Question

\(K_{\text {sp }}\) for copper(II) hydroxide, \(\mathrm{Cu}(\mathrm{OH})_{2},\) has a value \(2.2 \times 10^{-20}\) at 25 " \(\mathrm{C}\). Calculate the solubility of copper(II) hydroxide in \(\mathrm{mol} / \mathrm{L}\) and \(\mathrm{g} / \mathrm{L}\) at \(25 \mathrm{C}\)

Short answer

The solubility of copper(II) hydroxide, Cu(OH)₂, at 25°C is approximately \(1.7 \times 10^{-7} \mathrm{mol/L}\) and \(1.66 \times 10^{-5} \mathrm{g/L}\).

Step-by-step solution

Write down the chemical equation and the dissociation equilibrium expression for Cu(OH)₂

First, let's write down the chemical equation for the dissociation of Cu(OH)₂ into its constituent ions: \[ \mathrm{Cu(OH)_2 \rightleftharpoons Cu^{2+} + 2OH^-} \] Now we can write the equilibrium expression for this reaction. As we are given the Kₛₚ value for Cu(OH)₂, the expression will be: \[ K_{sp} = [\mathrm{Cu^{2+}}][\mathrm{OH^-}]^2 \]

Define variables for the solubility of Cu(OH)₂ in mol/L

Since we want to calculate the solubility of Cu(OH)₂, let's assume its solubility in water is x mol/L. After dissociation, x mol of Cu²⁺ will be formed, and 2x mol of OH⁻ will be present in the solution.

Plug in the variables into the Kₛₚ expression and solve for x

Now, let's substitute these variables into the Kₛₚ expression: \[ K_{sp} = [\mathrm{Cu^{2+}}][\mathrm{OH^-}]^2 = (x)(2x)^2 \] Given that Kₛₚ for Cu(OH)₂ is \(2.2 \times 10^{-20}\), let's solve the equation for x: \[ 2.2 \times 10^{-20} = x(4x^2) = 4x^3 \] \[ x^3 = \frac{2.2 \times 10^{-20}}{4} \] \[ x^3 = 5.5 \times 10^{-21} \] \[ x = \sqrt[3]{5.5 \times 10^{-21}} \] \[ x \approx 1.7 \times 10^{-7} \mathrm{mol / L} \]

Convert x into grams per liter

The solubility of Cu(OH)₂ in mol/L is approximately \(1.7 \times 10^{-7} \mathrm{mol/L}\). To convert this into grams per liter, we need to multiply this value by the molar mass of Cu(OH)₂. The molar mass of Cu(OH)₂ is: Cu = 63.55 g/mol O = 2 × 16.00 g/mol = 32.00 g/mol H = 2 × 1.01 g/mol = 2.02 g/mol Molar mass of Cu(OH)₂ = 63.55 + 32.00 + 2.02 = 97.57 g/mol Now, let's convert the solubility into grams per liter by multiplying the solubility in mol/L by the molar mass of Cu(OH)₂: \[ 1.7 \times 10^{-7} \mathrm{mol/L} \times 97.57 \mathrm{g/mol} \approx 1.66 \times 10^{-5} \mathrm{g/L} \] So, the solubility of Cu(OH)₂ at 25 °C is approximately \(1.7 \times 10^{-7} \mathrm{mol/L}\) and \(1.66 \times 10^{-5} \mathrm{g/L}\).

Key concepts

Chemical Equilibrium

When a chemical reaction reaches a state where its rate of forward reaction equals its rate of reverse reaction, it is said to be at equilibrium. At this point, the concentrations of reactants and products remain constant over time. In the case of copper(II) hydroxide, Cu(OH)₂, equilibrium is reached when it dissociates into its ions: Cu²⁺ and OH⁻. The expression that describes this equilibrium is the solubility product (K_{sp}).

• The formula for the dissociation is:\[ \text{Cu(OH)}_2 \rightleftharpoons \text{Cu}^{2+} + 2\text{OH}^- \]
• The equilibrium expression is:\[ K_{sp} = [\text{Cu}^{2+}][\text{OH}^-]^2 \]

The K_{sp} value is very small, indicating low solubility of Cu(OH)₂ in water.

Copper Hydroxide

Copper hydroxide, Cu(OH)₂, is an insoluble blue-green compound found in various chemical processes. It forms when copper ions react with hydroxide ions in a solution. In this exercise, the objective is to find its solubility in water.

• It's important to understand that Cu(OH)₂ is slightly soluble, meaning only a small amount will dissolve.
• Its K_{sp} value of \(2.2 \times 10^{-20}\) indicates its tendency to remain undissolved.

This property is crucial in applications where controlled copper release is needed, such as fungicides.

Molar Solubility

Molar solubility refers to the number of moles of a substance that can dissolve per liter of solution to reach saturation. For Cu(OH)₂, it's defined as 'x' in the reaction:

• When Cu(OH)₂ dissociates, it forms x mol of Cu²⁺ and 2x mol of OH⁻.
• To find x, use the equation: \[ K_{sp} = (x)(2x)^2 = 4x^3 \]
• Solve for x: \[ x = \sqrt[3]{5.5 \times 10^{-21}} \approx 1.7 \times 10^{-7} \text{ mol/L} \]

This calculation provides insight into the concentration of ions at equilibrium.

Molar Mass Calculation

Molar mass is critical for converting solubility from moles per liter to grams per liter. To convert Cu(OH)₂'s solubility:

• Determine molar mass: Combine the atomic masses:\[\text{Cu: } 63.55 \text{ g/mol} \\text{O: } 2 \times 16.00 \text{ g/mol} = 32.00 \text{ g/mol} \\text{H: } 2 \times 1.01 \text{ g/mol} = 2.02 \text{ g/mol} \\text{Total: } 97.57 \text{ g/mol}\]
• Multiply solubility by molar mass:\[ 1.7 \times 10^{-7} \text{ mol/L} \times 97.57 \text{ g/mol} \approx 1.66 \times 10^{-5} \text{ g/L} \]

This results in the solubility of Cu(OH)₂ in grams per liter, a useful measure for practical applications.