*K*

_{sp}, the solubility product constant, is an equilibrium constant that can be used to calculate molar solubilities.

When a substance dissolves, it breaks apart into individual molecules or ions. As a solute is added to a solvent, more and more of the solute molecules dissolve until eventually the solution reaches its
A saturated solution of an ionic solid MA is in equilibrium with its aqueous ion components, M
For example, an aqueous solution of silver bromide, AgBr, can be represented by the following equilibrium equation:
As always, the equilibrium constant is expressed as a product of the concentrations (denoted by square brackets) of the nonsolid species. For aqueous solutions, the equilibrium constant for a solid is called the
As usual for equilibrium constants,
Clearly silver chloride is more soluble than silver bromide because
While the value of
Let $s=\lbrack{\rm{CrO}_4}^{2-}\rbrack$. The reaction equation shows that for each mole of CrO
So even though

**saturation point**, which is the concentration at which no more solute dissolves in a solution for a given temperature and pressure. An**unsaturated solution**contains less than the maximum amount of dissolved solute normally possible at a certain temperature. A**saturated solution**contains the maximum amount of dissolved solute normally possible at a certain temperature. A**supersaturated solution**contains more than the maximum amount of dissolved solute normally possible at a certain temperature. Thus, if solid solute is added to a saturated aqueous solution, the solute will be present in both solid and aqueous forms. Put another way, in a saturated solution, the solid and aqueous forms of the solute are in equilibrium with one another. This relationship can therefore be represented by the following equilibrium equation for an ionic solid represented by the notation MA.${{{\rm{M}}_m}{{\rm{A}}_a}(s)}\rightleftharpoons{m{{\rm{M}}^{a+}}({aq})+a{{\rm{A}}^{m-}}(aq)}$

#### Saturated Solution in Equilibrium

^{+}and A

^{-}

${\rm{AgBr}}(s)\rightleftharpoons{\rm{Ag}^+}(aq)+{\rm{Br}^-}(aq)$

**solubility product constant (**. For AgBr, the solubility product constant is*K*_{sp})$K_{\rm{sp}}=[\rm{Ag}^+][\rm{Br}^-]=5.35\times10^{-13}$

*K*_{sp}has no units. Additionally, the AgBr(*s*) is not included in the calculation of*K*_{sp}because solids are never included in the calculation of equilibrium constants.*K*_{sp}provides data on how soluble a solid compound is in water. For AgBr, the small value of*K*_{sp}reveals that AgBr is only slightly soluble in water because the concentration of ions at saturation is so small. For silver chloride, AgCl, the equilibrium equation and*K*_{sp}are:$\begin{gathered}{\rm{AgCl}}(s)\rightleftharpoons{\rm{Ag}^+}(aq)+{\rm{Cl}^-}(aq) \\ K_{\rm{sp}}=[{\rm{Ag}}^+][{\rm{Cl}}^-]=1.77\times10^{-10}\end{gathered}$

*K*_{sp}shows that there is a 1,000 times greater concentration of ions at the saturation point. The molarity of a compound in a saturated solution is known as its**molar solubility**, and it can be calculated from the compound's*K*_{sp}. Calculating molar solubility is useful because it indicates the maximum number of moles of solute that can be expected to dissolve per liter of solution. For example, let the variable*s*represent the molar solubility of AgCl. The reaction equation shows that for every mole of AgCl dissolved, 1 mole of Ag^{+}and 1 mole of Cl^{–}are produced. Thus, $s=\lbrack\rm{Ag}^+\rbrack=\lbrack\rm{Cl}^-\rbrack$. The molar solubility of AgCl can therefore be calculated from the solubility product constant:$\begin{aligned}K_{\rm{sp}}&= s\cdot s=s^2\\s&=\sqrt{K_{\rm{sp}}}\\&= \sqrt{1.77\times10^{-10}}\\&= 1.33\times10^{-5}\;\rm M\end{aligned}$

*K*_{sp}is related to the molar solubility of a compound,*K*_{sp}can only be used to directly compare the solubilities of two solutes if they are of the same type; that is, if both are MA or M_{2}A or MA_{2}, etc. For compounds of the same type, the*K*_{sp}values have the same mathematical relationship to the solubilities. Recall that*K*_{sp}is an equilibrium constant, so it is the product concentrations multiplied together divided by the reactant concentrations multiplied together, with each term raised to the power of its coefficient in the balanced equation. The molar solubility therefore depends on the concentrations of the ionic products. When the solutes are not of the same type, the relationships of*K*_{sp}to solubility are not the same and must be calculated. For example,*K*_{sp}for silver chromate, Ag_{2}CrO_{4}, is $1.12\times10^{-12}$, but the equilibrium equation is${\rm{Ag}_2\rm{CrO}_4}(s) \rightleftharpoons{2\rm{Ag}^+}(aq)+{{\rm{CrO}_4}^{2-}}(aq)$

$K_{\rm{sp}}=\lbrack\rm{Ag}^+\rbrack^2\lbrack\rm{CrO_4}^{2-}\rbrack=1.12\times10^{-12}$

_{4}^{2–}that is produced, two moles of Ag^{+}are produced, so $\rm{[Ag}^+\rm{]}=2s$, and for Ag_{2}CrO_{4},$\begin{aligned}K_{\rm{sp}}&=(2s)^2s=4s^3=1.12\times10^{-12}\\{\lbrack}{\rm{CrO}_{4}}^{2-}\rbrack&=s=\sqrt[3]{\frac{1.12\times10^{-12}}4}=6.54\times10^{-5}\;\rm{M}\\{\lbrack}\rm{Ag}^+\rbrack&=2s=1.31\times10^{-4}\;\rm{M}\end{aligned}$

*K*_{sp}for AgCl is greater than*K*_{sp}for Ag_{2}CrO_{4}, the molar solubility of Ag^{+}is less for AgCl, which means AgCl is less soluble than Ag_{2}CrO_{4.}