Problem 58 - Chemistry The Central Science In Si Units

Question

At \(25^{\circ} \mathrm{C}\), the reaction $$ \mathrm{CaCrO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CrO}_{4}^{2-}(a q) $$ has an equilibrium constant \(K_{c}=7.1 \times 10^{-4}\). What are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{CrO}_{4}^{2-}\) in a saturated solution of \(\mathrm{CaCrO}_{4} ?\)

Step-by-Step Solution

Answer

The equilibrium concentrations are approximately 0.0266 M for both \( \mathrm{Ca}^{2+} \) and \( \mathrm{CrO}_{4}^{2-} \).

Step 1: Write the Equilibrium Expression

The reaction given is \( \mathrm{CaCrO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + \mathrm{CrO}_{4}^{2-}(aq) \). Since this is an equilibrium reaction, we use the equilibrium constant expression to find \( K_{c} \). For the reaction, \( K_{c} = [\mathrm{Ca}^{2+}][\mathrm{CrO}_{4}^{2-}] \). The concentration of the solid, \( \mathrm{CaCrO}_4(s) \), is not included in the \( K_c \) expression.

Step 2: Assume Equal Concentrations at Equilibrium

Since \( \mathrm{CaCrO}_4 \) dissociates into one \( \mathrm{Ca}^{2+} \) ion and one \( \mathrm{CrO}_{4}^{2-} \) ion, their concentrations at equilibrium are equal. Denote the concentration of \( \mathrm{Ca}^{2+} \) as \( x \), therefore, \( [\mathrm{Ca}^{2+}] = [\mathrm{CrO}_{4}^{2-}] = x \).

Step 3: Set Up the Equation

Substitute the concentrations into the equilibrium expression: \( K_{c} = x \times x = x^2 \). Given \( K_{c} = 7.1 \times 10^{-4} \), set up the equation: \( x^2 = 7.1 \times 10^{-4} \).

Step 4: Solve for \( x \)

To find \( x \), take the square root of both sides: \( x = \sqrt{7.1 \times 10^{-4}} \). Calculating gives \( x \approx 0.0266 \).

Step 5: Write the Equilibrium Concentrations

Since \( [\mathrm{Ca}^{2+}] = x \) and \( [\mathrm{CrO}_{4}^{2-}] = x \), the equilibrium concentrations are both approximately \( 0.0266 \) M.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), is a pivotal concept in chemical equilibrium that lets us understand the balance of reactions. For reactions that occur in a continuous and reversible fashion, the equilibrium constant represents the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients.

For the reaction \( \mathrm{CaCrO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq) + \mathrm{CrO}_{4}^{2-}(aq) \), the expression for the equilibrium constant is given by:

• \( K_{c} = [\mathrm{Ca}^{2+}][\mathrm{CrO}_{4}^{2-}] \)

Notably, solids like \( \mathrm{CaCrO}_4(s) \) do not appear in the \( K_c \) expression because their concentration remains constant as they do not enter the solution phase. The value of \( K_c \) offers insight into the state of equilibrium:

• A small \( K_c \), as in this exercise, indicates that at equilibrium, the concentration of the reactants remains higher compared to the products.

Saturated Solution

A saturated solution is one where no more solute can dissolve at a given temperature. In the context of this exercise, the saturated solution contains dissolved ions in equilibrium with the undissolved solid. For \( \mathrm{CaCrO}_4 \), it reaches a saturation point where an equilibrium is established between its solid form and the ions present in the solution.

• At \( 25^{\circ} \mathrm{C} \), this equilibrium is characterized by the maximum possible concentration of \( [\mathrm{Ca}^{2+}] \) and \( [\mathrm{CrO}_{4}^{2-}] \) ions for that temperature.
• The concentrations of these ions do not change, provided the temperature stays constant, the defining point of a saturated solution.

The solid form will precipitate if additional solute is added, maintaining the balance of this equilibrium.

Concentration Calculations

Concentration calculations are vital for quantitatively describing the state of a solution at equilibrium. They help us understand to what extent a reaction proceeds.

In our exercise equilibrium problem, we express the initial concentrations of \( [\mathrm{Ca}^{2+}] \) and \( [\mathrm{CrO}_{4}^{2-}] \) as \( x \), based on the assumption that the dissociation of \( \mathrm{CaCrO}_4 \) in a saturated solution yields equal concentrations of the ions.

• This gives \( x^2 = K_c \), where \( K_c = 7.1 \times 10^{-4} \).
• The solution requires us to solve the equation \( x^2 = 7.1 \times 10^{-4} \), leading to \( x = \sqrt{7.1 \times 10^{-4}} \).

The calculation of \( x \) equates to approximately \( 0.0266 \) M, providing the equilibrium concentrations in a saturated solution. Each of these calculation steps ensures precision and understanding of how components in a system behave when at equilibrium.