# Solubility Product and Precipitation Reactions This section explores the concept of solubility product and its application in calculating the concentrations of ions in saturated solutions of sparingly soluble ionic compounds, a topic of significant analytical importance. ## 3.3.1 Solubility Product ($\mathrm{K_{sp}}$) When an excess of a slightly soluble ionic compound is mixed with water, a small amount dissolves, and the remaining undissolved solid settles at the bottom. A dynamic equilibrium is established between the undissolved solid compound and its ions in the saturated solution. * **Example: Calcium Fluoride ($\mathrm{CaF_2}$)** Consider the dissolution of calcium fluoride in water: $$ \mathrm{CaF}_{2(\mathrm{s})} \rightleftharpoons \mathrm{Ca}^{2+}_{(\mathrm{aq})} + 2\mathrm{F}^{-}_{(\mathrm{aq})} $$ The equilibrium constant, $\mathrm{K}_{\mathrm{c}}$, for this reaction can be written as: $$ \mathrm{K}_{\mathrm{c}} = \frac{[\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^{2}}{[\mathrm{CaF}_{2}]} $$ Since $\mathrm{CaF_2}$ is a slightly soluble salt, the concentration of the undissolved solid ($[\mathrm{CaF_2}]$) essentially remains constant. Therefore, we can rearrange the expression: $$ \mathrm{K}_{\mathrm{c}}[\mathrm{CaF}_{2}] = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^{2} $$ The product of the constant $\mathrm{K}_{\mathrm{c}}$ and the constant concentration of the solid $[\mathrm{CaF}_{2}]$ is also a constant, which we define as the **solubility product constant ($\mathrm{K_{sp}}$)**: $$ \mathrm{K_{sp}} = [\mathrm{Ca}^{2+}][\mathrm{F}^{-}]^{2} $$ * **Definition of $\mathrm{K_{sp}}$** The solubility product constant ($\mathrm{K_{sp}}$) is defined as the product of the molar concentrations of the ions in a saturated solution, each raised to a power equal to its stoichiometric coefficient in the balanced chemical equation for the dissolution of the ionic compound. * **General Expression for $\mathrm{K_{sp}}$** For any slightly soluble ionic compound, $\mathrm{A}_{\mathrm{m}}\mathrm{B}_{\mathrm{n}}$, that dissociates according to the reaction: $$ \mathrm{A_mB_n(s)} \rightleftharpoons m\mathrm{A}^{+n}_{(\mathrm{aq})} + n\mathrm{B}^{-m}_{(\mathrm{aq})} $$ The solubility product constant, $\mathrm{K_{sp}}$, is given by: $$ K_{sp} = [\mathrm{A}^{+n}]^{m}[\mathrm{B}^{-m}]^{n} $$ This means $\mathrm{K_{sp}}$ is the product of the equilibrium concentration of ions, each raised to a power equal to the number of such ions in the formula unit of the compound. * **Important Note:** * $\mathrm{K_{sp}}$ is used for solutes that are only slightly soluble and do not completely dissolve in solution. * The **higher the $\mathrm{K_{sp}}$ value** of a compound, the **more soluble** it is in water. ## 3.3.2 Calculating Concentrations of Ions The $\mathrm{K_{sp}}$ expression allows us to calculate the concentrations of ions in a saturated solution of a slightly soluble salt. Since these salts dissociate only to a small extent, their ion concentrations are typically low. **Steps to calculate the concentration of ions from $\mathrm{K_{sp}}$:** 1. Write the balanced dissociation equation for the slightly soluble salt. 2. Set up the $\mathrm{K_{sp}}$ expression based on the dissociation equation. 3. Define the solubility of the salt (often denoted as $S$ or $x$) and express the equilibrium concentrations of the ions in terms of $S$, considering the stoichiometry. 