Ksp and Solubility

Calculation of K_sp from Equilibrium Concentrations
Fluorite, CaF₂, is a slightly soluble solid that dissolves according to the equation:

$$CaF_2(s)⇌Ca^{2+}(aq)+2F^-(aq)$$

The concentration of Ca²⁺ in a saturated solution of CaF₂ 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 CaF₂ solution is equal to twice its calcium ion molarity:

$$[F^-]=(\text{2 mol }F^-/\text{1 mol }Ca^{2+})=(2)(2.15×10^{-4}\;M)=4.30×10^{-4}\;M$$

Substituting the ion concentrations into the K_sp expression gives

$$K_{sp}=[Ca^{2+}][F^-]=(2.15×10^{-4})(4.30×10^{-4})^2=3.98×10^{-11}$$

Check Your LearningIn a saturated solution of Mg(OH)₂, the concentration of Mg²⁺ 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

$$K_{sp}=[Cu^+][Br^-]$$ $$6.3×10^{-9}=(x)(x)=x^2$$ $$x=\sqrt{(6.3×10^{-9})}=7.9×10^{-5}\;M$$

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 $7.9×10^{-5}\;M$.

Check Your Learning
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_spThe 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

$$K_{sp}=[Ca^{2+}][OH^-]^2$$ $$1.3×10^{-6}=(x)(2x)^2=(x)(4x^2)=4x^3$$ $$x=\sqrt[3]{\frac{1.3×10^{-6}}{4}}=6.9×10^{-3}\;M$$

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 $6.9×10^{-3}\;M$.

Check Your Learning
The K_sp of PbI₂ 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, PbCrO₄, is $4.6×10^{-6}\;$g/L. Determine the solubility product for PbCrO₄.

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 [Pb²⁺] and $[CrO_4^{2-}]$ are equal to the molar solubility of PbCrO₄:

$$[Pb^{2+}]=[CrO_4^{2-}]=1.4×10^{-8}\;M$$ $$K_{sp}=[Pb^{2+}][CrO_4^{2-}]=(1.4×10^{-8})(1.4×10^{-8})=2.0×10^{-16}$$

Check Your LearningThe solubility of TlCl [thallium(I) chloride], an intermediate formed when thallium is being isolated from ores, is 3.12 grams per liter at 20 °C. What is its solubility product?

Answer:

$1.69×10^{-4}$

Calculating the Solubility of Hg₂Cl₂
Calomel, Hg₂Cl₂, 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:

$$Hg_2Cl_2(s)⇌Hg_2^{2+}(aq)+2Cl^-(aq)\qquad K_{sp}=1.1×10^{-18}$$

Calculate the molar solubility of Hg₂Cl₂.

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 Hg₂Cl₂ 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 Hg₂Cl₂ is equal to $[Hg_2^{2+}]$, or $6.5×10^{-7}\;M$.

Check Your LearningDetermine the molar solubility of MgF₂ from its solubility product: K_sp = $6.4×10^{-9}$.

Answer:

$1.2×10^{-3}\;M$