*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:

$$M_pX_q(s)⇌pM^{m+}(aq)+qX^{n-}(aq)$$

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.

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

Fluorite, CaF_{2}, is a slightly soluble solid that dissolves according to the equation:

The concentration of Ca^{2+} in a saturated solution of CaF_{2} is $2.15×10^{-4}\;M$. What is the solubility product of fluorite?

Solution

According to the stoichiometry of the dissolution equation, the fluoride ion molarity of a CaF_{2} solution is equal to twice its calcium ion molarity:

Substituting the ion concentrations into the *K*_{sp} expression gives

_{2}, the concentration of Mg

^{2+}is $1.31×10^{-4}\;M$. What is the solubility product for Mg(OH)

_{2}? $$Mg(OH)_2(s)⇌Mg^{2+}(aq)+2OH^-(aq)$$

## Answer:

$$8.99×10^{-12}$$

Determination of Molar Solubility from *K*_{sp}

The *K*_{sp} of copper(I) bromide, CuBr, is $6.3×10^{-9}$. Calculate the molar solubility of copper bromide.

SolutionThe dissolution equation and solubility product expression are

$$CuBr(s)⇌Cu^+(aq)+Br^-(aq)$$ $$K_{sp}=[Cu^+][Br^-]$$Following the ICE approach to this calculation yields the table

Substituting the equilibrium concentration terms into the solubility product expression and solving for *x* yields

Since the dissolution stoichiometry shows one mole of copper(I) ion and one mole of bromide ion are produced for each moles of Br dissolved, the molar solubility of CuBr is $7.9×10^{-5}\;M$.

The

*K*

_{sp}of AgI is $1.5×10^{-16}$. Calculate the molar solubility of silver iodide.

## Answer:

$1.2×10^{-8}\;M$

Determination of Molar Solubility from *K*_{sp}The *K*_{sp} of calcium hydroxide, Ca(OH)_{2}, is $1.3×10^{-6}$. Calculate the molar solubility of calcium hydroxide.

SolutionThe dissolution equation and solubility product expression are

$$Ca(OH)_2(s)⇌Ca^{2+}(aq)+2OH^-(aq)$$ $$K_{sp}=[Ca^{2+}][OH^-]^2$$The ICE table for this system is

Substituting terms for the equilibrium concentrations into the solubility product expression and solving for *x* gives

As defined in the ICE table, *x* is the molarity of calcium ion in the saturated solution. The dissolution stoichiometry shows a 1:1 relation between moles of calcium ion in solution and moles of compound dissolved, and so, the molar solubility of Ca(OH)_{2} is $6.9×10^{-3}\;M$.

The

*K*

_{sp}of PbI

_{2}is $1.4×10^{-8}$. Calculate the molar solubility of lead(II) iodide.

## Answer:

$1.5×10^{-3}\;M$

Determination of *K*_{sp} from Gram SolubilityMany of the pigments used by artists in oil-based paints 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

Before calculating the solubility product, the provided solubility must be converted to molarity:

$$[PbCrO_4]=\frac{4.6×10^{-6}\;g\;PbCrO_4}{1\;L}\times\frac{1\;mol\;PbCrO_4}{323.2\;g\;PbCrO_4}$$ $$=\frac{1.4×10^{-8}\;mol\;PbCrO_4}{1\;L}$$ $$=1.4×10^{-8}\;M$$The dissolution equation for this compound is

$$PbCrO_4(s)⇌Pb^{2+}(aq)+CrO_4^{2-}(aq)$$The dissolution stoichiometry shows a 1:1 relation between the molar amounts of compound and its two ions, and so both [Pb^{2+}] and $[CrO_4^{2-}]$ are equal to the molar solubility of PbCrO_{4}:

## Answer:

$1.69×10^{-4}$

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

Calomel, Hg_{2}Cl_{2}, is a compound composed of the diatomic ion of mercury(I), $Hg_2^{2+}$, 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 has a very low solubility, as suggested by its very small *K*_{sp}:

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

SolutionThe dissolution stoichiometry shows a 1:1 relation between the amount of compound dissolved and the amount of mercury(I) ions, and so the molar solubility of Hg_{2}Cl_{2} is equal to the concentration of $Hg_2^{2+}$ ions

Following the ICE approach results in

Substituting the equilibrium concentration terms into the solubility product expression and solving for x gives

$$K_{sp}=[Hg_2^{2+}][Cl^-]^2$$ $$1.1×10^{-18}=(x)(2x)^2$$ $$4x^3=1.1×10^{-18}$$ $$x=\sqrt[3]{\frac{1.1×10^{-18}}{4}}=6.5×10^{-7}\;M$$ $$[Hg_2^{2+}]=6.5×10^{-7}\;M=6.5×10^{-7}\;M$$ $$[Cl^-]=2x=2(6.5×10^{-7})=1.3×10^{-6}\;M$$The dissolution stoichiometry shows the molar solubility of Hg_{2}Cl_{2} is equal to $[Hg_2^{2+}]$, or $6.5×10^{-7}\;M$.

_{2}from its solubility product:

*K*

_{sp}= $6.4×10^{-9}$.

## Answer:

$1.2×10^{-3}\;M$

Various types of medical imaging techniques are used to aid diagnoses of illnesses in a noninvasive manner. One such technique utilizes the ingestion of a barium compound before taking an X-ray image. A suspension of barium sulfate, a chalky powder, is ingested by the patient. Since the *K*_{sp} of barium sulfate is $2.3×10^{-8}$, very little of it dissolves as it coats the lining of the patient’s intestinal tract. Barium-coated areas of the digestive tract then appear on an X-ray as white, allowing for greater visual detail than a traditional X-ray ([link]).

Medical imaging using barium sulfate can be used to diagnose acid reflux disease, Crohn’s disease, and ulcers in addition to other conditions.

Visit this website for more information on how barium is used in medical diagnoses and which conditions it is used to diagnose.