Question

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

Short answer

The molar solubility of \(\mathrm{Mg(OH)_2}\) is \(1.34 \times 10^{-4}\,\text{mol/L}\).

Step-by-step solution

Write the dissolution equation

In this step, we will write the balanced chemical equation for the dissolution of \(\mathrm{Mg(OH)_2}\) in water: \[\mathrm{Mg(OH)_2 (s) \rightleftharpoons Mg^{2+} (aq) + 2OH^{-} (aq)}\] #step 2: Write \(K_{\mathrm{sp}}\) expression#

Write the expression for \(K_{\mathrm{sp}}\)

Based on the dissolution equation, we can write the \(K_{\mathrm{sp}}\) expression as: \[K_{\mathrm{sp}} = [\mathrm{Mg^{2+}}][\mathrm{OH^{-}}]^2\] #step 3: Define the molar solubility and equilibrium concentrations#

Define the molar solubility and equilibrium concentrations

Let's assume that the molar solubility of \(\mathrm{Mg(OH)_2}\) is \(s\) moles per litre. Then, at equilibrium, the concentrations of the ions will be: \([\mathrm{Mg^{2+}}] = s\) and \([\mathrm{OH^{-}}] = 2s\) #step 4: Substitute the equilibrium concentrations into the \(K_{\mathrm{sp}}\) expression#

Substitute the equilibrium concentrations into the \(K_{\mathrm{sp}}\) expression

Using the equilibrium concentrations, \([\mathrm{Mg^{2+}}] = s\) and \([\mathrm{OH^{-}}] = 2s\), we can rewrite the \(K_{\mathrm{sp}}\) expression as: \[K_{\mathrm{sp}} = s(2s)^2\] #step 5: Solve for the molar solubility, s#

Solve for the molar solubility (s)

To find the molar solubility (s), we first plug in the given \(K_{\mathrm{sp}}\) value, \(8.9 \times 10^{-12}\), and solve for \(s\): \[8.9 \times 10^{-12} = s(2s)^2\] This simplifies to: \[8.9 \times 10^{-12} = 4s^3\] Now solving for \(s\): \[s = \sqrt[3]{\frac{8.9 \times 10^{-12}}{4}} = 1.34 \times 10^{-4}\,\text{mol/L}\] So, the molar solubility of \(\mathrm{Mg(OH)_2}\) is \(1.34 \times 10^{-4}\,\text{mol/L}\).

Key concepts

Ksp calculation

Calculating the solubility product constant, often referred to as \(K_{sp}\), is a critical concept in understanding how solubility works at equilibrium. It helps determine how much of a substance can dissolve in water to form an equilibrium state. In this regard, \(K_{sp}\) is not just a number. It specifically represents the product of the molar concentrations of the ions each raised to the power of their respective coefficients from the balanced dissolution equation.

For example, if a substance dissolves and dissociates into magnesium ions \([\text{Mg}^{2+}]\) and hydroxide ions \([\text{OH}^-]\), as shown in the dissolution reaction, the expression for \(K_{sp}\) would be \([\text{Mg}^{2+}][\text{OH}^-]^2\). This calculation includes each ion's concentration at equilibrium, underlining the importance of understanding "equilibrium concentrations," which we will cover later.

Understanding \(K_{sp}\) can later be used to predict the behavior of ionic compounds in saturated solutions and is vital for applications ranging from environmental science to pharmaceuticals.

Chemical equilibrium

Chemical equilibrium refers to the state of a chemical reaction where the rate of the forward reaction is equal to the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products over time. It’s a dynamic situation where reactions continue to occur, but because they are happening at the same speed in both directions, concentrations remain constant.

For dissolution reactions like that of \(\text{Mg(OH)}_2\), achieving chemical equilibrium means that the rate of \(\text{Mg(OH)}_2\) dissolving into magnesium and hydroxide ions is equal to the rate at which these ions can recombine and precipitate out as solid \(\text{Mg(OH)}_2\).

• At equilibrium, the concentration of each ion in solution doesn't change.
• The equilibrium state is captured quantitatively through the \(K_{sp}\), which represents the maximum product of the ion concentrations under equilibrium conditions.

Grasping chemical equilibrium is essential to predict how changes in conditions (like temperature or concentration) can shift the equilibrium, which in turn can affect the solubility of compounds.

Dissolution equation

A dissolution equation represents the process by which an ionic compound dissolves in a solvent to form its constituent ions. It's the backbone of understanding how substances interact in solutions. For \(\text{Mg(OH)}_2\), the equation is:\[ \text{Mg(OH)}_2(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq) \]

• On the left, you have the solid: \(\text{Mg(OH)}_2\), indicating it's initially undissolved.
• On the right, you see the ions \([\text{Mg}^{2+}]\) and \([\text{OH}^-]^2\), showing how they separate in water.
• The double arrow signifies that the equation is reversible, emphasizing the concept of equilibrium.

The dissolution equation is crucial for writing the \(K_{sp}\) expression, as it shows exactly how many ions are produced per formula unit of compound, directly impacting the calculation of equilibrium concentrations. Understanding this equation is also a stepping stone for more complex chemical reactions, illustrating the balance and interplay between reactants and products.

Equilibrium concentrations

Equilibrium concentrations are the concentrations of reactants and products when a reaction is at equilibrium. For complete mastery of solubility concepts, recognizing how to calculate these concentrations is foundational.

In our case study with \(\text{Mg(OH)}_2\), we assume the molar solubility \(s\) is how much \(\text{Mg(OH)}_2\) dissolves to form ions. Thus, at equilibrium:

• \([\text{Mg}^{2+}] = s\)
• \([\text{OH}^-] = 2s\)

This is because each unit of \(\text{Mg(OH)}_2\) generates one \([\text{Mg}^{2+}]\) ion and two \([\text{OH}^-]\) ions when dissolved. Substituting these values into the \(K_{sp}\) expression allows us to find the specific value of \(s\) and determine the molar solubility.

Understanding and calculating equilibrium concentrations is not only critical for solving exercises like these but also offers a window into how various systems behave under equilibrium conditions, enabling predictions about solubility in different environments.