*K*_{sp} and Solubility

The *K _{sp}* of a slightly soluble ionic compound may be simply related to its measured solubility provided the dissolution process involves only dissociation and solvation, for example:

\({\text{M}}_{p}{\text{X}}_{q}\left(s\right)⇌p{\text{M}}^{\text{m+}}\left(aq\right)+q{\text{X}}^{\text{n−}}\left(aq\right)\)

For cases such as these, one may derive *K _{sp}* values from provided solubilities, or vice-versa. Calculations of this sort are most conveniently performed using a compound’s molar solubility, measured as moles of dissolved solute per liter of saturated solution.

## Example

### Calculation of *K*_{sp} from Equilibrium Concentrations

We began the tutorial with an informal discussion of how the mineral fluorite (see introductory lesson) is formed. Fluorite, CaF_{2}, is a slightly soluble solid that dissolves according to the equation:

\({\text{CaF}}_{2}\left(s\right)⇌{\text{Ca}}^{\text{2+}}\left(aq\right)+{\text{2F}}^{\text{−}}\left(aq\right)\)

The concentration of Ca^{2+} in a saturated solution of CaF_{2} is 2.15 \(×\) 10^{–4}*M*; therefore, that of F^{–} is 4.30 \(×\) 10^{–4}*M*, that is, twice the concentration of Ca^{2+}. What is the solubility product of fluorite?

### Solution

First, write out the *K*_{sp} expression, then substitute in concentrations and solve for *K*_{sp}:

\({\text{CaF}}_{2}\left(s\right)⇌{\text{Ca}}^{\text{2+}}\left(aq\right)+{\text{2F}}^{\text{−}}\left(aq\right)\)

A saturated solution is a solution at equilibrium with the solid. Thus:

\(K_{\text{sp}} = [\text{Ca}^{2+}][\text{F}^{-}]^2 = (2.1\;\times\;10^{-4})(4.2\;\times\;10^{-4})^2 = 3.7\;\times\;10^{-11}\)

As with other equilibrium constants, we do not include units with *K*_{sp}.

## Example

### Determination of Molar Solubility from *K*_{sp}The *K*_{sp} of copper(I) bromide, Cu

Br, is 6.3 \(×\) 10^{–9}. Calculate the molar solubility of copper bromide.

### Solution

The solubility product of copper(I) bromide is 6.3 \(×\) 10^{–9}.

The reaction is:

\(\text{CuBr}\left(s\right)⇌{\text{Cu}}^{\text{+}}\left(aq\right)+{\text{Br}}^{\text{−}}\left(aq\right)\)

First, write out the solubility product expression:

\({K}_{\text{sp}}=\left[{\text{Cu}}^{\text{+}}\right]\left[{\text{Br}}^{\text{−}}\right]\)

Create an ICE table (as introduced in the tutorial on fundamental equilibrium concepts), leaving the CuBr column empty as it is a solid and does not contribute to the *K*_{sp}:

At equilibrium:

\(K_{\text{sp}} = [\text{Cu}^{+}][\text{Br}^{-}]\)

\(6.3\;×\;{10}^{-9}=\left(x\right)\left(x\right)={x}^{2}\)

\(x=\sqrt{(6.3\;×\;{10}^{-9})}=\text{7.9}\;×\;{10}^{-5}\)

Therefore, the molar solubility of CuBr is 7.9 \(×\) 10^{–5}*M*.

## Example

### Determination of Molar Solubility from *K*_{sp}, Part II

The *K*_{sp} of calcium hydroxide, Ca(OH)_{2}, is 1.3 \(×\) 10^{–6}. Calculate the molar solubility of calcium hydroxide.

### Solution

The solubility product of calcium hydroxide is 1.3 \(×\) 10^{–6}.

