What exactly is a solubility product, and what is its use?

I've come across this term while reading, and don't have access to anyone who can help me understand this concept.


If we take small particles of a sparingly soluble salt, say AgCl, then one or two particles may dissolve in water and a saturated solution is obtained pretty soon. Upon adding extra AgCl to the solution precipitation of AgCl starts. At this stage, you'll find an equilibrium is established between the dissolved and the undissolved AgCl.

$$\ce{AgCl(s) <=> Ag+ (aq.) + Cl- (aq.)}$$

And for this equilibrium if you're to find the equilibrium constant, let's call it $K_{sp}$, you get:

$$ \ce{K_{sp} = [Ag+] \cdot [Cl-]} $$

(Both the concentrations are the ones taken at equilibrium.)

This $K_{sp}$ is called solubility product. It's useful to predict whether or not more solute can be dissolved in your solution or not. We do this by comparing it to something which is called the ionic product.

  • $\begingroup$ I think that a sentence about the meaning of 'saturated solution ' can be useful $\endgroup$ – user1420303 Apr 18 '17 at 13:27

The solubility product is an equilibrium constant for the dissolution of a sparingly soluble but highly ionised electrolyte. It does not apply in principle to strong soluble electrolytes or to compounds that exist in an unionised state in solution, such as $\ce{HgCl2}$.

In a reaction such as $$\ce{AgCl <=>Ag^+ + Cl^-}$$ which must have reached equilibrium, which means that some solid is present together with the solution (and so the solution is saturated), the solubility product is $$\ce{[Ag^+][Cl^-]=}K_s $$

In the general case $\ce{A_nB_M <=> nA + mB }$ then $\ce{[A]^n[B]^m=}K_s$

(The thermodynamic derivation of $K_s$ depends on using activities which may be replaced by concentration when the concentration is low as it is in the case of sparingly soluble / insoluble salts)

The solubility product illustrates a quantitative application of Le Chatelier’s principle. If, for example, the concentration of $\ce{Cl^-}$ is increased by adding sodium chloride some AgCl is precipitated until equilibrium is re-established, i.e. $\ce{[Ag^+]}$ is reduced until $\ce{[Ag^+][Cl^-]=$K_s$} $. Similarly if $\ce{[Ag^+]}$ in solution is reduced by forming a soluble complex, such as by adding ammonia, then more $\ce{Cl-}$ will be produced as more solid AgCl dissolves until equilibrium is established.

The solubility product $K_s$ can be used to calculate the concentration of an ion in solution and hence its total amount in a solution. In the case where there is a common ion it can be used to determine how solubility varies with common ion concentration. Thus it can be used to answer questions such as 'How much $\ce{Ag^+}$ is there in 100 ml of a solution of 0.1 M potassium chromate saturated with silver chromate'.


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