April 24, 2024, 07:54:06 AM
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Topic: Add different size of KCl (s) into saturated KCl (aq), what is the result?  (Read 1775 times)

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Offline orthotraces

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Hello everyone,

The question I'd like to ask is that:
If we have a cup of saturated KCl (aq) and then add some extra KCl (s) with different powder sizes, what will happen?


Will the smaller powder KCl dissolve and the larger powder KCl grow much larger?

Or, the larger powder KCl dissolve and the smaller powder KCl grow larger, and the solid KCl powder eventually become to be the similar size?

Or, the powder remain their original sizes? (or they remain the same distribution of sizes?)


I wonder what is the result in the saturated solution with extra solute added.
(the competition between dissolution and crystallization?)

Thanks for all of your answer and discussion!

Offline Hunter2

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If the solution is saturated, nothing will dissolve, doesnt matter of crystal size.
If the solution is not saturated which one will  dissolve faster until the solution is saturated, the smaller or larger size crystals? And why?

Offline mjc123

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Even when the solution is saturated, there is a dynamic solution-precipitation equilibrium, even if the concentration in solution doesn't increase.

Offline Enthalpy

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I expect an effect only at nanoscopic grain sizes, similarly to evaporation and condensation.

An effect at the edges, softening or sharpening, must appear more easily.

Offline jeffmoonchop

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Ostwalds ripening. Smaller particles are more soluble than larger particles due to their inherent instability in comparison to larger particles (higher surface tension). The molecules in the smaller particles will break off and migrate to the most stable structure, the larger particles (lower surface tension). So if the solution remains saturated for extended periods (and doesn't become supersaturated by loss of solvent or change in T), then the smaller particles will eventually dissolve and the larger particles will grow.

Offline orthotraces

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Thanks for all of your help and response!

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