Question

Calculate the molar solubility of \(\mathrm{Co}(\mathrm{OH})_{3}, K_{\mathrm{sp}}=2.5 \times 10^{-43}\).

Short answer

The molar solubility of Co(OH)₃, with a Ksp value of \(2.5 × 10^{-43}\), is approximately \(3.01 × 10^{-11}\) mol/L.

Step-by-step solution

Write the balanced dissolution equation

Co(OH)₃(s) → Co³⁺(aq) + 3OH⁻(aq)

Set up expression for Ksp

Ksp is the product of the equilibrium concentrations of the ions, raised to the power of their stoichiometric coefficients. For this reaction: Ksp = [Co³⁺] * [OH⁻]³

Define molar solubility (s)

Since the dissolution is 1:1 for Co³⁺ and 1:3 for OH⁻, we can represent their concentrations in terms of the molar solubility (s). [Co³⁺] = s [OH⁻] = 3s

Substitute concentrations into Ksp expression

Plug the expressions for [Co³⁺] and [OH⁻] into the Ksp expression: Ksp = s * (3s)³

Plug in given Ksp value and solve for s

We were given the Ksp value: Ksp = 2.5 × 10⁻⁴³ 2.5 × 10⁻⁴³ = s * (3s)³ Rearranging to solve for s: \(s^4 = \frac{2.5 × 10^{-43}}{27}\) Take the 4th root of both sides: \(s = \sqrt[4]{\frac{2.5 × 10^{-43}}{27}}\)

Find the molar solubility

Compute the numerical value of s: s = \( \sqrt[4]{\frac{2.5 × 10^{-43}}{27}} \approx 3.01 × 10^{-11}\) The molar solubility of Co(OH)₃ is approximately 3.01 × 10⁻¹¹ mol/L.

Key concepts

Solubility Product Constant (Ksp)

The Solubility Product Constant, commonly denoted as \( K_{sp} \), is a fundamental concept in chemistry that quantifies the solubility of a sparingly soluble compound. When a compound is added to a solvent and reaches a dynamic equilibrium, not all of it dissolves. The part that dissolves forms a saturated solution, reaching a point where the rate of dissolving is equal to the rate of precipitation. Here, \( K_{sp} \) comes into play.

The \( K_{sp} \) is the product of the molar concentrations of the resulting ions, with each concentration raised to the power of its respective coefficient in the balanced equation. Taking the dissolution of \( \mathrm{Co(OH)_3} \) as an example, the equation is:

• \( \mathrm{Co(OH)_3(s) \rightarrow Co^{3+}(aq) + 3OH^{-}(aq)} \)

For this reaction, the solubility product expression is set as:

• \( K_{sp} = [\mathrm{Co}^{3+}] \times [\mathrm{OH}^{-}]^3 \)

This equation signifies that the \( K_{sp} \) helps us predict the extent to which a compound will dissolve in water. If too high or too low, it can prevent the reaction from proceeding to a significant extent.

Ionic Equilibrium

Understanding ionic equilibrium is crucial for calculating the molar solubility of any compound. Ionic equilibrium refers to the state where the concentrations of all chemical species in a saturated solution are constant over time. It involves the equilibrium established between the dissolved ions and the undissolved solid.

When \( \mathrm{Co(OH)_3} \) dissolves, it forms \( \mathrm{Co^{3+}} \) ions and \( \mathrm{OH^{-}} \) ions in solution. The process achieves equilibrium when these ions exist at constant concentrations. In our case, we write the concentrations in terms of molar solubility \( s \):

• \([\mathrm{Co}^{3+}] = s\)
• \([\mathrm{OH}^{-}] = 3s\)

To find out how these ions behave under equilibrium conditions, substitute these into the expression for \( K_{sp} \). This substitution will give insights into how the dissolution process will balance itself out at equilibrium.

Stoichiometry

Stoichiometry is the study of the quantitative relationships in chemical reactions, particularly focusing on the ratios of reactants and products. In ionic solutions, understanding stoichiometry allows us to accurately set up equations that depict how much of each ion will result from the dissolution process.

In the dissolution reaction of \( \mathrm{Co(OH)_3} \) into \( \mathrm{Co^{3+}} \) and \( \mathrm{OH^{-}} \) ions, stoichiometry dictates that for every mole of \( \mathrm{Co(OH)_3} \) that dissolves, one mole of \( \mathrm{Co^{3+}} \) and three moles of \( \mathrm{OH^{-}} \) are produced. This relationship is crucial for setting up the expression in terms of molar solubility \( s \).

Using stoichiometry, we expressed these concentrations as follows:

• \([\mathrm{Co}^{3+}] = s\)
• \([\mathrm{OH}^{-}] = 3s\)

Recognizing this stoichiometric ratio is vital for calculating the correct molar solubility using the given \( K_{sp} \). It ensures that the reaction equations are balanced and that all calculated concentrations reflect reality in the system we’re analyzing.