Problem 31 - Chemistry An Atoms First Approach

Question

Calculate the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=8.9 \times 10^{-12}\).

Step-by-Step Solution

Answer

The molar solubility of \(\mathrm{Mg(OH)_2}\) is \(1.3 \times 10^{-4}\,\text{mol/L}\).

Step 1: Write the dissolution equation for \(\mathrm{Mg(OH)_2}\)

The dissolution equation for \(\mathrm{Mg(OH)_2}\) is: \[ \mathrm{Mg(OH)_{2}(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^{-}(aq)} \]

Step 2: Define the solubility product expression

The solubility product expression uses the product of the concentrations of dissolved ions raised to their stoichiometric coefficients. For this dissolution reaction, the expression is: \[ K_{sp} = [\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2 \]

Step 3: Express the concentrations of ions in terms of molar solubility

Let's call the molar solubility of \(\mathrm{Mg(OH)_2}\) "x". This means that x moles of \(\mathrm{Mg(OH)_2}\) dissolve per liter of solution. Dissolving x moles of \(\mathrm{Mg(OH)_2}\) will also produce x moles of \(\mathrm{Mg^{2+}}\) ions and 2x moles of \(\mathrm{OH^-}\) ions. We can then express the concentrations as follows: \[ [\mathrm{Mg^{2+}}] = x \quad \text{and} \quad [\mathrm{OH^-}]=2x \]

Step 4: Substitute the concentrations into the solubility product expression and solve for x

Substitute the concentrations of ions in terms of x into the solubility product expression: \[ K_{sp} = x(2x)^2 \] Now, plug in the given \(K_{sp}=8.9 \times 10^{-12}\) and solve for x: \[ 8.9 \times 10^{-12} = x(4x^2) \] \[ 8.9 \times 10^{-12} = 4x^3 \] \[ x^3=\frac{8.9 \times 10^{-12}}{4} \] \[ x^3=2.225 \times 10^{-12} \] \[ x=\sqrt[3]{2.225 \times 10^{-12}} \] \[ x=1.3 \times 10^{-4} \] So, the molar solubility of \(\mathrm{Mg(OH)_2}\) is \(1.3 \times 10^{-4}\,\text{mol/L}\).

Key Concepts

Dissolution Equation

When it comes to understanding how a solid dissolves in a liquid, the dissolution equation plays a vital role. It's an expression that shows how a compound breaks down into its ions in solution. For example, let's consider the compound magnesium hydroxide, \(\mathrm{Mg(OH)_2}\). When this compound dissolves in water, it forms magnesium ions and hydroxide ions. The equation for this process is represented as:\[\mathrm{Mg(OH)_{2}(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^{-}(aq)}\]This equation shows a dynamic equilibrium between the solid \(\mathrm{Mg(OH)_{2}}\) and the ions in solution. It's important to note that in dissolution, the compound does not fully break apart unless enough water is present. It’s like a see-saw where the dissolution and crystallization of the solid occur at the same rates, reaching a stable state known as chemical equilibrium.

Solubility Product

The solubility product, represented as \(K_{sp}\), is an equilibrium constant that helps us understand the solubility of a compound. It tells us the maximum concentration of ions that can be present in solution before a precipitate forms. The solubility product expression for a given dissolution is written based on the ions that form from the solute. In our magnesium hydroxide example, we have:\[K_{sp} = [\mathrm{Mg^{2+}}][\mathrm{OH^-}]^2\]This shows that the solubility product is the product of the concentrations of the ions, each raised to the power of their stoichiometric coefficients in the dissolution equation. In essence, \(K_{sp}\) acts as a threshold to predict whether a compound will dissolve or precipitate based on the ionic concentrations in a solution.

Ion Concentration

Understanding ion concentration is key to predicting solubility. It indicates how much of each ion is present in a solution. When magnesium hydroxide dissolves, it produces magnesium ions and twice as many hydroxide ions, as the chemical equation shows.If "x" is the molar solubility of \(\mathrm{Mg(OH)_2}\), this corresponds to:

• \((x)\) moles of \(\mathrm{Mg^{2+}}\)
• \((2x)\) moles of \(\mathrm{OH^-}\)

These concentrations can be plugged into the solubility product expression to determine the solubility or if given the solubility, to calculate \(K_{sp}\). This is crucial as the concentration of ions affects the reaction’s ability to remain in equilibrium and can lead to precipitation if exceeded.

Chemical Equilibrium

Chemical equilibrium in the context of solubility refers to the balance between the dissolution of a compound and its precipitation from solution. It’s the state where the rate of solute dissolving equals the rate of solute precipitating. For reversible reactions like the dissolution of magnesium hydroxide:\[\mathrm{Mg(OH)_{2}(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^{-}(aq)}\]The system reaches equilibrium when the forward rate of \(\mathrm{Mg(OH)_2}\) dissolving equals the reverse rate of \(\mathrm{Mg^{2+}}\) and \(\mathrm{OH^-}\) ions forming back into the solid. Equilibrium ensures that the concentrations remain constant over time unless disturbed by changing conditions such as temperature or pressure. This balance ensures that the solid and the ions coexist in a dynamic but stable state, a concept that is essential in understanding reactions and solubility behaviors.