Question

At \(25^{\circ} \mathrm{C}\), the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) in water is \(2.86 \times 10^{-9} \mathrm{M}\). What are the equilibrium concentrations of the cation and the anion in a saturated solution?

Short answer

In a saturated solution of Al(OH)3 at 25°C with solubility \(2.86 \times 10^{-9} \mathrm{M}\), the equilibrium concentrations of Al3+ and OH- are \(2.86 \times 10^{-9} \mathrm{M}\) and \(8.58 \times 10^{-9} \mathrm{M}\), respectively.

Step-by-step solution

Write the balanced chemical equation for Al(OH)3's dissolution

Al(OH)3 (s) ↔ Al3+ (aq) + 3 OH- (aq)

Set up the expression for Ksp

For a balanced chemical equation 'aA(s) ↔ bB(aq) + cC(aq)', the solubility product constant (Ksp) can be defined as: Ksp = [B]^b × [C]^c Using the balanced chemical equation for the dissolution of Al(OH)3, Ksp = [Al3+] × [OH-]^3

Calculate the equilibrium concentrations of cation and anion

Since the solubility of Al(OH)3 is given as \(2.86 \times 10^{-9}\) M, we have the same solubility for Al3+ as it is in a 1:1 stoichiometry with Al(OH)3. Therefore, [Al3+] = \(2.86 \times 10^{-9}\) M. Now, the balanced chemical equation shows that there is a 1:3 stoichiometry between Al3+ and OH-. Hence, for each Al3+ ion, there will be three OH- ions in the solution. Thus, to find [OH-], we multiply the given molar concentration of Al3+ by 3: [OH-] = 3 × [Al3+] = 3 × \(2.86 \times 10^{-9}\) M = \(8.58 \times 10^{-9}\) M Therefore, the equilibrium concentrations of Al3+ and OH- are \(2.86 \times 10^{-9}\) M and \(8.58 \times 10^{-9}\) M, respectively.

Key concepts

Equilibrium Concentrations

When we talk about equilibrium concentrations, we're referring to the concentrations of ions or molecules in a solution at equilibrium, where the rates of the forward and backward reactions are equal. In the case of solubility, this involves a solid dissolving in water until no more can dissolve, leading to a saturated solution. At this point, the concentration of dissolved ions reaches a stable state, which we call the equilibrium concentration.

For Al(OH)₃, its dissolution reaches equilibrium when it turns into Al³⁺ ions and OH⁻ ions in water. The given solubility of Al(OH)₃ tells us how many moles of Al(OH)₃ can dissolve per liter of water, reaching this stable state. Using the solubility value, we can directly find the equilibrium concentration of Al³⁺, because the solid dissociates to produce one Al³⁺ ion for each formula unit that dissolves.

Calculating for OH⁻ involves considering the stoichiometry of the reaction, as each Al(OH)₃ produces three OH⁻ ions. Knowing how to determine these concentrations is key to understanding the solubility product, Ksp.

Stoichiometry

Stoichiometry helps us understand the quantitative relationships in chemical reactions. It is based on the balanced chemical equation and allows us to convert between different chemical species. In the dissolution reaction of Al(OH)₃, stoichiometry plays a crucial role in figuring out how much of each ion is produced.

The balanced equation for the dissolution is: Al(OH)₃ (s) ↔ Al³⁺ (aq) + 3 OH⁻ (aq).

This tells us that one mole of Al(OH)₃ produces one mole of Al³⁺ and three moles of OH⁻ ions. Thus, if the solubility of Al³⁺ is given, as in our example, we multiply this value by three to find the OH⁻ concentration, because there are three hydroxide ions for every one aluminum ion formed. This stoichiometric relationship is central to interconverting the concentrations of solute particles.

Understanding stoichiometry helps clarify how much of each ion will be in the solution once equilibrium is reached.

Dissolution Reaction

A dissolution reaction involves a solid substance dissolving in a solvent to form a solution, and in the context of salts like Al(OH)₃, it's a reversible process. This means equilibrium can be achieved, as both the dissolution and precipitation processes may occur simultaneously until a balance is reached.

In the example of Al(OH)₃, when it dissolves in water, it dissociates into its constituent ions: one Al³⁺ ion and three OH⁻ ions. At the molecular level, you can envision solid Al(OH)₃ particles dissociating into ions that move into the solution. It's at dissolution that solubility characteristics of a compound are defined, and a balanced equation is key to understanding how many ions of each type are produced.

This balanced process defines how a material dissolves and allows us to express the equilibrium state through the solubility product constant, or Ksp, which quantifies the saturated concentration of ions in the solution at equilibrium.