<h2>Ksp and Solubility</h2>
The Ksp of a slightly soluble <a data-lesson="ionic-solids" data-subject="chemistry" data-tutorial="liquids-and-solids" href="https://app.ulearngo.com/chemistry/liquids-and-solids/ionic-solids">ionic</a> compound may be simply related to its measured solubility provided <a data-lesson="summarizing-the-dissolution-process" data-subject="chemistry" data-tutorial="solutions-and-colloids" href="https://app.ulearngo.com/chemistry/solutions-and-colloids/summarizing-the-dissolution-process">the dissolution process</a> 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 <a data-lesson="saturated-and-unsaturated-structures" data-subject="chemistry" data-tutorial="organic-molecules" href="https://app.ulearngo.com/chemistry/organic-molecules/saturated-and-unsaturated-structures">saturated</a> solution.
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<h2>Example</h2>
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<h3>Calculation of Ksp from Equilibrium Concentrations</h3>
We began the tutorial with an informal discussion of how the mineral <a data-lesson="introducing-equilibria-of-other-reaction-classes" data-subject="chemistry" data-tutorial="equilibria-other-reaction-classes" href="https://app.ulearngo.com/chemistry/equilibria-other-reaction-classes/introducing-equilibria-of-other-reaction-classes">fluorite</a> (<a data-page="91" href="https://app.ulearngo.com/tutorials/equilibria-of-other-reaction-classes/introducing-equilibria-of-other-reaction-classes">see introductory lesson</a>) is formed. Fluorite, CaF2, is a slightly soluble <a data-lesson="melting-and-freezing" data-subject="chemistry" data-tutorial="liquids-and-solids" href="https://app.ulearngo.com/chemistry/liquids-and-solids/melting-and-freezing">solid</a> 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 <a data-lesson="the-solubility-product" data-subject="chemistry" data-tutorial="equilibria-other-reaction-classes" href="https://app.ulearngo.com/chemistry/equilibria-other-reaction-classes/the-solubility-product">the solubility product</a> of fluorite?
<h3>Solution</h3>

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.
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<h2>Example</h2>
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<h3>Determination of Molar Solubility from KspThe Ksp of copper(I) bromide, Cu</h3>
Br, is 6.3 \(×\) 10–9. Calculate the molar solubility of copper bromide.
<h3>Solution</h3>

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:
<img alt="This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, “C u B r equilibrium arrow C u superscript positive sign plus B r superscript negative sign.” Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, x, 0 plus x equals x. The third column has the following 0, x, 0 plus x equals x." src="https://data.ulearngo.com/assets/865fb3ce-8116-420e-aace-8ee66140937f"/>
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.
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<h2>Example</h2>
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<h3>Determination of Molar Solubility from Ksp, Part II</h3>

The Ksp of calcium hydroxide, Ca(OH)2, is 1.3 \(×\) 10–6. Calculate the molar solubility of calcium hydroxide.

