, it is supersaturated in CO2, but it is not at equilibrium and CO2 effervesces.

I'll add a little on to DavePhD's answer.

A change in general has two things you need to worry about: the system, or the place that the reaction takes place, and the surroundings, a.k.a. everything else in the universe. So lets start with that $100\deg\pu{Celsius}$ water that you have (just for example). From the Wikipedia page we can see that the solubility is:

$\pu{70.1 g/100 mL}$ ($20$ °C)

$\pu{231 g/100 mL}$ ($100$ °C)

You then dissolve $23$ grams of thiosulfate in $10$ grams of water (just for example). Then, you cool the solution to $20\deg \pu{C}$ using the ice bath. In fact, you cool the solution of sodium thiosulfate beyond its freezing point, something called supercooling or undercooling.

Image of current state

You can think of it as a ball, rolling down a hill coming to a rest because of not having enough kinetic energy to roll all the way down the slope. Our supersaturated solution is currently at (1). It has to overcome an activation energy (2) to reach crystallization (3). Because it wants to crystallize until the solution is $\pu{70.1 g/mL}$, simply providing a surface for it to crystallize on causes it to crystallize rapidly.

I found a resource here that may help.

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You might have some honey in the house. Is it clear? Or has it begun to crystallize? Honey is mostly fructose and glucose with a little sucrose and about 15% water. It is supersaturated and will eventually crystallize out (but you can reheat and cool to clarity, as long as all the crystals have dissolved and cannot act as seed crystals). Sugars are complex molecules with many hydroxyl groups that interfere with crystallization.

The difficulty of getting all the molecules aligned in the preferred order may permit supersaturation for a time; sometimes intermediate alignments are fairly stable but do not sustain orderly growth. Then the partial crystal must partially redissolve and try again; when everything is in perfect alignment, the crystal grows another layer.

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