Question

Use the following data to calculate the \(K_{\text {sp }}\) value for each solid. a. The solubility of \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) is \(4.8 \times 10^{-5} \mathrm{~mol} / \mathrm{L}\). b. The solubility of \(\mathrm{BiI}_{3}\) is \(1.32 \times 10^{-5} \mathrm{~mol} / \mathrm{L}\).

Short answer

The calculated \(K_{\text {sp}}\) values for the two solids are: a. For \(\mathrm{CaC}_{2} \mathrm{O}_{4}\): \(K_{\text {sp}} = 1.1066 \times 10^{-11}\) b. For \(\mathrm{BiI}_{3}\): \(K_{\text {sp}} = 5.4383 \times 10^{-17}\)

Step-by-step solution

Write the balanced dissolution equation

For each solid, we will write down the dissolution equation. a. For \(\mathrm{CaC}_{2} \mathrm{O}_{4}\): \[\mathrm{CaC}_{2} \mathrm{O}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + 2 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(aq)\] b. For \(\mathrm{BiI}_{3}\): \[\mathrm{BiI}_{3}(s) \rightleftharpoons \mathrm{Bi}^{3+}(aq) + 3 \mathrm{I}^{-}(aq)\]

Write the expression for solubility product (\(K_{\text {sp}}\))

Now, we will write the equilibrium expression for each solid. a. For \(\mathrm{CaC}_{2} \mathrm{O}_{4}\): \[K_{\text {sp}} = [\mathrm{Ca}^{2+}][\mathrm{C}_{2}\mathrm{O}_{4}^{2-}]^{2}\] b. For \(\mathrm{BiI}_{3}\): \[K_{\text {sp}} = [\mathrm{Bi}^{3+}][\mathrm{I}^{-}]^{3}\]

Substitute the given solubility data and calculate the \(K_{\text {sp}}\) values

We will now substitute the given solubility data into the expressions and calculate \(K_{\text {sp}}\) for each solid. a. The solubility of \(\mathrm{CaC}_{2}\mathrm{O}_{4}\) is \(4.8 \times 10^{-5}\ \mathrm{mol\cdot L^{-1}}\). So, \([\mathrm{Ca}^{2+}] = 4.8 \times 10^{-5}\) and \([\mathrm{C}_{2}\mathrm{O}_{4}^{2-}] = 2(4.8 \times 10^{-5})\). Substituting these values into the equilibrium expression: \[K_{\text {sp}} = (4.8 \times 10^{-5})(2(4.8 \times 10^{-5}))^{2} = 1.1066 \times 10^{-11}\] b. The solubility of \(\mathrm{BiI}_{3}\) is \(1.32 \times 10^{-5}\ \mathrm{mol\cdot L^{-1}}\). So, \([\mathrm{Bi}^{3+}] = 1.32 \times 10^{-5}\) and \([\mathrm{I}^{-}] = 3(1.32 \times 10^{-5})\). Substituting these values into the equilibrium expression: \[K_{\text {sp}} = (1.32 \times 10^{-5})(3(1.32 \times 10^{-5}))^{3} = 5.4383 \times 10^{-17}\]

Report the \(K_{\text {sp}}\) values

The calculated \(K_{\text {sp}}\) values for the two solids are: a. For \(\mathrm{CaC}_{2} \mathrm{O}_{4}\): \(K_{\text {sp}} = 1.1066 \times 10^{-11}\) b. For \(\mathrm{BiI}_{3}\): \(K_{\text {sp}} = 5.4383 \times 10^{-17}\)

Key concepts

Chemical Equilibrium and Solubility Product Constant

Understanding chemical equilibrium is essential when studying the solubility product constant, often represented as the symbol \( K_{sp} \). At chemical equilibrium, the rate of the forward reaction (dissolution of a solid into its constituent ions) is equal to the rate of the reverse reaction (reformation of the solid from its ions). This balance doesn't mean the reactants and products are in the same concentrations, but rather that their concentrations have stabilized and are no longer changing over time.

For sparingly soluble ionic compounds, this equilibrium between the undissolved solid and the dissolved ions can be quantified by the solubility product constant. \( K_{sp} \) is a special type of equilibrium constant that applies to these solutes which dissolve to a small extent. It depicts the maximum product of the ionic concentrations of the solute at a given temperature. Different compounds have unique \( K_{sp} \) values, which are useful for predicting whether a precipitate will form in a solution and to what extent a compound will dissolve.

Dissolution Equations

A dissolution equation represents the process by which a solid ionic compound breaks apart into its individual ions in aqueous solution. Writing a correct balanced dissolution equation is critical for the determination of the solubility product constant. Each dissolution equation is specific to the solid compound at hand.

Take, for example, the dissolution of calcium oxalate, \( CaC_{2}O_{4} \): in water, it dissociates into one calcium ion \( Ca^{2+} \) and two oxalate ions \( C_{2}O_{4}^{2-} \). Mathematically, this is represented as:\[ CaC_{2}O_{4}(s) \rightleftharpoons Ca^{2+}(aq) + 2 C_{2}O_{4}^{2-}(aq) \]
Similarly, for bismuth iodide \( BiI_{3} \), the dissolution equation is:\[ BiI_{3}(s) \rightleftharpoons Bi^{3+}(aq) + 3 I^{-}(aq) \]
Note how the stoichiometry of the equation impacts the exponents in the \( K_{sp} \) expression. This reflects the principle that equilibrium constants are products of the concentrations of the products, each raised to the power of its stoichiometric coefficient in the balanced equation.

Concentration Calculations

When calculating the solubility product constant, the task essentially revolves around performing accurate concentration calculations. The solubility provided in a problem is typically the molar solubility, which indicates the number of moles of a solute that can dissolve in a liter of solution to reach saturation without precipitating.

Given the molar solubility, you can calculate individual ion concentrations by considering the stoichiometry of the compound's dissolution. For example, if the solubility of calcium oxalate is \( 4.8 \times 10^{-5} \) moles per liter, we find that the concentration of calcium ions \( [Ca^{2+}] \) is the same as the solubility, while the concentration of oxalate ions \( [C_{2}O_{4}^{2-}] \) is twice this value because each formula unit of calcium oxalate produces two oxalate ions.

These calculated concentrations are then substituted into the \( K_{sp} \) expression to find the equilibrium constant's value. Consistently employing units (such as moles per liter for concentration) and ensuring that the ion stoichiometry is correctly factored into the calculations are elemental to obtaining the correct \( K_{sp} \).