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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 (in aqueous solutions):

$\ce{aA(s) <=> cC(aq) + dD(aq)} \nonumber$

To solve for the $$K_{sp}$$ it is necessary to take the molarities or concentrations of the products ($$\ce{cC}$$ and $$\ce{dD}$$) 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:

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

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 extent that a solid that can dissolved in solution.

Exercise 1: Magnesium Floride

What is the solubility product constant expression for $$MgF_2$$?

Solution

The relavant equilibrium is

$MgF_{2(s)} \rightleftharpoons Mg^{2+}_{(aq)} + 2F^-_{(aq)} \nonumber$

so the associated equilibrium constant is

$K_{sp} = [Mg^{2+}][F^-]^2\nonumber$

Exercise 2: Silver Chromate

What is the solubility product constant expression for $$Ag_2CrO_4$$?

Solution

The relavant equilibrium is

$Ag_2CrO_{4(s)} \rightleftharpoons 2Ag^+_{(aq)} + CrO^{2-}_{4(aq)}\nonumber$

so the associated equilibrium constant is

$K_{sp} = [Ag^{+}]^2[CrO_4^{2-}]\nonumber$

## 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.

## References

• Clark, Roy W.; Bonicamp, Judith M. " Solubility and Solubility Products (about J. Chem. Educ. 1998, 75, 1179-1181 and J. Chem. Educ. 1998, 75, 1182-1185)." J. Chem. Educ. 2000 77 1558.
• Euler, William B.; Kirschenbaum, Louis J.; Ruekberg, Ben. "Determination of $$K_{sp}$$, ΔG0, ΔH0, and ΔS0" J. Chem. Educ. 2000 77 1039.
• Martin, R. Bruce. "Solubility and Solubility Products (about J. Chem. Educ. 1998, 75, 1179-1181 and J. Chem. Educ. 1998, 75, 1182-1185)." J. Chem. Educ. 2000 77 1558.
• Petrucci, Ralph H., et al. General Chemistry: Principles and Modern Applications. Upper Saddle River, NJ: Prentice Hall 2007.
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