4. Substitute these expressions into the $\mathrm{K_{sp}}$ equation and solve for $S$. 5. Use the value of $S$ to find the concentration of each ion. ## Worked Examples ### Example 1: Calculating ion concentrations for AgBr **Problem:** Given the $\mathrm{K_{sp}}$ for $\mathrm{AgBr}$ ($\mathrm{K_{sp}} = 1.60 \times 10^{-10}$), calculate the concentrations of each ion in the solution. 1. **Given values:** $\mathrm{K_{sp}} = 1.60 \times 10^{-10}$ for $\mathrm{AgBr}$. 2. **Dissociation reaction:** $$ \mathrm{AgBr}_{(\mathrm{s})} \rightleftharpoons \mathrm{Ag}^{+}_{(\mathrm{aq})} + \mathrm{Br}^{-}_{(\mathrm{aq})} $$ 3. **$\mathrm{K_{sp}}$ expression:** $$ K_{sp} = [\mathrm{Ag}^{+}][\mathrm{Br}^{-}] $$ 4. **Concentration Calculation:** Let $x$ represent the molar solubility of $\mathrm{AgBr}$ (in $\mathrm{mol/dm^3}$). From the stoichiometry of the dissociation, if $x$ moles of $\mathrm{AgBr}$ dissolve, then $x$ moles of $\mathrm{Ag}^{+}$ and $x$ moles of $\mathrm{Br}^{-}$ are produced. So, $[\mathrm{Ag}^{+}] = x$ and $[\mathrm{Br}^{-}] = x$. Substitute these into the $\mathrm{K_{sp}}$ expression: $$ 1.60 \times 10^{-10} = (x)(x) $$ $$ x^{2} = 1.60 \times 10^{-10} $$ Solve for $x$: $$ x = \sqrt{1.60 \times 10^{-10}} $$ $$ x = 1.26 \times 10^{-5}\,\mathrm{M} $$ Therefore, the concentrations of the ions are: $$ [\mathrm{Ag}^{+}] = 1.26 \times 10^{-5}\,\mathrm{M} $$ $$ [\mathrm{Br}^{-}] = 1.26 \times 10^{-5}\,\mathrm{M} $$ ### Example 2: Calculate the concentration of Silver chloride (AgCl) in its aqueous solution 1. **Given values:** $\mathrm{K_{sp}}$ for $\mathrm{AgCl} = 1.8 \times 10^{-10}$ at $25^{\circ}\mathrm{C}$. 2. **Dissociation reaction:** $$ \mathrm{AgCl}_{(\mathrm{s})} \rightleftharpoons \mathrm{Ag}^{+}_{(\mathrm{aq})} + \mathrm{Cl}^{-}_{(\mathrm{aq})} $$ 3. **$\mathrm{K_{sp}}$ expression:** $$ \mathrm{K_{sp}} = [\mathrm{Ag}^{+}][\mathrm{Cl}^{-}] $$ 4. **Concentration Calculation:** Let the solubility of $\mathrm{AgCl}$ be $x\,\mathrm{mol/dm^3}$. Since each mole of $\mathrm{AgCl}$ gives one mole of $\mathrm{Ag}^{+}$ and one mole of $\mathrm{Cl}^{-}$ ions, then: $[\mathrm{Ag}^{+}] = x$ and $[\mathrm{Cl}^{-}] = x$. Substitute these into the $\mathrm{K_{sp}}$ expression: $$ \mathrm{K_{sp}} = [x][x] $$ $$ 1.8 \times 10^{-10} = x^{2} $$ $$ x^{2} = 1.8 \times 10^{-10} $$ Solve for $x$: $$ x = \sqrt{1.8 \times 10^{-10}} $$ $$ x = 1.34 \times 10^{-5}\,\mathrm{mol/dm^3} $$ Thus, the solubility (or concentration of $\mathrm{Ag}^{+}$ and $\mathrm{Cl}^{-}$) in water is approximately: $$ [\mathrm{Ag}^{+}] = 1.34 \times 10^{-5}\,\mathrm{mol/dm^3} $$ $$ [\mathrm{Cl}^{-}] = 1.34 \times 10^{-5}\,\mathrm{mol/dm^3} $$ ## Possible Questions/Answers ### CONCEPT ASSESSMENT EXERCISE 3.1 * **Q1:** Write the dissociation reaction and the solubility product constant expression for $\mathrm{PbSO_4}$. **A1:** * Dissociation: $$ \mathrm{PbSO}_{4(\mathrm{s})} \rightleftharpoons \mathrm{Pb}^{2+}_{(\mathrm{aq})} + \mathrm{SO_4}^{2-}_{(\mathrm{aq})} $$ * $\mathrm{K_{sp}}$ expression: $$ \mathrm{K_{sp}} = [\mathrm{Pb}^{2+}][\mathrm{SO_4}^{2-}] $$ * **Q2:** Write the dissociation reaction and the solubility product constant expression for $\mathrm{Al(OH)_3}$. **A2:** * Dissociation: $$ \mathrm{Al(OH)}_{3(\mathrm{s})} \rightleftharpoons \mathrm{Al}^{3+}_{(\mathrm{aq})} + 3\mathrm{OH}^{-}_{(\mathrm{aq})} $$ * $\mathrm{K_{sp}}$ expression: $$ \mathrm{K_{sp}} = [\mathrm{Al}^{3+}][\mathrm{OH}^{-}]^{3} $$ * **Q3:** Write the dissociation reaction and the solubility