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When evaporating a saturated solution, does the concentration remain the same?

My thinking is that because the solution is saturated, and the volume is decreased, the excess solute would precipitate out of the solution. Therefore maintaining the original concentration.

On the other hand, would the precipitate be included when measuring the concentration? And therefore evaporation of a saturated solution would lead to an increase in concentration?

For example: if you have a 100mL of 2.5M saturated NaCl solution and evaporate it until the volume is 50mL, what would the final concentration be?

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When a solution is saturated, it contains the maximum amount of solute which can be dissolved at that particular temperature. So effectively it is at its "maximum concentration" (for a given temperature).

This means that if your solvent is evaporating and the temperature remains constant, then solute will precipitate out of solution. The concentration of the solution will still be at its maximum as you are concerned with the amount of solute dissolved.

If you are heating up the solution in order to evaporate the solvent then the increase in temperature might mean that your solute solubility increases. This will depend on what solvent/solute you have.

However, a saturated solution which cools down without being disturbed may form a super-saturated solution whereby its concentration has gone beyond the solubility of the solute at that temperature.

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    $\begingroup$ As a heated saturated solution cools, excess solute may not necessarily fall out of solution, instead forming a supersaturated solution. However, the supersaturated solution will be unstable, and insertion of a "seed" crystal or rough object will cause immediate crystallization. Glassware that has a scratch or rough area on the inside will prevent this from happening though, as the rough/scratched portion provides a surface for crystals to form on, preventing supersaturation. It's quite interesting to watch: youtube.com/watch?v=2Kud9eVbmRY&app=desktop $\endgroup$ – KeatonB Aug 20 '16 at 1:51
  • $\begingroup$ @Ki11akd0g Thank you, I did not consider that, edited my answer to reflect that. $\endgroup$ – Georgeos Hardo Aug 20 '16 at 1:57
  • $\begingroup$ This one is a little more clear youtube.com/watch?v=HnSg2cl09PI&app=desktop $\endgroup$ – KeatonB Aug 20 '16 at 1:57

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