Question

The \(K_{\text {sp }}\) values for \(\mathrm{MnCO}_{3}\) and \(\mathrm{Mn}(\mathrm{OH})_{2}\) are \(1.8 \times 10^{-11}\) and \(4.6 \times 10^{-14}\), respectively. In saturated solutions of \(\mathrm{MnCO}_{3}\) and \(\mathrm{Mn}(\mathrm{OH})_{2},\) which has the higher manganese(II) ion concentration?

Short answer

Mn(OH)₂ has the higher manganese(II) ion concentration in its saturated solution.

Step-by-step solution

Identify Solubility Product Formulas

The solubility product ( \(K_{ ext{sp}} \)) for a salt is given by the expression of the concentrations of the ions it dissociates into, raised to the power of their stoichiometric coefficients. For \(\mathrm{MnCO}_{3}\), it dissociates into \(\text{Mn}^{2+}\) and \(\text{CO}_3^{2-}\). Similarly, for \(\mathrm{Mn(OH)}_2\), it dissociates into \(\text{Mn}^{2+}\) and \(\text{OH}^-\).

Write K_sp Expressions

For \(\mathrm{MnCO}_3\), the solubility product is \(K_{ ext{sp}} = [\mathrm{Mn}^{2+}][\mathrm{CO}_3^{2-}]\) and for \(\mathrm{Mn(OH)}_2\), it is \(K_{ ext{sp}} = [\mathrm{Mn}^{2+}][\mathrm{OH}^-]^2\).

Express Ion Concentrations with Solubility (S)

Assuming \(S\) is the solubility, for \(\mathrm{MnCO}_3\), \([\mathrm{Mn}^{2+}] = [\mathrm{CO}_3^{2-}] = S\). For \(\mathrm{Mn(OH)}_2\), \([\mathrm{Mn}^{2+}] = S\) and \([\mathrm{OH}^-] = 2S\).

Calculating S for MnCO3

Substitute into the expression: \(K_{\text{sp}} = (S)(S) = S^2 = 1.8 \times 10^{-11}\) Solve for \(S\) to find \(\sqrt{1.8 \times 10^{-11}}\). Calculating, \(S = 1.34 \times 10^{-6} \, \text{M}\).

Calculating S for Mn(OH)2

Substitute into the expression: \(K_{\text{sp}} = (S)(2S)^2 = 4S^3 = 4.6 \times 10^{-14}\) Divide both sides by 4 and solve for \(S^3\), \(S^3 = 1.15 \times 10^{-14}\). Take the cube root to find \(S\). Calculating, \(S = 2.27 \times 10^{-5} \, \text{M}\).

Compare Manganese Ion Concentrations

The concentration of \(\mathrm{Mn}^{2+}\) from \(\mathrm{MnCO}_3\) is \(1.34 \times 10^{-6} \, \text{M}\) and from \(\mathrm{Mn(OH)}_2\) is \(2.27 \times 10^{-5} \, \text{M}\). Thus, \(\mathrm{Mn(OH)}_2\) has a higher concentration of \(\mathrm{Mn}^{2+}\) ions in its saturated solution.

Key concepts

Solubility Product

The solubility product constant, abbreviated as \( K_{\text{sp}} \), is a critical concept in understanding the solubility of sparingly soluble salts. \( K_{\text{sp}} \) represents the level at which a solute dissolves in a solvent to form a saturated solution, where the dissolution and precipitation of the solute are at equilibrium.

For any ionic compound dissociating into its ions, \( K_{\text{sp}} \) can be defined by the product of the concentrations of the ions, each raised to the power of their respective stoichiometric coefficients. This constant varies for different compounds and provides a measure of how much solute will dissolve in the solution.

- **Mnemonic to Remember**: "More \( K_{\text{sp}} \), more soluble!" As the \( K_{\text{sp}} \) value increases, so does the solubility.- **Example**: For \( \mathrm{MnCO}_3 \) dissociating into \( \mathrm{Mn}^{2+} \) and \( \mathrm{CO}_3^{2-} \), the \( K_{\text{sp}} \) expression is \( [\mathrm{Mn}^{2+}][\mathrm{CO}_3^{2-}] \).

Understanding \( K_{\text{sp}} \) helps us calculate and predict the ion concentrations in a solution, which is crucial in various applications such as pharmaceuticals, environmental science, and chemical engineering.

Manganese Ion Concentration

The concentration of manganese ions \( (\mathrm{Mn}^{2+}) \) in solution is an essential factor in determining the solubility characteristics of manganese-containing compounds. This concentration is directly linked to the solubility product \( K_{\text{sp}} \) as it affects the degree of the compound's ionic dissociation.

To calculate the manganese ion concentration, we initially assume that solubility \( S \) represents the molarity of the ions at equilibrium. For a compound like \( \mathrm{MnCO}_3 \), its ionic equation \( \mathrm{MnCO}_3 \rightarrow \mathrm{Mn}^{2+} + \mathrm{CO}_3^{2-} \) indicates that the concentration of \( \mathrm{Mn}^{2+} \) is equal to \( S \), the solubility.

- **Calculation and Assumption**: When given the \( K_{\text{sp}} \), substituting \( S \) for each ion in the \( K_{\text{sp}} \) expression allows us to solve for \( S \), thereby determining \( \mathrm{Mn}^{2+} \) concentration.- For compounds like \( \mathrm{Mn(OH)}_2 \), the dissociation contributes additional hydroxide ions. As per the formula \( \mathrm{Mn(OH)}_2 \rightarrow \mathrm{Mn}^{2+} + 2\mathrm{OH}^- \), we note the factors that affect \( \mathrm{Mn}^{2+} \) ion concentration differ due to additional ions affecting solubility.

Using these calculations, it is evident that \( \mathrm{Mn(OH)}_2 \), with its larger solubility, results in a higher \( \mathrm{Mn}^{2+} \) concentration, illustrating how solubility directly influences ion availability in solutions.

Ionic Dissociation

Ionic dissociation in chemical equilibrium refers to the process by which a compound breaks down into its constituent ions when dissolved in a solvent. This forms the basis for solubility and dictates the presence of ions in a solution.

- **Process**: When salts like \( \mathrm{MnCO}_3 \) or \( \mathrm{Mn(OH)}_2 \) are introduced into water, they dissociate into \( \mathrm{Mn}^{2+} \) ions and their respective anions (carbonates or hydroxides). The extent of this dissociation affects the concentration of ions and is represented by solubility product expressions.

- **Example Expressions**: For \( \mathrm{MnCO}_3 \), dissociation results in \( [\mathrm{Mn}^{2+}][\mathrm{CO}_3^{2-}]= K_{\text{sp}} \), capturing the equilibrium state.

- **Influence on Solubility**: Ionic dissociation is a dynamic process that depends on several parameters, such as temperature and the common ion effect. The equilibrium between the dissolved ions and the undissolved solid adjusts accordingly, maintaining the solubility product balance.

By mastering ionic dissociation, one can predict and control the behavior of solutions, enhancing fields like chemistry and material science through the computation and application of solubility and ion concentration. This provides the foundational knowledge needed to solve complex problems in chemical reactions and solution dynamics.