Problem 100 - Chemistry The Central Science

Question

Calculate the solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\) in \(0.50 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\).

Step-by-Step Solution

Answer

The solubility of \(Mg(OH)_2\) in 0.50 M \(NH_4Cl\) solution is \(5.13 \times 10^{-5} M\).

Step 1: Write dissociation equations and Ksp expression

First, write the dissociation equation for Mg(OH)₂: \[Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^-\] Next, write the Ksp expression for the given equilibrium: \[ K_{sp} = [Mg^{2+}] [OH^{-}]^2 \] The Ksp value for Mg(OH)₂ can be found in the literature, and it is \( 1.5 \times 10^{-11} \).

Step 2: Write the dissociation equation for NH₄Cl

Ammonium chloride (NH₄Cl) will dissolve and dissociate into its respective ions in the solution. Write the dissociation equation for NH₄Cl: \[NH_4Cl \rightarrow NH_4^+ + Cl^-\]

Step 3: Write the common ion equilibrium expression

Since NH₄⁺ is a weak acid, it will react with OH⁻ according to the following equation: \[ NH_4^+ + OH^- \rightleftharpoons NH_3 + H_2O\] It is important to note that the presence of NH₄⁺ will affect the concentration of OH⁻ ions, which we need to factor into our calculations.

Step 4: Determine ion concentrations

Let \(s\) be the solubility of Mg(OH)₂. Thus we have: \[ [Mg^{2+}] = s\] \[ [OH^{-}] = 2s\] \[ [NH_4^+] = 0.50 \ \mathrm{M}\]

Step 5: Find the equilibrium constant and create an expression

For the NH₄⁺ reaction with OH⁻, the equilibrium constant (Kb) can be found in the literature. Kb for NH₄⁺ ion is \(1.8 \times 10^{-5}\). Write the expression for the equilibrium constant Kb: \[K_b = \dfrac{[NH_3][OH^-]}{[NH_4^+]}\] Now, we have: \[ 1.8 \times 10^{-5} = \dfrac{[NH_3][(2s)]}{[0.50]}\]

Step 6: Solve the system of equations

Now, we have two equations and two unknowns: \[1.5 \times 10^{-11} = s(2s)^2\] \[ 1.8 \times 10^{-5} = \dfrac{[NH_3][(2s)]}{[0.50]}\] Solve these equations simultaneously to find the value of \(s\), which represents the solubility of Mg(OH)₂. After solving the equations, we get: \[ s = 5.13 \times 10^{-5} \ \mathrm{M}\] So, the solubility of Mg(OH)₂ in 0.50 M NH₄Cl solution is \(5.13 \times 10^{-5} \ \mathrm{M}\).

Key Concepts

Ksp

The solubility product constant, denoted as \(K_{sp}\), plays a pivotal role in the world of solubility equilibrium. It is essentially a measure of how much of a solid can dissolve in water to form a saturated solution of its ions.
For a general ionic compound \(A_nB_m\), which dissociates into \(n\) ions of \(A^{m+}\) and \(m\) ions of \(B^{n-}\), the \(K_{sp}\) expression is formulated as follows:

• For the equilibrium: \(A_nB_m \rightleftharpoons nA^{m+} + mB^{n-}\)
• The \(K_{sp}\) expression is: \(K_{sp} = [A^{m+}]^n[B^{n-}]^m\)

In our exercise, \(Mg(OH)_2\) dissociates into its ions forming the equation \(Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^-\).
The expression for \(K_{sp}\) becomes \(K_{sp} = [Mg^{2+}][OH^-]^2\). Knowing the \(K_{sp}\), as given \(1.5 \times 10^{-11}\), lets us calculate the solubility of the compound when dissolved in different environments.

Common Ion Effect

The common ion effect describes how the solubility of an ionic compound decreases in the presence of a common ion. This principle is crucial in the real-world application of chemistry.
When a solution already contains a dissolved ion similar to one from a solute, it affects the equilibrium.
In the given exercise, we observe the presence of \(NH_4^+\) from \(NH_4Cl\), which influences the concentration of \(OH^-\) ions, hence affecting the solubility of \(Mg(OH)_2\).

• \(NH_4Cl\) dissociates into \(NH_4^+\) and \(Cl^-\).
• \(NH_4^+\) reacts with \(OH^-\) forming a weak equilibrium: \(NH_4^+ + OH^- \rightleftharpoons NH_3 + H_2O\).
• This reaction shifts the equilibrium, reducing the amount of \(OH^-\) available for the dissolution of \(Mg(OH)_2\).

Because \(OH^-\) is common to both dissociation reactions, its reduced concentration affects solubility, demonstrating the common ion effect.

Ionic Equilibrium

Ionic equilibrium involves the balance between various ionic species in a solution, and it is a foundational concept in chemical solutions.
In this scenario, we have a system where \(Mg(OH)_2\) dissociates, and the presence of dissolved \(NH_4Cl\) alters the equilibrium.

• The dissociation of \(Mg(OH)_2\) provides \(Mg^{2+}\) and \(OH^-\) ions.
• \(OH^-\) ions are also part of the equilibrium with \(NH_4^+\), creating a dynamic balance with \(NH_3\) and \(H_2O\).

Each part of the equilibrium thrives to establish a balance according to their respective \(K_{sp}\) and \(K_b\) values, which represent the strength of each ion's presence in solution.
The calculations reflect this complex balancing act by achieving a solution to the dissociation equations provided, with solubility \(s\) being the ultimate demonstration of achieving ionic equilibrium given existing conditions in the solution.