Problem 23 - Chemistry Matter and Change

Question

Calculate the solubility of \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\left(K_{\mathrm{sp}}=2.6 \times 10^{-18}\right)\)

Step-by-Step Solution

Answer

The solubility of \(Ag_{3}PO_{4}\) in water is approximately \(6.34 \times 10^{-5}\) mol/L.

Step 1: Write the dissolution equilibrium equation

To solve this problem, we first need to write the balanced equation for the dissolution of silver phosphate in water: \[Ag_{3}PO_{4}(s) \rightleftharpoons 3Ag^{+}(aq) + PO_{4}^{3-}(aq)\]

Step 2: Define solubility and equilibrium expressions

Let's define solubility (s) as the amount of Ag₃PO₄ that dissolves in water. When one unit of Ag₃PO₄ dissolves, it forms three units of Ag⁺ ions and one unit of PO₄³⁻ ions. Therefore, the concentration of Ag⁺ at equilibrium is 3s, and the concentration of PO₄³⁻ is s. Write the equilibrium expression for Ksp: \[K_{sp} = [Ag^{+}]^{3}[PO_{4}^{3-}]\] Substitute the concentrations of Ag⁺ and PO₄³⁻ in terms of s: \[K_{sp} = (3s)^{3}(s)\]

Step 3: Substitute Ksp and solve for s

Given that the Ksp of Ag₃PO₄ is 2.6 × 10⁻¹⁸, we can now substitute this value into the equation and solve for s: \[2.6 \times 10^{-18} = (3s)^{3}(s)\] Simplify and solve for s: \[2.6 \times 10^{-18} = 27s^{4}\] \[s^{4} = \frac{2.6 \times 10^{-18}}{27}\] \[s^{4} = 9.63 \times 10^{-20}\] \[s = \sqrt[4]{9.63 \times 10^{-20}}\] \[s = 6.34 \times 10^{-5}\]

Step 4: Calculate the solubility of Ag₃PO₄

The value of s represents the solubility of Ag₃PO₄ in water. Therefore, the solubility of Ag₃PO₄ is 6.34 × 10⁻⁵ mol/L.

Key Concepts

Dissolution Equilibrium

In the context of chemistry, dissolution equilibrium is a fascinating concept. It describes the state when the process of a solid dissolving in a liquid and the reverse process of precipitation are occurring at equal rates. This dynamic balance allows a substance to be in a state where no net change in concentration occurs with time. Consider the dissolution of silver phosphate, \(Ag_{3}PO_{4}\). When it dissolves in water, the bonds in the solid structure break apart. This results in the formation of silver ions (\(Ag^{+}\)) and phosphate ions (\(PO_{4}^{3-}\)). At equilibrium, the dissolution and reformation of Ag₃PO₄ are perfectly balanced. This means the rate at which \(Ag_{3}PO_{4}\) dissolves is equal to the rate it precipitates.

• This equilibrium can be represented in a chemical equation as follows: \(Ag_{3}PO_{4}(s) \rightleftharpoons 3Ag^{+}(aq) + PO_{4}^{3-}(aq)\).
• The double-headed arrow indicates the reversible nature of the reaction.
• Understanding this equilibrium is key to solving solubility problems.

Solubility Product Constant (Ksp)

The solubility product constant, denoted as \(K_{sp}\), serves as a numerical representation of the solubility of a compound. It is a specific type of equilibrium constant that calculates the product of the ion concentrations at which a sparingly soluble compound is at equilibrium. In the case of \(Ag_{3}PO_{4}\), the \(K_{sp}\) value is \(2.6 \times 10^{-18}\). This extremely small value signifies that Ag₃PO₄ poorly dissolves in water.

• The expression of \(K_{sp}\) for \(Ag_{3}PO_{4}\) can be written as \(K_{sp} = [Ag^{+}]^{3}[PO_{4}^{3-}]\).
• This formula shows the relationship between the molar concentrations of the dissociated ions.
• The \(K_{sp}\) equation is crucial for solving and understanding the solubility of a compound.

Chemical Equilibrium Expressions

Chemical equilibrium expressions are vital in predicting the concentrations of reactants and products in a chemical reaction at equilibrium. For the dissolution of compounds like \(Ag_{3}PO_{4}\), these expressions are derived directly from the balanced dissolution equation.
To obtain the equilibrium expression for silver phosphate:

• First, state the dissolution equilibrium: \(Ag_{3}PO_{4}(s) \rightarrow 3Ag^{+}(aq) + PO_{4}^{3-}(aq)\).
• Then, write the expression using the concentrations of ions: \(K_{sp} = [Ag^{+}]^{3}[PO_{4}^{3-}]\).
• Remember: only aqueous ions and molecules appear in this expression, as the concentration of a pure solid remains constant.

Substituting the concentration terms into the equilibrium expression, where \([Ag^{+}] = 3s\) and \([PO_{4}^{3-}] = s\) (with \(s\) being solubility), makes it possible to solve for \(s\). This is particularly useful for determining how much of a solute can dissolve in a solvent under specific conditions, providing practical insights into the behavior of sparingly soluble compounds.