Question

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties. a. \(\mathrm{PbI}_{2}, K_{\mathrm{sp}}=1.4 \times 10^{-8}\) b. \(\mathrm{CdCO}_{3}, K_{\text {sp }}=5.2 \times 10^{-12}\) c. \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}, K_{\text {sp }}=1 \times 10^{-31}\)

Short answer

The solubilities of the given compounds are as follows: a. \(\mathrm{PbI}_{2}\): approximately \(9.83 \times 10^{-4}\) M b. \(\mathrm{CdCO}_{3}\): approximately \(2.28 \times 10^{-6}\) M c. \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}\): approximately \(3.66 \times 10^{-7}\) M

Step-by-step solution

Part a: Solubility of \(\mathrm{PbI}_{2}\)

1. Write the balanced dissolution equation for \(\mathrm{PbI}_{2}\): \[\mathrm{PbI}_{2(s)} \rightleftharpoons \mathrm{Pb}^{2+}_{(aq)} + 2\mathrm{I}^{-}_{(aq)}\] 2. Write the \(K_{sp}\) expression for the dissolution equation: \[K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{I}^{-}]^2\] 3. Let \(x\) be the solubility of \(\mathrm{PbI}_{2}\) in moles per liter. This means that at equilibrium, the concentration of \(\mathrm{Pb}^{2+}\) ions is \(x\) and the concentration of \(\mathrm{I}^{-}\) ions is \(2x\). 4. Substitute the concentrations into the \(K_{sp}\) expression: \[1.4 \times 10^{-8} = (x)(2x)^2\] 5. Simplify and solve for \(x\): \[1.4 \times 10^{-8} = 4x^3\] \[x = \sqrt[3]{\frac{1.4 \times 10^{-8}}{4}}\] \[x \approx 9.83 \times 10^{-4}\,\text{M}\] The solubility of \(\mathrm{PbI}_{2}\) is approximately \(9.83 \times 10^{-4}\) M.

Part b: Solubility of \(\mathrm{CdCO}_{3}\)

1. Write the balanced dissolution equation for \(\mathrm{CdCO}_{3}\): \[\mathrm{CdCO}_{3(s)} \rightleftharpoons \mathrm{Cd}^{2+}_{(aq)} + \mathrm{CO}_{3}^{2-}_{(aq)}\] 2. Write the \(K_{sp}\) expression for the dissolution equation: \[K_{sp} = [\mathrm{Cd}^{2+}][\mathrm{CO}_{3}^{2-}]\] 3. Let \(x\) be the solubility of \(\mathrm{CdCO}_{3}\) in moles per liter. This means that at equilibrium, the concentration of \(\mathrm{Cd}^{2+}\) ions is \(x\) and the concentration of \(\mathrm{CO}_{3}^{2-}\) ions is \(x\). 4. Substitute the concentrations into the \(K_{sp}\) expression: \[5.2 \times 10^{-12} = (x)(x)\] 5. Simplify and solve for \(x\): \[x^2 = 5.2 \times 10^{-12}\] \[x \approx 2.28 \times 10^{-6}\,\text{M}\] The solubility of \(\mathrm{CdCO}_{3}\) is approximately \(2.28 \times 10^{-6}\) M.

Part c: Solubility of \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

1. Write the balanced dissolution equation for \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}\): \[\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2(s)} \rightleftharpoons 3\mathrm{Sr}^{2+}_{(aq)} + 2\mathrm{PO}_{4}^{3-}_{(aq)}\] 2. Write the \(K_{sp}\) expression for the dissolution equation: \[K_{sp} = [\mathrm{Sr}^{2+}]^3[\mathrm{PO}_{4}^{3-}]^2\] 3. Let \(x\) be the solubility of \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in moles per liter. This means that at equilibrium, the concentration of \(\mathrm{Sr}^{2+}\) ions is \(3x\) and the concentration of \(\mathrm{PO}_{4}^{3-}\) ions is \(2x\). 4. Substitute the concentrations into the \(K_{sp}\) expression: \[1\times10^{-31} = (3x)^3(2x)^2\] 5. Simplify and solve for \(x\): \[1\times10^{-31} = 108x^5\] \[x = \sqrt[5]{\frac{1\times10^{-31}}{108}}\] \[x \approx 3.66 \times 10^{-7}\,\text{M}\] The solubility of \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is approximately \(3.66 \times 10^{-7}\) M.

Key concepts

Solubility Product Constant

Understanding the solubility product constant, commonly referred to as Ksp, is fundamental in the study of solubility in chemistry. The Ksp is a mathematical expression that quantifies the maximum amount of a solid substance that can dissolve in a liquid to form a saturated solution at a given temperature. It is specific to ionic compounds that dissociate into their constituent ions in a solution.

The value of Ksp is constant for a given substance at a specific temperature and is determined experimentally. A high Ksp value indicates that more of the substance can dissolve, making it more soluble, whereas a low Ksp means less of the substance can dissolve, rendering it less soluble. Note that Ksp does not give the actual concentration of the ions in the solution but rather the maximum possible product of these concentrations under saturated conditions.

Ksp Expression

The Ksp expression is crafted from the equilibrium equation for the dissolution of an ionic compound. When an ionic solid dissolves in water, it separates into its component ions. For example, the dissolution of calcium fluoride, CaF2, in water can be represented as
\[ \text{CaF}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{F}^-(aq) \]
From this equilibrium, the Ksp expression for calcium fluoride would be written as
\[ K_{sp} = [\text{Ca}^{2+}][\text{F}^-]^2 \]
Where \( [\text{Ca}^{2+}] \) is the molar concentration of calcium ions and \( [\text{F}^-] \) is the molar concentration of fluoride ions at equilibrium. The coefficients in the dissolution equation become exponents in the Ksp expression, accounting for the stoichiometry of the reaction.

Molar Solubility

Molar solubility is defined as the number of moles of a solute that can be dissolved per liter of solution to reach saturation. It is often represented by the symbol 'x' in solubility calculations. Calculating the molar solubility involves using the Ksp value and the stoichiometry of the dissolution reaction.

For instance, if an ionic compound AB2 has a Ksp expression written as \( K_{sp} = [A^{2+}][B^-]^2 \), and 'x' is the molar solubility of AB2 (i.e., [A^{2+}] = x and [B^-] = 2x), we can set up an equation like \( K_{sp} = x(2x)^2 \) to find the value of 'x'.

The calculated molar solubility not only indicates how much of a compound can dissolve but also allows the determination of the concentrations of the individual ions in a saturated solution. Understanding molar solubility is essential for predicting the outcome of reactions in solution, such as precipitation reactions.