<h3>Solution</h3>

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:
<img alt="This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), and Equilibrium concentration ( M ). The second column has the header of, “C a ( O H ) subscript 2 equilibrium arrow C a superscript 2 positive sign plus 2 O H superscript negative sign.” Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, x, and 0 plus x equals x. The third column has the following 0, 2 x, and 0 plus 2 x equals 2 x." src="https://data.ulearngo.com/assets/0fc5ecb5-c963-4f4d-afdc-db8cdd4785f2"/>
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.
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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 <a data-lesson="bond-order" data-subject="chemistry" data-tutorial="covalent-bonding-continued" href="https://app.ulearngo.com/chemistry/covalent-bonding-continued/bond-order">order</a> to use it in the solubility product expression. The example below shows how to perform those unit conversions before determining the solubility product equilibrium.
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<h2>Example</h2>
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<h3>Determination of Ksp from Gram Solubility</h3>
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.
<div class="wp-caption alignnone" style="width: 650px"><img alt="A photograph is shown of a portion of an oil painting which reveals colors of orange, brown, yellow, green, blue, and purple colors in its strokes. A few water droplets rest on the surface." height="433" src="https://data.ulearngo.com/assets/bc626a84-bdef-47b4-95b1-c56c4ac80ac0" width="650"/>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)</div>
<h3>Solution</h3>
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:
<img alt="This figure shows four horizontally oriented rectangles. The first three from the left are shaded green and the last one at the right is shaded white. Right pointing arrows between the rectangles are labeled “1,” “2,” and “3” moving left to right across the diagram. The first rectangle is labeled “Solubility of P b C r O subscript 4, in g divdided by L.” The second rectangle is labeled “[ P b C r O subscript 4 ], in m o l divided by L.” The third is labeled “[ P b superscript 2 plus] and [ C r O subscript 4 superscript 2 negative ].” The fourth rectangle is labeled “K subscript s p.”" src="https://data.ulearngo.com/assets/95dd8158-2a46-4ad7-a066-2bf0341be168"/>
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Use the <a data-lesson="molar-mass" data-subject="chemistry" data-tutorial="quantitative-aspects-chemical-change" href="https://app.ulearngo.com/chemistry/quantitative-aspects-chemical-change/molar-mass">molar mass</a> 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:
\(\begin{array}{}\\ \\ \left[{\text{PbCrO}}_{4}\right]=\cfrac{4.6\;×\;{10}^{-6}{\text{g PbCrO}}_{4}}{1\;\text{L}}\phantom{\rule{0.4em}{0ex}}\;×\;\phantom{\rule{0.4em}{0ex}}\cfrac{1{\;\text{mol PbCrO}}_{4}}{323.2{\;\text{g PbCrO}}_{4}}\;\\ =\;\cfrac{1.4\;×\;{10}^{-8}{\;\text{mol PbCrO}}_{4}}{1\;\text{L}}\;\\ =\;1.4\;×\;{10}^{-8}M\end{array}\)
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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:
\(\left[{\text{Pb}}^{\text{2+}}\right]=\left[{\text{CrO}}_{4}{}^{\text{2−}}\right]=\text{1.4}\;×\;{10}^{-8}M\)
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Solve. Ksp = [Pb2+]\(\left[{\text{CrO}}_{4}{}^{\text{2−}}\right]\) = (1.4 \(×\) 10–8)(1.4 \(×\) 10–8) = 2.0 \(×\) 10–16
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<h2>Example</h2>
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<h3>Calculating the Solubility of Hg2Cl2</h3>

Calomel, Hg2Cl2, is a compound composed of the diatomic ion of mercury(I), \({\text{Hg}}_{2}{}^{\text{2+}},\) and chloride ions, Cl–. Although most <a data-lesson="group-12" data-subject="chemistry" data-tutorial="metals-metalloids-nonmetals" href="https://app.ulearngo.com/chemistry/metals-metalloids-nonmetals/group-12">mercury</a> <a data-lesson="compounds-2" data-subject="chemistry" data-tutorial="quantitative-aspects-chemical-change" href="https://app.ulearngo.com/chemistry/quantitative-aspects-chemical-change/compounds-2">compounds</a> 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.
<h3>Solution</h3>

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:<img alt="This figure shows four horizontally oriented light green rectangles. Right pointing arrows are placed between them. The first rectangle is labeled “Determine the direction of change.” The second rectangle is labeled “Determine x and the equilibrium concentrations.” The third is labeled “Solve for x and the equilibrium concentrations.” The fourth rectangle is labeled “Check the math.”" src="https://data.ulearngo.com/assets/9e534552-2511-4703-a3f6-701e088fbd23"/>
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Determine the direction of change. Before any Hg2Cl2 dissolves, Q is zero, and the reaction will <a data-lesson="effect-of-pressure-on-equilibrium" data-subject="chemistry" data-tutorial="chemical-equilibrium" href="https://app.ulearngo.com/chemistry/chemical-equilibrium/effect-of-pressure-on-equilibrium">shift</a> to the right to reach equilibrium.
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Determine x and equilibrium concentrations. Concentrations and changes are given in the following ICE table:
<img alt="This table has two main columns and four rows. The first row for the first column does not have a heading and then has the following in the first column: Initial concentration ( M ), Change ( M ), Equilibrium concentration ( M ). The second column has the header of, “H g subscript 2 C l subscript 2 equilibrium arrow H g subscript 2 superscript 2 positive sign plus 2 C l superscript negative sign.” Under the second column is a subgroup of three rows and three columns. The first column is blank. The second column has the following: 0, x, 0 plus x equals x. The third column has the following: 0, 2 x, 0 plus 2 x equals 2 x." src="https://data.ulearngo.com/assets/f2410f0b-3be5-409b-9fd3-28ea1f3fdfe8"/>
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.
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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.
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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.
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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 <a data-lesson="conversion-temperature-units" data-subject="chemistry" data-tutorial="chemistry-essential-ideas" href="https://app.ulearngo.com/chemistry/chemistry-essential-ideas/conversion-temperature-units">temperature</a> until Q equals Ksp.

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Ksp and Solubility

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<em>K</em><sub>sp</sub> and Solubility

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