Question

The solubility product for \(\mathrm{Zn}(\mathrm{OH})_{2}\) is \(3.0 \times 10^{-16}\). The formation constant for the hydroxo complex, \(\mathrm{Zn}(\mathrm{OH})_{4}{ }^{2-},\) is \(4.6 \times 10^{17}\). What concentration of \(\mathrm{OH}^{-}\) is required to dissolve 0.015 mol of \(\mathrm{Zn}(\mathrm{OH})_{2}\) in a liter of solution?

Short answer

To dissolve 0.015 mol of Zn(OH)₂ in a liter of solution, the concentration of OH⁻ required can be calculated using the given solubility product and formation constant, as well as the equilibrium equations for the dissolution of Zn(OH)₂ and the formation of Zn(OH)₄²⁻ complex. After solving the system of non-linear equations for the amount of Zn(OH)₄²⁻ formed (x), the concentration of OH⁻ required can be determined using the relation: \(OH^{-} = 0.030 - 2x\).

Step-by-step solution

Write the balanced equation for the dissolution of Zn(OH)₂

For the dissolution of Zn(OH)₂, the balanced equation is: \[Zn(OH)_{2} \rightleftharpoons Zn^{2+} + 2OH^{-}\]

Write the balanced equation for the formation of Zn(OH)₄²⁻ complex

For the formation of the Zn(OH)₄²⁻ complex, the balanced equation is: \[Zn^{2+} + 4OH^{-} \rightleftharpoons Zn(OH)_{4}^{2-}\]

Write the expressions for the solubility product and the formation constant

The solubility product (Ksp) expression for Zn(OH)₂ is: \[K_{sp} = [Zn^{2+}][OH^{-}]^{2}\] And the formation constant (Kf) expression for the Zn(OH)₄²⁻ complex is: \[K_{f} = \frac{[Zn(OH)_{4}^{2-}]}{[Zn^{2+}][OH^{-}]^{4}}\]

Use the given data to set the equilibrium equations

The dissolution process will produce 0.015 mol of Zn(OH)₂, 1 mol of which produces 2 mol of OH⁻. The formation of Zn(OH)₄²⁻ will consume some OH⁻ ions. We can use x as the amount of Zn(OH)₄²⁻ formed and set up the relation between the concentrations of the species involved: \[Zn^{2+} = 0.015 - x\] \[OH^{-} = 0.030 - 2x\] \[Zn(OH)_{4}^{2-} = x\]

Substitute the given values into the equations and solve for x

Substitute the values of Ksp, Kf, and concentrations into the equilibrium expressions: \(K_{sp} = (0.015 - x)(0.030 - 2x)^{2} = 3.0 \times 10^{-16}\) \(K_{f} = \frac{x}{(0.015 - x)(0.030 - 2x)^{4}} = 4.6 \times 10^{17}\) Now, we can solve these equations simultaneously for x, which is the amount of Zn(OH)₄²⁻ formed. Your best option is to use a graphing calculator or online software to solve this system of non-linear equations to find the value of x.

Determine the concentration of OH⁻ required

After solving for x, the concentration of OH⁻ required can be calculated using the relation: \[OH^{-} = 0.030 - 2x\] After substituting the value of x, you will get the concentration of OH⁻ required to dissolve 0.015 mol of Zn(OH)₂ in a liter of solution.

Key concepts

Solubility Product Constant (Ksp)

Understanding the solubility product constant, commonly referred to as \(K_{sp}\), is crucial in solubility equilibria. It is the equilibrium constant for a solid substance dissolving in an aqueous solution. In simple terms, it's a measure of how much of the compound can dissolve in water. For a general reaction where a weakly soluble salt \(AB\) dissolves to form its ions, the expression is:

• \(AB \rightleftharpoons A^+ + B^-\)

The \(K_{sp}\) expression would be \([A^+][B^-]\). The solubility product is specific to a particular compound.

In the exercise given, the solubility product for \(Zn(OH)_{2}\) is \(3.0 \times 10^{-16}\). This very low \(K_{sp}\) value indicates that \(Zn(OH)_{2}\) is not very soluble in water. Calculating \(K_{sp}\) involves understanding the concentrations of the products of the dissolution reaction at equilibrium:

• \(K_{sp} = [Zn^{2+}][OH^{-}]^{2}\)

This equation tells us how the ions are distributed in solution. Since \(K_{sp}\) multiplied by \(OH^-\) squared yields a constant, it becomes pivotal when predicting the solubility of the compound under various conditions.

Complex Ion Formation

Complex ion formation involves the combination of a metal ion with ligands (typically ions or molecules) to form a complex ion. This process significantly influences the solubility of compounds by changing equilibria. When a complex forms, it can stabilize the metal ion in solution, often increasing the solubility of the original compound.

In this particular exercise, the complex ion formation is represented by the equation:

• \(Zn^{2+} + 4OH^{-} \rightleftharpoons Zn(OH)_{4}^{2-}\)

The formation constant \(K_{f}\) reflects how strongly the complex ion forms; a high \(K_{f}\)—like the \(4.6 \times 10^{17}\) in this case—indicates a highly stable complex.

This equilibrium shows us that as \(Zn^{2+}\) ions in the solution combine with the \(OH^-\) ions, they form the \(Zn(OH)_{4}^{2-}\) complex, thereby reducing the concentration of \(Zn^{2+}\) ions in the solution. The shift makes more \(Zn(OH)_{2}\) dissolve, altering the original solubility equilibrium.

Equilibrium Calculations

Equilibrium calculations help predict the concentrations of substances involved in solubility equilibria at a given state. With the exercise at hand, the goal is to calculate the concentration of \(OH^-\) required to dissolve a certain amount of \(Zn(OH)_{2}\).

In this scenario, we approached the problem step-by-step:

• Identifying variables: the dissolution of \(Zn(OH)_{2}\) and the formation of \(Zn(OH)_{4}^{2-}\).
• Using expressions for \(K_{sp}\) and \(K_{f}\), connect the solubility and complex formation processes.

The system involves simultaneous equations derived from these equilibria:

• \(K_{sp} = (0.015 - x)(0.030 - 2x)^{2}\)
• \(K_{f} = \frac{x}{(0.015 - x)(0.030 - 2x)^{4}}\)

Here, \(x\) represents the amount of complex formed. Solving these equations simultaneously provides the value of \(x\),

Once \(x\) is found, substitute it back to determine the remaining \(OH^-\) concentration. This exact method ensures that the desired quantities of reactants and products are achieved, maintaining equilibrium.