The reaction is:

\({\text{Ca(OH)}}_{2}\left(s\right)⇌{\text{Ca}}^{\text{2+}}\left(aq\right)+{\text{2OH}}^{\text{−}}\left(aq\right)\)

First, write out the solubility product expression:

\({K}_{\text{sp}}=\left[{\text{Ca}}^{\text{2+}}\right]\left[{\text{OH}}^{\text{−}}\right]^{2}\)

Create an ICE table, leaving the Ca(OH)_{2} column empty as it is a solid and does not contribute to the *K*_{sp}:

At equilibrium:

\({K}_{\text{sp}}=\left[{\text{Ca}}^{\text{2+}}\right]\left[{\text{OH}}^{\text{−}}\right]^{2}\)

\(1.3\;\times\;10^{-6} = (x)(2x)^2 = (x)(4x^2) = 4x^3\)

\(x=\sqrt[3]{\;\cfrac{1.3\;×\;{10}^{-6}}{4}\;}=\text{6.9}\;×\;{10}^{-3}\)

Therefore, the molar solubility of Ca(OH)_{2} is 6.9 \(×\) 10^{–3}*M*.

Note that solubility is not always given as a molar value. When the solubility of a compound is given in some unit other than moles per liter, we must convert the solubility into moles per liter (i.e., molarity) in order to use it in the solubility product expression. The example below shows how to perform those unit conversions before determining the solubility product equilibrium.

## Example

### Determination of *K*_{sp} from Gram Solubility

Many of the pigments used by artists in oil-based paints (see the figure below) are sparingly soluble in water. For example, the solubility of the artist’s pigment chrome yellow, PbCrO_{4}, is 4.6 \(×\) 10^{–6} g/L. Determine the solubility product for PbCrO_{4}.

### Solution

We are given the solubility of PbCrO_{4} in grams per liter. If we convert this solubility into moles per liter, we can find the equilibrium concentrations of Pb^{2+} and \({\text{CrO}}_{4}{}^{\text{2−}},\) then *K*_{sp}:

*Use the molar mass of PbCrO*\(\left(\cfrac{323.2\;\text{g}}{1\;\text{mol}}\right)\)_{4}*to convert the solubility of PbCrO*_{4}in grams per liter into moles per liter:\(\begin{array}{}\\ \\ \left[{\text{PbCrO}}_{4}\right]=\cfrac{4.6\;×\;{10}^{-6}{\text{g PbCrO}}_{4}}{1\;\text{L}}\phantom{\rule{0.4em}{0ex}}\;×\;\phantom{\rule{0.4em}{0ex}}\cfrac{1{\;\text{mol PbCrO}}_{4}}{323.2{\;\text{g PbCrO}}_{4}}\;\\ =\;\cfrac{1.4\;×\;{10}^{-8}{\;\text{mol PbCrO}}_{4}}{1\;\text{L}}\;\\ =\;1.4\;×\;{10}^{-8}M\end{array}\)

*The chemical equation for the dissolution indicates that 1 mol of PbCrO*\({\text{CrO}}_{4}{}^{\text{2−}}\left(aq\right)\text{:}\)_{4}gives 1 mol of Pb^{2+}(aq) and 1 mol of\({\text{PbCrO}}_{4}\left(s\right)⇌{\text{Pb}}^{\text{2+}}\left(aq\right)+{\text{CrO}}_{4}{}^{\text{2−}}\left(aq\right)\)

Thus, both [Pb

^{2+}] and \(\left[{\text{CrO}}_{4}{}^{\text{2−}}\right]\) are equal to the molar solubility of PbCrO_{4}:\(\left[{\text{Pb}}^{\text{2+}}\right]=\left[{\text{CrO}}_{4}{}^{\text{2−}}\right]=\text{1.4}\;×\;{10}^{-8}M\)

*Solve. K*_{sp}= [Pb^{2+}]\(\left[{\text{CrO}}_{4}{}^{\text{2−}}\right]\) = (1.4 \(×\) 10^{–8})(1.4 \(×\) 10^{–8}) = 2.0 \(×\) 10^{–16}

