Question

The \(K_{\text {sp }}\) values for \(\mathrm{CaCO}_{3}\) and \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) are \(3.8 \times 10^{-9}\) and \(2.3 \times 10^{-9}\), respectively. In saturated solutions of \(\mathrm{CaCO}_{3}\) and \(\mathrm{CaC}_{2} \mathrm{O}_{4}\), which has the higher calcium ion concentration?

Short answer

The calcium ion concentration is higher in the saturated solution of \( \text{CaCO}_3 \).

Step-by-step solution

Understand Solubility Product

The solubility product constant ( K_{sp} ) of a compound is a measure of how much of the compound can dissolve in solution to form its constituent ions. It is unique for each compound.

Write Dissolution Equations

Write the dissolution equations for each compound: For \( \text{CaCO}_3 \text{,} \rightarrow \text{Ca}^{2+} + \text{CO}_3^{2-} \) For \( \text{CaC}_2\text{O}_4 \text{,} \rightarrow \text{Ca}^{2+} + \text{C}_2\text{O}_4^{2-} \)

Express Solubility Product

Express the K_{sp} for each compound:\( K_{sp, \, CaCO_3} = [\text{Ca}^{2+}][\text{CO}_3^{2-}] \) \( K_{sp, \, CaC_2O_4} = [\text{Ca}^{2+}][\text{C}_2O_4^{2-}] \)

Substitute Concentrations

Since one mole of \( \text{CaCO}_3 \) or \( \text{CaC}_2\text{O}_4 \) produces one mole each of \( \text{Ca}^{2+} \) and\( \text{CO}_3^{2-} \) or \( \text{C}_2O_4^{2-} \), let the concentration of \( \text{Ca}^{2+} \) be \( x \): \( K_{sp, \, CaCO_3} = x^2 \) and \( K_{sp, \, CaC_2O_4} = x^2 \)

Solve for Calcium Ion Concentration

Calculate \( x \) for each compound:\( x = \sqrt{K_{sp, \, CaCO_3}} = \sqrt{3.8 \times 10^{-9}} \approx 6.16 \times 10^{-5} \: \text{M} \) for \( \text{CaCO}_3 \) \( x = \sqrt{K_{sp, \, CaC_2O_4}} = \sqrt{2.3 \times 10^{-9}} \approx 4.79 \times 10^{-5} \: \text{M} \) for \( \text{CaC}_2O_4 \)

Compare Ion Concentrations

The calculated \( \text{Ca}^{2+} \) ion concentration for \( \text{CaCO}_3 \) is \( 6.16 \times 10^{-5} \: \text{M} \), whereas for \( \text{CaC}_2O_4 \) it is \( 4.79 \times 10^{-5} \: \text{M} \). Therefore, \( \text{CaCO}_3 \) has a higher calcium ion concentration in solution.

Key concepts

Calcium Ion Concentration

To understand which of the two compounds has a higher calcium ion concentration in their saturated solutions, you first need to know what calcium ion concentration refers to. The concentration of ions in a solution is basically how many of those ions are present in a given volume of solution. In this context, we are interested in the concentration of calcium ions (Ca^{2+}) that have dissolved into the solution from the solid salts: calcium carbonate (\(CaCO_3\)) and calcium oxalate (\(CaC_2O_4\)).

• Higher calcium ion concentration indicates that more of the solid salt has dissolved in the solution.
• Comparing the concentrations of calcium ions in two different solutions can tell us which solution contains more dissolved calcium ions.

The concentration of calcium ions depends on the solubility of the salt and is directly linked to the solubility product constant, or \(K_{sp}\). Let's now dive into how this concept is used to find out which salt provides a higher calcium ion concentration when dissolved.

Dissolution Equations

To find the calcium ion concentration, we first write down the dissolution equations for the compounds. Dissolution equations show how a solid compound breaks down into its ions in a solution. For calcium carbonate (\(CaCO_3\)) and calcium oxalate (\(CaC_2O_4\)), their dissolution can be represented as follows:

• \(CaCO_3 (s) \rightarrow Ca^{2+} (aq) + CO_3^{2-} (aq)\)
• \(CaC_2O_4 (s) \rightarrow Ca^{2+} (aq) + C_2O_4^{2-} (aq)\)

In both cases, the solid dissolves into one calcium ion (Ca^{2+}) and one accompanying anion per formula unit. This means that, stoichiometrically, each mole of dissolved compound contributes one mole of calcium ions to the solution.
Simply put, the dissolution equation tells us that the concentration of calcium ions in the solution will be equal to the concentration of the entire compound in the solution. This forms the basis for using the solubility product constant (\(K_{sp}\)) to determine ion concentrations.

Ksp Calculation

The solubility product constant, or \(K_{sp}\), is a crucial concept for calculating how much of a compound can dissolve in a solution. This constant values is distinct for each substance and can be used to predict the concentration of ions in a saturated solution. Here's how we use \(K_{sp}\) to compute the concentration of calcium ions for our compounds.
Start by expressing \(K_{sp}\) based on the dissolution equations:

• For \(CaCO_3\), \(K_{sp, \, CaCO_3} = [Ca^{2+}][CO_3^{2-}]\)
• For \(CaC_2O_4\), \(K_{sp, \, CaC_2O_4} = [Ca^{2+}][C_2O_4^{2-}]\)

Since the stoichiometry of dissolution gives a 1:1 ratio of calcium ions to anions, both ions appear in equal amounts in solution. Therefore, we can express \(K_{sp}\) for both compounds as \(x^2\), where \(x\) is the concentration of the calcium ions, which equals the concentration of the entire compound in the solution.
To find \(x\), take the square root of each \(K_{sp}\):

• \(x = \sqrt{3.8 \times 10^{-9}} = 6.16 \times 10^{-5} \, M\) for \(CaCO_3\)
• \(x = \sqrt{2.3 \times 10^{-9}} = 4.79 \times 10^{-5} \, M\) for \(CaC_2O_4\)

By comparing these values, we can conclude that \(CaCO_3\) offers a higher calcium ion concentration in its saturated solution because its \(K_{sp}\) value is larger, allowing more ions to be present in solution.