Chemistry » Equilibria of Other Reaction Classes » Precipitation and Dissolution

# Ksp and Solubility

## Ksp and Solubility

The Ksp 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 Ksp 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 Ksp from Equilibrium Concentrations

We began the tutorial with an informal discussion of how the mineral fluorite (see introductory lesson) is formed. Fluorite, CaF2, 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 Ca2+ in a saturated solution of CaF2 is 2.15 $$×$$ 10–4M; therefore, that of F is 4.30 $$×$$ 10–4M, that is, twice the concentration of Ca2+. What is the solubility product of fluorite?

### Solution

First, write out the Ksp expression, then substitute in concentrations and solve for Ksp:

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

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 Ksp.

## Example

### Determination of Molar Solubility from KspThe Ksp 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 Ksp:

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 $$×$$ 10–5M.

## Example

### Determination of Molar Solubility from Ksp, Part II

The Ksp 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 Ksp:

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 $$×$$ 10–3M.

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 Ksp 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, PbCrO4, is 4.6 $$×$$ 10–6 g/L. Determine the solubility product for PbCrO4.

Oil paints contain pigments that are very slightly soluble in water. In addition to chrome yellow (PbCrO4), examples include Prussian blue (Fe7(CN)18), the reddish-orange color vermilion (HgS), and green color veridian (Cr2O3). (credit: Sonny Abesamis)

### Solution

We are given the solubility of PbCrO4 in grams per liter. If we convert this solubility into moles per liter, we can find the equilibrium concentrations of Pb2+ and $${\text{CrO}}_{4}{}^{\text{2−}},$$ then Ksp:

1. Use the molar mass of PbCrO4$$\left(\cfrac{323.2\;\text{g}}{1\;\text{mol}}\right)$$to convert the solubility of PbCrO4 in grams per liter into moles per liter:

$$\begin{array}{}\\ \\ \left[{\text{PbCrO}}_{4}\right]=\cfrac{4.6\;×\;{10}^{-6}{\text{g PbCrO}}_{4}}{1\;\text{L}}\phantom{\rule{0.4em}{0ex}}\;×\;\phantom{\rule{0.4em}{0ex}}\cfrac{1{\;\text{mol PbCrO}}_{4}}{323.2{\;\text{g PbCrO}}_{4}}\;\\ =\;\cfrac{1.4\;×\;{10}^{-8}{\;\text{mol PbCrO}}_{4}}{1\;\text{L}}\;\\ =\;1.4\;×\;{10}^{-8}M\end{array}$$

2. The chemical equation for the dissolution indicates that 1 mol of PbCrO4 gives 1 mol of Pb2+(aq) and 1 mol of$${\text{CrO}}_{4}{}^{\text{2−}}\left(aq\right)\text{:}$$

$${\text{PbCrO}}_{4}\left(s\right)⇌{\text{Pb}}^{\text{2+}}\left(aq\right)+{\text{CrO}}_{4}{}^{\text{2−}}\left(aq\right)$$

Thus, both [Pb2+] and $$\left[{\text{CrO}}_{4}{}^{\text{2−}}\right]$$ are equal to the molar solubility of PbCrO4:

$$\left[{\text{Pb}}^{\text{2+}}\right]=\left[{\text{CrO}}_{4}{}^{\text{2−}}\right]=\text{1.4}\;×\;{10}^{-8}M$$

3. Solve. Ksp = [Pb2+]$$\left[{\text{CrO}}_{4}{}^{\text{2−}}\right]$$ = (1.4 $$×$$ 10–8)(1.4 $$×$$ 10–8) = 2.0 $$×$$ 10–16

## Example

### Calculating the Solubility of Hg2Cl2

Calomel, Hg2Cl2, 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 Hg2Cl2.

### Solution

The molar solubility of Hg2Cl2 is equal to the concentration of $${\text{Hg}}_{2}{}^{\text{2+}}$$ ions because for each 1 mol of Hg2Cl2 that dissolves, 1 mol of $${\text{Hg}}_{2}{}^{\text{2+}}$$ forms:

1. Determine the direction of change. Before any Hg2Cl2 dissolves, Q is zero, and the reaction will shift to the right to reach equilibrium.

2. Determine x and equilibrium concentrations. Concentrations and changes are given in the following ICE table:

Note that the change in the concentration of Cl (2x) 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. Hg2Cl2 is a pure solid, so it does not appear in the calculation.

3. Solve for x and the equilibrium concentrations. We substitute the equilibrium concentrations into the expression for Ksp 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 Hg2Cl2 is equal to $$[{\text{Hg}}_{\text{2}}{}^{\text{2+}}],$$ or 6.5 $$×$$ 10–7M.

4. Check the work. At equilibrium, Q = Ksp:

$$Q = [\text{Hg}_2^{\;\;2+}][\text{Cl}^{-}]^2 = (6.5\;\times\;10^{-7})(1.3\;\times\;10^{-6})^2 = 1.1\;\times\;10^{-18}$$

The calculations check.

Tabulated Ksp 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 Ksp at equilibrium; if Q is less than Ksp, the solid will dissolve until Q equals Ksp; if Q is greater than Ksp, precipitation will occur at a given temperature until Q equals Ksp.

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