product constant expression for $\mathrm{BaSO_4}$. **A3:** * Dissociation: $$ \mathrm{BaSO}_{4(\mathrm{s})} \rightleftharpoons \mathrm{Ba}^{2+}_{(\mathrm{aq})} + \mathrm{SO_4}^{2-}_{(\mathrm{aq})} $$ * $\mathrm{K_{sp}}$ expression: $$ \mathrm{K_{sp}} = [\mathrm{Ba}^{2+}][\mathrm{SO_4}^{2-}] $$ ### CONCEPT ASSESSMENT EXERCISE 3.2 * **Q1:** $\mathrm{K_{sp}}$ for $\mathrm{PbF_2}$ is $4 \times 10^{-8}$. Calculate the concentration of $\mathrm{Pb^{2+}}$ and $\mathrm{F^{-}$ ions in solution. **A1:** 1. **Dissociation:** $$ \mathrm{PbF}_{2(\mathrm{s})} \rightleftharpoons \mathrm{Pb}^{2+}_{(\mathrm{aq})} + 2\mathrm{F}^{-}_{(\mathrm{aq})} $$ 2. **$\mathrm{K_{sp}}$ expression:** $$ \mathrm{K_{sp}} = [\mathrm{Pb}^{2+}][\mathrm{F}^{-}]^{2} $$ 3. **Concentration Calculation:** Let the solubility of $\mathrm{PbF_2}$ be $x\,\mathrm{mol/dm^3}$. Then, $[\mathrm{Pb}^{2+}] = x$ and $[\mathrm{F}^{-}] = 2x$. Substitute into the $\mathrm{K_{sp}}$ expression: $$ 4 \times 10^{-8} = (x)(2x)^{2} $$ $$ 4 \times 10^{-8} = x(4x^{2}) $$ $$ 4 \times 10^{-8} = 4x^{3} $$ $$ x^{3} = \frac{4 \times 10^{-8}}{4} = 1 \times 10^{-8} $$ $$ x = \sqrt[3]{1 \times 10^{-8}} $$ $$ x \approx 2.15 \times 10^{-3}\,\mathrm{mol/dm^3} $$ 4. **Result:** $$ [\mathrm{Pb}^{2+}] = x = 2.15 \times 10^{-3}\,\mathrm{mol/dm^3} $$ $$ [\mathrm{F}^{-}] = 2x = 2(2.15 \times 10^{-3}\,\mathrm{mol/dm^3}) = 4.30 \times 10^{-3}\,\mathrm{mol/dm^3} $$ * **Q2:** $\mathrm{K_{sp}}$ for $\mathrm{BaSO_4}$ is $1 \times 10^{-10}$. Write the equation and the equilibrium expression for the dissolving of barium sulphate. Also, calculate the concentration of each ion. **A2:** 1. **Dissociation:** $$ \mathrm{BaSO}_{4(\mathrm{s})} \rightleftharpoons \mathrm{Ba}^{2+}_{(\mathrm{aq})} + \mathrm{SO_4}^{2-}_{(\mathrm{aq})} $$ 2. **$\mathrm{K_{sp}}$ expression:** $$ \mathrm{K_{sp}} = [\mathrm{Ba}^{2+}][\mathrm{SO_4}^{2-}] $$ 3. **Concentration Calculation:** Let the solubility of $\mathrm{BaSO_4}$ be $x\,\mathrm{mol/dm^3}$. Then, $[\mathrm{Ba}^{2+}] = x$ and $[\mathrm{SO_4}^{2-}] = x$. Substitute into the $\mathrm{K_{sp}}$ expression: $$ 1 \times 10^{-10} = (x)(x) $$ $$ x^{2} = 1 \times 10^{-10} $$ $$ x = \sqrt{1 \times 10^{-10}} $$ $$ x = 1 \times 10^{-5}\,\mathrm{mol/dm^3} $$ 4. **Result:** $$ [\mathrm{Ba}^{2+}] = 1 \times 10^{-5}\,\mathrm{mol/dm^3} $$ $$ [\mathrm{SO_4}^{2-}] = 1 \times 10^{-5}\,\mathrm{mol/dm^3} $$ ## Summary * **Solubility Product Constant ($\mathrm{K_{sp}}$):** Represents the equilibrium constant for the dissolution of a sparingly soluble ionic compound in a saturated solution. * **General Equation:** For a compound $\mathrm{A_mB_n}$ dissolving, the equilibrium is: $$ \mathrm{A_mB_n(s)} \rightleftharpoons m\mathrm{A}^{+n}_{(\mathrm{aq})} + n\mathrm{B}^{-m}_{(\mathrm{aq})} $$ And its $\mathrm{K_{sp}}$ expression is: $$ K_{sp} = [\mathrm{A}^{+n}]^{m}[\mathrm{B}^{-m}]^{n} $$ * **Solubility Relationship:** A higher $\mathrm{K_{sp}}$ value indicates greater solubility of the compound. * **Applications:** $\mathrm{K_{sp}}$ values are crucial for calculating the concentrations of ions in saturated solutions of sparingly soluble salts. These calculations are fundamental in understanding and predicting **precipitation reactions**, which are important processes in analytical chemistry, environmental chemistry, and industrial applications (e.g., water treatment, mineral processing).

3.3 Solubility Product And Precipitation Reactions | Chemical Equilibria | Chemistry 12