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# Solubility Product Constant (Ksp)

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Solubility Product Constant (Ksp)
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No more weighing

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 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 g per liter of water at 25°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 × 10 -4  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 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.

$\text{AgCl}(s) \rightleftarrows \text{Ag}^+(aq)+\text{Cl}^-(aq)$

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.

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

This equilibrium constant is called the solubility product constant $(K_{sp})$ 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  $K_{sp}$ expression. For example the formula of calcium phosphate is Ca 3 (PO 4 ) 2 . The dissociation equation and  $K_{sp}$ expression are shown below:

$\text{Ca}_3(\text{PO}_4)_2(s) \rightleftarrows 3\text{Ca}^{2+}(aq)+2\text{PO}^{3-}_4(aq) \quad K_{sp}=[\text{Ca}^{2+}]^3[\text{PO}^{3-}_4]^2$

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

 Compound $K_{sp}$ Compound $K_{sp}$ AgBr 5.0 × 10 -13 CuS 8.0 × 10 -37 AgCl 1.8 × 10 -10 Fe(OH) 2 7.9 × 10 -16 Al(OH) 3 3.0 × 10 -34 Mg(OH) 2 7.1 × 10 -12 BaCO 3 5.0 × 10 -9 PbCl 2 1.7 × 10 -5 BaSO 4 1.1 × 10 -10 PbCO 3 7.4 × 10 -14 CaCO 3 4.5 × 10 -9 PbI 2 7.1 × 10 -9 Ca(OH) 2 6.5 × 10 -6 PbSO 4 6.3 × 10 -7 Ca 3 (PO 4 ) 2 1.2 × 10 -26 Zn(OH) 2 3.0 × 10 -16 CaSO 4 2.4 × 10 -5 ZnS 3.0 × 10 -23

#### Summary

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

#### Practice

Read the material at the link below and solve the problems at the end of the reading:

#### Review

1. What does the  $K_{sp}$ tell us?
2. Which of the lead salts listed in the table above is the most soluble?
3. What is the exponent for an ion in the equation?