Problem 115 - Chemistry The Science in Context

Question

When silver oxide dissolves in water, the following reaction occurs: $$ \mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{Ag}^{+}(a q)+2 \mathrm{OH}^{-}(a q) $$ If a saturated aqueous solution of silver oxide is \(1.6 \times 10^{-4} M\) in hydroxide ion, what is the \(K_{\mathrm{sp}}\) of silver oxide?

Step-by-Step Solution

Answer

Answer: The solubility product constant (Kₛₚ) for silver oxide is approximately \(1.0 \times 10^{-11}\).

Step 1: Write the balanced chemical equation for the reaction

Given the balanced chemical equation for the dissolution of silver oxide in water: $$ \mathrm{Ag}_{2}\mathrm{O}(s)+\mathrm{H}_{2}\mathrm{O}(\ell) \rightarrow 2\mathrm{Ag}^{+}(a q)+2\mathrm{OH}^{-}(a q) $$

Step 2: Write the expression for Kₛₚ

The expression for the solubility product constant, Kₛₚ, can be written as follows: $$ K_{sp} = [\mathrm{Ag}^{+}]^{2}[\mathrm{OH}^{-}]^{2} $$

Step 3: Identify the relationship between the given concentration and the equilibrium concentrations

Since the reaction produces 2 moles of Ag⁺ and 2 moles of OH⁻ for each mole of Ag₂O that dissolves, we can deduce that: $$ [\mathrm{Ag}^{+}] = 2[\mathrm{OH}^{-}] $$

Step 4: Substitute the given concentration and relationship into the Kₛₚ expression

We substitute the given concentration of hydroxide ions, which is \(1.6 \times 10^{-4} M\), into the Kₛₚ expression and use the relationship between the Ag⁺ and OH⁻ concentrations: $$ K_{sp} = (2 \times 1.6 \times 10^{-4})^{2} \times (1.6 \times 10^{-4})^{2} $$

Step 5: Calculate the value of Kₛₚ

Now, we simply need to compute the value for Kₛₚ: $$ K_{sp} = (3.2 \times 10^{-4})^{2}(1.6 \times 10^{-4})^{2} = 1.024 \times 10^{-11} $$ So, the solubility product constant (Kₛₚ) for silver oxide is approximately \(1.0 \times 10^{-11}\).

Key Concepts

Silver Oxide Dissolution

Silver oxide (\( \mathrm{Ag}_2\mathrm{O} \)) dissolves in water by breaking down into silver ions and hydroxide ions. This process is known as dissolution, and it involves a chemical reaction where the solid substance is converted into ions in an aqueous solution. In this case, the balanced equation is:

• \( \mathrm{Ag}_2\mathrm{O}(s) + \mathrm{H}_2\mathrm{O}(\ell) \rightarrow 2\mathrm{Ag}^{+}(aq) + 2\mathrm{OH}^{-}(aq) \)

Understanding this reaction is crucial because it reveals how the solid form of silver oxide is transformed into ions when mixed with water. As the equation shows, each unit of silver oxide produces two silver ions (\( \mathrm{Ag}^{+} \)) and two hydroxide ions (\( \mathrm{OH}^{-} \)). This balanced chemical equation is foundational to calculating other properties, such as the solubility product constant (Kₛₚ).

Ksp Expression

Solubility product constant, or Kₛₚ, is a measure of the solubility of a compound in a solution. The Kₛₚ for a dissolution reaction is expressed using the concentrations of the ions produced by the dissolution. In the case of silver oxide dissolving in water, the expression for \(K_{sp} \) is:

• \( K_{sp} = [\mathrm{Ag}^{+}]^{2}[\mathrm{OH}^{-}]^{2} \)

This formula indicates that the solubility product is determined by multiplying the concentrations of the silver ions and hydroxide ions, each raised to the power of their stoichiometric coefficients from the balanced chemical equation. When the solution is saturated, the concentrations of these ions will remain constant, reflecting equilibrium within the solution. Calculating \( K_{sp} \) helps predict whether a precipitate will form under a given set of conditions.

Equilibrium Concentrations

Equilibrium concentrations refer to the concentrations of the reactants and products when the chemical reaction has reached a state where the rates of the forward and reverse reactions are equal. In the context of silver oxide dissolution, these concentrations can be calculated using the stoichiometry from the balanced chemical equation.With the given information, we know the concentration of hydroxide ions (\( \mathrm{OH}^{-} \)) is \( 1.6 \times 10^{-4} \, \text{M} \). Using the equation relationship, we find:

• \( [\mathrm{Ag}^{+}] = 2[\mathrm{OH}^{-}] = 2 \times 1.6 \times 10^{-4} = 3.2 \times 10^{-4} \, \text{M} \)

These concentrations are critical because they are used in the Kₛₚ expression to determine the solubility product constant. The equilibrium state shows the levels at which ions exist in solution without further net change, an essential aspect for predicting solubility and conditions that lead to precipitation.