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ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Equilibria > Solubility > Solubility Product Constant, Ksp

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Solubility Product Constant, Ksp

The solubility product constant, \(K_{sp}\)​, is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. The more soluble a substance is, the higher the \(K_{sp}\) value it has.

Consider the general dissolution reaction below:

\[ aA (s) \rightleftharpoons cC (aq) + dD (aq) \]

To solve for the \(K_{sp}\) it is necessary to take the molarities or concentrations of the products (cC and dD) and multiply them. If there are coefficients in front of any of the products, it is necessary to raise the product to that coefficient power(and also multiply the concentration by that coefficient). This is shown below:

\[ K_{sp} = [C]^c [D]^d \]

Note that the reactant, aA, is not included in the \(K_{sp}\) equation. Solids are not included when calculating equilibrium constant expressions, because their concentrations do not change the expression; any change in their concentrations are insignificant, and therefore omitted.

Hence, \(K_{sp}\) represents the maximum amount of solid that can be dissolved in the aqueous solution. The answer will have the units of molarity, mol L-1, a measure of concentration.

Important effects

  • For highly soluble ionic compounds the ionic activities must be found instead of the concentrations that are found in slightly soluble solutions.
  • Common Ion Effect: The solubility of the reaction is reduced by the common ion. For a given equilibrium, a reaction with a common ion present has a lower  \(K_{sp}\)​, and the reaction without the ion has a greater \(K_{sp}\).
  • Salt Effect (diverse ion effect): Having an opposing effect on the \(K_{sp}\) value compared to the common ion effect, uncommon ions increase the \(K_{sp}\) value. Uncommon ions are ions other than those involved in equilibrium.
  • Ion Pairs: With an ionic pair (a cation and an anion), the \(K_{sp}\) value calculated is less than the experimental value due to ions involved in pairing. To reach the calculated \(K_{sp}\) value, more solute must be added.

Example problems:

  • What is the solubility product constant expression for \(MgF_2\)?

MgF2(s) ↔ Mg2+(aq) + 2F-(aq) so the \(K_{sp}\) = [Mg2+][F-]2

  • What is the solubility product constant expression for \(Ag_2CrO_4\)?

Ag2CrO4(s) ↔ 2Ag+(aq) + CrO42-(aq) so the \(K_{sp}\) = [Ag]+2[CrO42-]


  1. Petrucci, Ralph H., et al. General Chemistry: Principles and Modern Applications. Upper Saddle River, NJ: Prentice Hall 2007.


  • Kathryn Rashe, Lisa Peterson

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Last modified
08:31, 20 Oct 2014



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