# 19.11: Solubility Product Constant $$\left( K_\text{sp} \right)$$

At one time, a major analytical technique was gravimetric analysis. An ion would be precipitated out of solution, purified, and weighed to determine the amount of that ion in the original material. As an example, measurement of $$\ce{Ca^{2+}}$$ involved dissolving the sample in water, precipitating the calcium as calcium oxalate, purifying the precipitate, drying it, and weighing the final product. Although this approach can be very accurate (atomic weights for many elements were determined this way), the process is slow, tedious, and prone to a number of errors in technique. Newer methods are now available that measure minute amounts of calcium ions in solution without the long, involved gravimetric approach.

### Solubility Product Constant

Ionic compounds have widely differing solubilities. Sodium chloride has a solubility of about $$360 \: \text{g}$$ per liter of water at $$25^\text{o} \text{C}$$. Salts of alkali metals tend to be quite soluble. On the other end of the spectrum the solubility of zinc hydroxide is only $$4.2 \times 10^{-4} \text{g/L}$$ of water at the same temperature. Many ionic compounds containing hydroxide are relatively insoluble.

Most ionic compounds that are considered to be insoluble will still dissolve to a small extent in water. These "mostly insoluble" compounds are considered to be strong electrolytes because whatever portion of the compounds are considered to be strong electrolytes because whatever portion of the compound that dissolved also dissociates. As an example, silver chloride dissociates to a small extent into silver ions and chloride ions upon being added to water.

$\ce{AgCl} \left( s \right) \rightleftharpoons \ce{Ag^+} \left( aq \right) + \ce{Cl^-} \left( aq \right)$

The process is written as an equilibrium because the dissociation occurs only to a small extent. Therefore, an equilibrium expression can be written for the process. Keep in mind that the solid silver chloride does not have a variable concentration and so is not included in the expression.

$K_\text{sp} = \left[ \ce{Ag^+} \right] \left[ \ce{Cl^-} \right]$

This equilibrium constant is called the solubility product constant $$\left( K_\text{sp} \right)$$ and is equal to the mathematical product of the ions each raised to the power of the coefficient of the ion in the dissociation equation.

The stoichiometry of the formula of the ionic compound dictates the form of the $$K_\text{sp}$$ expression. For example the formula of calcium phosphate is $$\ce{Ca_3(PO_4)_2}$$. The dissociation equation and $$K_\text{sp}$$ expression are shown below:

$\ce{Ca_3(PO_4)_2} \left( s \right) \rightleftharpoons 3 \ce{Ca^{2+}} \left( aq \right) + 2 \ce{PO_4^{3-}} \left( aq \right) \: \: \: K_\text{sp} = \left[ \ce{Ca^{2+}} \right]^3 \left[ \ce{PO_4^{3-}} \right]^2$

The table below lists solubility product constants for some common nearly insoluble ionic compounds.

 Table 19.11.1: Solubility Product Constants $$\left( 25^\text{o} \text{C} \right)$$ Compound $$K_\text{sp}$$ Compound $$K_\text{sp}$$ $$\ce{AgBr}$$ $$5.0 \times 10^{-13}$$ $$\ce{CuS}$$ $$8.0 \times 10^{-37}$$ $$\ce{AgCl}$$ $$1.8 \times 10^{-10}$$ $$\ce{Fe(OH)_2}$$ $$7.9 \times 10^{-16}$$ $$\ce{Al(OH)_3}$$ $$3.0 \times 10^{-34}$$ $$\ce{Mg(OH)_2}$$ $$7.1 \times 10^{-12}$$ $$\ce{BaCO_3}$$ $$5.0 \times 10^{-9}$$ $$\ce{PbCl_2}$$ $$1.7 \times 10^{-5}$$ $$\ce{BaSO_4}$$ $$1.1 \times 10^{-10}$$ $$\ce{PbCO_3}$$ $$7.4 \times 10^{-14}$$ $$\ce{CaCO_3}$$ $$4.5 \times 10^{-9}$$ $$\ce{PbI_2}$$ $$7.1 \times 10^{-9}$$ $$\ce{Ca(OH)_2}$$ $$6.5 \times 10^{-6}$$ $$\ce{PbSO_4}$$ $$6.3 \times 10^{-7}$$ $$\ce{Ca_3(PO_4)_2}$$ $$1.2 \times 10^{-26}$$ $$\ce{Zn(OH)_2}$$ $$3.0 \times 10^{-16}$$ $$\ce{CaSO_4}$$ $$2.4 \times 10^{-5}$$ $$\ce{ZnS}$$ $$3.0 \times 10^{-23}$$

### Summary

• The solubility product constant is defined.
• Calculations using solubility product constants are illustrated.

### Contributors

• CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.