Question

Chromium(III) hydroxide dissolves in water only to the extent of \(8.21 \times 10^{-5} M\) at 25 C. Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Cr}(\mathrm{OH})_{3}\) at this temperature.

Short answer

The solubility product constant (Ksp) for Chromium(III) hydroxide (Cr(OH)₃) at 25°C is \(1.00 \times 10^{-11}\).

Step-by-step solution

1. Write the dissolution reaction for Cr(OH)₃

The balanced dissolution reaction for Cr(OH)₃ in water is: \[ \mathrm{Cr(OH)_3(s) \rightleftharpoons Cr^{3+}(aq) + 3OH^-(aq)} \]

2. Determine the equilibrium concentrations of Cr³⁺ and OH⁻ ions

Given the solubility of Cr(OH)₃ at 25°C is \(8.21 \times 10^{-5}\) M, it means that each mole of Cr(OH)₃ will produce 1 mole of Cr³⁺ ions and 3 moles of OH⁻ ions when dissolved in water. So, the equilibrium concentrations of Cr³⁺ and OH⁻ ions are: \[ [\mathrm{Cr^{3+}}] = 8.21 \times 10^{-5} \mathrm{M} \] \[ [\mathrm{OH^-}] = 3 \times 8.21 \times 10^{-5} \mathrm{M} = 2.463 \times 10^{-4} \mathrm{M} \]

3. Calculate Ksp for Cr(OH)₃

The Ksp expression for Cr(OH)₃, based on the dissolution reaction, is: \[ \mathrm{K_{sp} = [Cr^{3+}][OH^-]^3} \] Now, plug the equilibrium concentrations of Cr³⁺ and OH⁻ into the Ksp expression: Ksp = \((8.21 \times 10^{-5})(2.463 \times 10^{-4})^3\) Calculate Ksp: Ksp = \(1.00 \times 10^{-11}\) So, the solubility product constant (Ksp) for Chromium(III) hydroxide (Cr(OH)₃) at 25°C is \(1.00 \times 10^{-11}\).

Key concepts

Chromium Hydroxide

Chromium hydroxide is a chemical compound often represented by the formula \(\text{Cr(OH)}_3\). It appears as a solid that is slightly soluble in water. Its limited solubility means that only a tiny amount of this compound will dissolve to form ions.
When \(\text{Cr(OH)}_3\) is placed in water, it partially dissociates into chromium ions \(\text{Cr}^{3+}\) and hydroxide ions \(\text{OH}^-\). These ions are crucial for understanding how chromium hydroxide behaves in aqueous solutions.
This property is important in various chemical processes and environmental contexts, especially in wastewater treatment, where the precipitation and dissolution of hydroxides can affect heavy metal removal.

Dissolution Reaction

A dissolution reaction describes how a compound breaks down in water. For chromium hydroxide, the dissolution reaction is as follows:
\[ \text{Cr(OH)}_3(s) \rightleftharpoons \text{Cr}^{3+}(aq) + 3\text{OH}^-(aq) \]
In this reaction, solid chromium hydroxide dissociates into chromium ions and hydroxide ions. This process establishes a balance called dynamic equilibrium between the dissolved ions and the undissolved solid.
Understanding dissolution reactions helps us predict how substances will behave in various solutions and conditions. It also aids in calculating certain properties, like solubility product constants, to understand the extent of a substance's solubility.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of ions in a solution when the dissolution reaction has reached a steady state, where no further change occurs unless the conditions are altered. For \(\text{Cr(OH)}_3\), the equilibrium concentrations are calculated from its solubility.
Given that the solubility of chromium hydroxide at a certain temperature is \(8.21 \times 10^{-5}\) M, the concentration of \(\text{Cr}^{3+}\) ions in solution is the same, \(8.21 \times 10^{-5} \text{ M}\). Meanwhile, the concentration of \(\text{OH}^-\) ions is three times as much because each \(\text{Cr(OH)}_3\) releases three hydroxide ions upon dissolution. Thus, \([\text{OH}^-] = 3 \times 8.21 \times 10^{-5} = 2.463 \times 10^{-4} \text{ M}\).
These concentrations are vital for further calculations and understanding reactions that depend on the presence of these ions.

Ksp Calculation

The solubility product constant, \(K_{sp}\), quantifies the solubility of sparingly soluble compounds. Calculating this constant involves using the equilibrium concentrations of the ions formed from the dissolution. For \(\text{Cr(OH)}_3\), the formula for \(K_{sp}\) is:
\[ K_{sp} = [\text{Cr}^{3+}][\text{OH}^-]^3 \]
By substituting the equilibrium concentrations \([\text{Cr}^{3+}] = 8.21 \times 10^{-5} \text{ M}\) and \([\text{OH}^-] = 2.463 \times 10^{-4} \text{ M}\) into the equation, we find:
\[ K_{sp} = \left(8.21 \times 10^{-5}\right)\left(2.463 \times 10^{-4}\right)^3 \]
Calculating this yields \(K_{sp} = 1.00 \times 10^{-11}\).
Understanding \(K_{sp}\) is crucial for predicting how a solid will dissolve under different conditions, which is vital for both academic and practical applications, such as designing chemical reactions and treating water.