Question

Use the following solubility data to calculate a value of \(K_{\mathrm{sp}}\) for each compound: (a) \(\operatorname{SrF}_{2} ; 1.03 \times 10^{-3} \mathrm{M}\) (b) \(\mathrm{CuI} ; 1.05 \times 10^{-6} \mathrm{M}\) (c) \(\mathrm{MgC}_{2} \mathrm{O}_{4} ; 0.094 \mathrm{~g} / \mathrm{L}\) (d) \(\mathrm{Zn}(\mathrm{CN})_{2} ; 4.95 \times 10^{-4} \mathrm{~g} / \mathrm{L}\)

Short answer

(a) \(4.34 \times 10^{-9}\), (b) \(1.10 \times 10^{-12}\), (c) \(7.01 \times 10^{-7}\), (d) \(2.98 \times 10^{-16}\).

Step-by-step solution

Determine the Dissociation Equation

Write the dissociation equations for each salt in water. (a) \( \text{SrF}_2 (s) \rightleftharpoons \text{Sr}^{2+} (aq) + 2 \text{F}^- (aq) \) (b) \( \text{CuI} (s) \rightleftharpoons \text{Cu}^+ (aq) + \text{I}^- (aq) \) (c) \( \text{MgC}_2\text{O}_4 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + \text{C}_2\text{O}_4^{2-} (aq) \) (d) \( \text{Zn(CN)}_2 (s) \rightleftharpoons \text{Zn}^{2+} (aq) + 2 \text{CN}^- (aq) \).

Calculate Molar Solubility from Given Data

Convert the solubility in grams per liter into molarity if necessary. (a) For \( \text{SrF}_2 \), molarity is \( 1.03 \times 10^{-3} \text{ M}\).(b) For \( \text{CuI} \), molarity is \( 1.05 \times 10^{-6} \text{ M} \).(c) For \( \text{MgC}_2\text{O}_4 \), convert \( 0.094 \text{ g/L} \) using its molar mass (\( 112.37 \text{ g/mol} \)) to get molarity: \( \frac{0.094}{112.37} \text{ M} = 0.000836 \text{ M} \).(d) For \( \text{Zn(CN)}_2 \), convert \( 4.95 \times 10^{-4} \text{ g/L} \) using its molar mass (\( 117.442 \text{ g/mol} \)) to get molarity: \( \frac{4.95 \times 10^{-4}}{117.442} \text{ M} = 4.21 \times 10^{-6} \text{ M} \).

Set Up Ksp Expression

From the dissociation equation, write the expression for the solubility product \( K_{sp} \) for each compound. (a) \( K_{sp} = [\text{Sr}^{2+}][\text{F}^-]^2 \) (b) \( K_{sp} = [\text{Cu}^+][\text{I}^-] \) (c) \( K_{sp} = [\text{Mg}^{2+}][\text{C}_2\text{O}_4^{2-}] \) (d) \( K_{sp} = [\text{Zn}^{2+}][\text{CN}^-]^2 \).

Calculate Ksp Using Molarity

Substitute the molarity values to calculate \( K_{sp} \):(a) \( [\text{Sr}^{2+}] = 1.03 \times 10^{-3} \), \( [\text{F}^-] = 2 \times 1.03 \times 10^{-3} \), \( K_{sp} = (1.03 \times 10^{-3})(2.06 \times 10^{-3})^2 = 4.34 \times 10^{-9} \).(b) \( [\text{Cu}^+] = [\text{I}^-] = 1.05 \times 10^{-6} \), \( K_{sp} = (1.05 \times 10^{-6})^2 = 1.10 \times 10^{-12} \).(c) \( [\text{Mg}^{2+}] = [\text{C}_2\text{O}_4^{2-}] = 0.000836 \), \( K_{sp} = (0.000836)^2 = 7.01 \times 10^{-7} \).(d) \( [\text{Zn}^{2+}] = 4.21 \times 10^{-6} \), \( [\text{CN}^-] = 2 \times 4.21 \times 10^{-6} \), \( K_{sp} = (4.21 \times 10^{-6})(8.42 \times 10^{-6})^2 = 2.98 \times 10^{-16} \).

Key concepts

Dissociation Equation

When a slightly soluble ionic compound dissolves in water, it dissociates into its constituent ions. This process is expressed using a dissociation equation. For example, when the compound strontium fluoride (\(\text{SrF}_2\)) dissolves, it separates into strontium ions (\(\text{Sr}^{2+}\)) and fluoride ions (\(\text{F}^-\)). The dissociation equation for strontium fluoride can be written as:

• \( \text{SrF}_2 (s) \leftrightarrows \text{Sr}^{2+} (aq) + 2 \text{F}^- (aq) \)

The numbers represent the stoichiometry of the ions produced. In this case, 1 mole of strontium fluoride produces 1 mole of strontium ions and 2 moles of fluoride ions.
Understanding the dissociation equation is crucial because it directly affects the calculation of the solubility product constant (\(K_{sp}\)), a value that quantifies the solubility of a compound under equilibrium conditions.

Molar Solubility

Molar solubility refers to the number of moles of a solute that can dissolve in a liter of solution before reaching saturation. It gives us a direct measure of how much solute can be dissolved, which is vital to finding the solubility product constant (\(K_{sp}\)). For some compounds, the solubility is given in grams per liter, requiring conversion to moles per liter using the substance's molar mass.
For instance, magnesium oxalate's solubility is provided as 0.094 g/L. With a molar mass of 112.37 g/mol, the molar solubility can be calculated using:

• Molarity = \( \frac{0.094}{112.37} \approx 0.000836 \text{ M} \)

Following these conversions ensures proper values are used in subsequent calculations of the solubility product, where each ion's concentration must be included.

Ksp Expression Calculation

The solubility product constant, \(K_{sp}\), is an expression that relates the concentrations of the ions in a saturated solution of a slightly soluble salt. The \(K_{sp}\) expression is derived from the dissociation equation and represents the equilibrium condition where the rates of dissolution and precipitation are balanced. \(K_{sp}\) can be calculated using the molar solubility to find the concentrations of ions.
Taking copper iodide (\(\text{CuI}\)) as an example, the \(K_{sp}\) expression is \([\text{Cu}^+][\text{I}^-]\). If the molarity for both ions is \(1.05 \times 10^{-6} \text{ M}\), the \(K_{sp}\) value is:

• \(K_{sp} = (1.05 \times 10^{-6})^2 = 1.10 \times 10^{-12}\)

This calculation uses the principle of multiplying the ion concentrations raised to their stoichiometric coefficients, making the \(K_{sp}\) expression a powerful tool for understanding solubility properties.