## Example

### Calculating the Solubility of Hg_{2}Cl_{2}

Calomel, Hg_{2}Cl_{2}, is a compound composed of the diatomic ion of mercury(I), \({\text{Hg}}_{2}{}^{\text{2+}},\) and chloride ions, Cl^{–}. Although most mercury compounds are now known to be poisonous, eighteenth-century physicians used calomel as a medication. Their patients rarely suffered any mercury poisoning from the treatments because calomel is quite insoluble:

\({\text{Hg}}_{2}{\text{Cl}}_{2}\left(s\right)⇌{\text{Hg}}_{2}{}^{\text{2+}}\left(aq\right)+{\text{2Cl}}^{\text{−}}\left(aq\right)\phantom{\rule{4em}{0ex}}{K}_{\text{sp}}=\text{1.1}\;×\;{10}^{-18}\)

Calculate the molar solubility of Hg_{2}Cl_{2}.

### Solution

The molar solubility of Hg_{2}Cl_{2} is equal to the concentration of \({\text{Hg}}_{2}{}^{\text{2+}}\) ions because for each 1 mol of Hg_{2}Cl_{2} that dissolves, 1 mol of \({\text{Hg}}_{2}{}^{\text{2+}}\) forms:

*Determine the direction of change.*Before any Hg_{2}Cl_{2}dissolves,*Q*is zero, and the reaction will shift to the right to reach equilibrium.*Determine*x*and equilibrium concentrations.*Concentrations and changes are given in the following ICE table:Note that the change in the concentration of Cl

^{–}(2*x*) is twice as large as the change in the concentration of \({\text{Hg}}_{2}{}^{\text{2+}}\) (*x*) because 2 mol of Cl^{–}forms for each 1 mol of \({\text{Hg}}_{2}{}^{\text{2+}}\) that forms. Hg_{2}Cl_{2}is a pure solid, so it does not appear in the calculation.*Solve for*x*and the equilibrium concentrations.*We substitute the equilibrium concentrations into the expression for*K*_{sp}and calculate the value of*x*:\({K}_{\text{sp}}=\left[{\text{Hg}}_{\text{2}}{}^{\text{2+}}\right]\left[{\text{Cl}}^{\text{−}}\right]^{2}\)

\(1.1\;×\;{10}^{-18}=\left(x\right)\left(2x\right)^{2}\)

\(4{x}^{3}=\text{1.1}\;×\;{10}^{-18}\)

\(x=\sqrt[3]{\left(\cfrac{1.1\;×\;{10}^{-18}}{4}\right)}=\text{6.5}\;×\;{10}^{-7}\;M\)

\(\left[{\text{Hg}}_{2}{}^{\text{2+}}\right]=\text{6.5}\;×\;{10}^{-7}\;M=\text{6.5}\;×\;{10}^{-7}\;M\)

\([{\text{Cl}}^{\text{−}}]=2x=\text{2(6.5}\;×\;{10}^{-7})=1.3\;×\;{10}^{-6}\;M\)

The molar solubility of Hg

_{2}Cl_{2}is equal to \([{\text{Hg}}_{\text{2}}{}^{\text{2+}}],\) or 6.5 \(×\) 10^{–7}*M*.*Check the work.*At equilibrium,*Q*=*K*_{sp}:\(Q = [\text{Hg}_2^{\;\;2+}][\text{Cl}^{-}]^2 = (6.5\;\times\;10^{-7})(1.3\;\times\;10^{-6})^2 = 1.1\;\times\;10^{-18}\)

The calculations check.

Tabulated *K*_{sp} values can also be compared to reaction quotients calculated from experimental data to tell whether a solid will precipitate in a reaction under specific conditions: *Q* equals *K*_{sp} at equilibrium; if *Q* is less than *K*_{sp}, the solid will dissolve until *Q* equals *K*_{sp}; if *Q* is greater than *K*_{sp}, precipitation will occur at a given temperature until *Q* equals *K*_{sp}.