Problem 31 - Chemistry Matter and Change

Question

Calculate The \(K_{\text { sp }}\) of magnesium carbonate \(\left(\mathrm{Mg} \mathrm{CO}_{3}\right)\) is \(2.6 \times 10^{-9} .\) What is the solubility of Mg CO \(_{3}\) in pure water?

Step-by-Step Solution

Answer

The solubility of MgCO₃ in pure water is approximately 5.1 × 10⁻⁵ mol/L.

Step 1: Understand Dissolution Reaction

Start by writing the dissolution reaction for magnesium carbonate (\( \text{MgCO}_3 \)) in water:\[\text{MgCO}_3(s) \rightleftharpoons \text{Mg}^{2+}(aq) + \text{CO}_3^{2-}(aq)\]This shows that one mole of \( \text{MgCO}_3 \) dissociates into one mole of \( \text{Mg}^{2+} \) ions and one mole of \( \text{CO}_3^{2-} \) ions.

Step 2: Expression for Ksp

The solubility product constant \( K_{sp} \) expression for the equilibrium in Step 1 is given by:\[K_{sp} = [\text{Mg}^{2+}][\text{CO}_3^{2-}]\]Since each mole of \( \text{MgCO}_3 \) produces one mole of \( \text{Mg}^{2+} \) and one mole of \( \text{CO}_3^{2-} \), we can denote the concentration of both ions as \( s \), where \( s \) is the solubility of the compound in moles per liter.

Step 3: Substitute and Solve for Solubility

Substitute \( s \) into the \( K_{sp} \) expression:\[K_{sp} = s^2\]We already know \( K_{sp} = 2.6 \times 10^{-9} \), so we can solve for \( s \):\[s^2 = 2.6 \times 10^{-9}\]Take the square root of both sides:\[s = \sqrt{2.6 \times 10^{-9}}\]Using a calculator, \( s \approx 5.1 \times 10^{-5} \).

Step 4: Interpret the Result

The solubility \( s \) represents the concentration of \( \text{MgCO}_3 \) that can dissolve in water to reach equilibrium with the solid compound. Therefore, the solubility of \( \text{MgCO}_3 \) in pure water is approximately \( 5.1 \times 10^{-5} \) mol/L.

Key Concepts

Dissolution Reaction

A dissolution reaction describes the process through which a solid ionic compound dissolves in a solvent, typically water, to form ions. In this case, magnesium carbonate (\( \text{MgCO}_3 \)) is the ionic compound that dissolves. Upon adding \( \text{MgCO}_3 \) to water, it breaks apart into its constituent ions. The chemical equation for this process is \( \text{MgCO}_3(s) \rightleftharpoons \text{Mg}^{2+}(aq) + \text{CO}_3^{2-}(aq) \).

• The solid state of magnesium carbonate (s) at the beginning of the reaction indicates it is not yet dissolved.
• The aqueous state (aq) of magnesium and carbonate ions shows that they are dissolved in water.

This equation is reversible, meaning that the process can achieve a state of equilibrium. Here, the forward reaction represents dissolution, and the reverse reaction represents the precipitation or reformation of the solid.

Magnesium Carbonate

Magnesium carbonate is a white solid chemical compound with the formula \( \text{MgCO}_3 \). It is a common compound found in nature, such as in the form of mineral magnesite. In chemistry, it is often used to demonstrate principles of solubility and equilibrium.

The chemical reaction and properties involve:

• Magnesium carbonate belongs to a group of carbonates that generally show limited solubility in water.
• When added to water, only a small portion dissolves to form magnesium ions (\( \text{Mg}^{2+} \)) and carbonate ions (\( \text{CO}_3^{2-} \)).

The solubility of magnesium carbonate in water is quite low, making it an ideal candidate for studying the concept of the solubility product constant, \( K_{sp} \). This property highlights its significance in applications requiring controlled solubility and neutrality in chemical behavior.

Solubility Calculation

Calculating the solubility of a compound like magnesium carbonate involves determining how much of it can dissolve in a solution to form a saturated solution. This is where the solubility product constant (\( K_{sp} \)) becomes important.

The steps to calculate solubility include:

• First, write the dissolution reaction and solubility product expression: \( K_{sp} = [\text{Mg}^{2+}][\text{CO}_3^{2-}] \).
• Next, set the concentrations of both ions as \( s \), where \( s \) represents the solubility in \( \text{mol/L} \).
• The expression becomes \( K_{sp} = s^2 \).
• Given \( K_{sp} = 2.6 \times 10^{-9} \), solve for \( s \) which is \( s = \sqrt{2.6 \times 10^{-9}} \).

The solution gives us \( s \approx 5.1 \times 10^{-5} \) mol/L, emphasizing the low solubility of magnesium carbonate in water. This calculation helps in understanding the extent to which a compound can dissolve before it reaches a state of equilibrium.

Chemical Equilibrium

Chemical equilibrium is a key concept in solubility and dissolution reactions. In this state, the rates of the forward dissolution reaction and the reverse precipitation reaction are equal, meaning no net change occurs in the concentration of reactants and products over time.

For magnesium carbonate's dissolution:

• The forward reaction: solid \( \text{MgCO}_3 \) becoming \( \text{Mg}^{2+} \) and \( \text{CO}_3^{2-} \).
• The reverse reaction: \( \text{Mg}^{2+} \) and \( \text{CO}_3^{2-} \) forming solid \( \text{MgCO}_3 \).

At equilibrium:

• The concentration of ions remains constant.
• Changes in temperature or pressure can shift the equilibrium, affecting solubility.
• The equilibrium is described by the solubility product constant \( K_{sp} \), a fixed value at a given temperature.

Understanding equilibrium helps in predicting how a change in conditions could affect the dissolution of magnesium carbonate, linking it to broader reactions and systems in chemistry.