Suppose I have a solution of a sparingly soluble compound, and consider its solubility in g/L. If I have less g/L of solution than that amount, it means the solution is unsaturated, but what happens in this case? Will all of the compound dissolve such that there is no solid left? And what happens to Ksp? Thank you.


2 Answers 2


No. In this case the product of concentrations will be less than $K_{sp}$.

$K_{sp}$ denotes the product of the concentration of the solute that has dissolved when the solution is saturated.

  • $\begingroup$ A better way to say it is that equilibrium constants apply when a system is at equilibrium. $\endgroup$
    – jimchmst
    Feb 10 at 19:49

No. In an unsaturated solution the product of concentration of ions gives ionic product(I). Ksp is the product of maximum concentration of ions.

So, if I < Ksp : then there will be no solid left (unsaturated solution) if I > Ksp : then the salt will start precipitating (super saturated solution)


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