Problem 55 - Chemistry The Central Science In Si Units

Question

Consider the reaction $$ \mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) $$ At \(25^{\circ} \mathrm{C}\), the equilibrium constant is \(K_{c}=2.4 \times 10^{-5}\) for this reaction. (a) If excess \(\operatorname{CaSO}_{4}(s)\) is mixed with water at \(25^{\circ} \mathrm{C}\) to produce a saturated solution of \(\mathrm{CaSO}_{4}\), what are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}{\underline{\phantom{xx}}}^{2-}\) ? (b) If the resulting solution has a volume of \(1.4 \mathrm{~L}\), what is the minimum mass of \(\operatorname{CaSO}_{4}(s)\) needed to achieve equilibrium?

Step-by-Step Solution

Answer

Equilibrium concentrations: \( [\text{Ca}^{2+}] = [\text{SO}_4^{2-}] = 4.9 \times 10^{-3} \) M; minimum mass of \( \text{CaSO}_4 = 0.93 \) g.

Step 1: Write the expression for Kc

The equilibrium constant expression for the given reaction is \[ K_c = [\text{Ca}^{2+}][\text{SO}_4^{2-}] \] because \( K_c \) only considers the concentrations of the dissolved ions and not solids, such as \( \text{CaSO}_4(s) \). We know \( K_c = 2.4 \times 10^{-5} \).

Step 2: Define the variables for equilibrium concentrations

Let's define the equilibrium concentration of \( \text{Ca}^{2+} \) as \( x \), then \( \text{SO}_4^{2-} \) will also be \( x \) since both ions come from the dissociation of \( \text{CaSO}_4 \). Thus, at equilibrium, we have \( [\text{Ca}^{2+}] = x \) and \( [\text{SO}_4^{2-}] = x \).

Step 3: Substitute into the Kc expression

Substituting the defined variables into the \( K_c \) expression, we have: \[ K_c = x \cdot x = x^2 = 2.4 \times 10^{-5} \].

Step 4: Solve for x

To find \( x \), solve the equation: \[ x^2 = 2.4 \times 10^{-5} \] Taking the square root of both sides, we get: \[ x = \sqrt{2.4 \times 10^{-5}} \approx 4.9 \times 10^{-3} \]. Thus, the equilibrium concentrations are \( [\text{Ca}^{2+}] = 4.9 \times 10^{-3} \) M and \( [\text{SO}_4^{2-}] = 4.9 \times 10^{-3} \) M.

Step 5: Calculate moles needed for equilibrium (Part b)

Using the equilibrium concentration of \( [\text{Ca}^{2+}] = 4.9 \times 10^{-3} \) M in a 1.4 L solution, calculate the moles of \( \text{Ca}^{2+} \): \[ \text{Moles of } \text{Ca}^{2+} = 4.9 \times 10^{-3} \text{ M} \times 1.4 \text{ L} \approx 6.86 \times 10^{-3} \text{ moles} \].

Step 6: Calculate the mass of CaSO4 needed

Calculate the mass of \( \text{CaSO}_4 \) using its molar mass (136.14 g/mol): \[ \text{Mass of } \text{CaSO}_4 = 6.86 \times 10^{-3} \text{ moles} \times 136.14 \text{ g/mol} \approx 0.93 \text{ g} \]. Hence, at least 0.93 g of \( \text{CaSO}_4 \) is required.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), is a pivotal concept in chemical equilibrium. It quantifies the ratio of the concentration of products to reactants for a reversible reaction at equilibrium. In our reaction of calcium sulfate (CaSO\(_4\)) dissociating into calcium ions (Ca\(^{2+}\)) and sulfate ions (SO\(_4^{2-}\)), the equilibrium constant is calculated using the formula:

• \[ K_c = [\text{Ca}^{2+}][\text{SO}_4^{2-}] \]
• For our exercise, \( K_c = 2.4 \times 10^{-5} \).

The value of \( K_c \) tells us how far the reaction proceeds towards the formation of products. A small \( K_c \), such as 2.4 \( \times 10^{-5} \), suggests that at equilibrium, only a small amount of calcium sulfate dissociates into its ions. Hence, the system prefers to remain mostly as solid CaSO\(_4\). Understanding \( K_c \) helps us predict the concentrations of the ions at equilibrium under given conditions.

Saturation

Saturation refers to the point at which a solution can no longer dissolve additional solute at a given temperature and pressure. In the case of our reaction, it describes the state where the dissolved ions of calcium and sulfate in water reach their maximum possible concentrations. At this stage, named saturation, adding more CaSO\(_4\) doesn't increase the concentration of dissolved Ca\(^{2+}\) or SO\(_4^{2-}\), because the solution cannot hold any more at the given conditions.When a saturated solution is reached, it means the rate of the forward reaction (CaSO\(_4\) dissolving) equals the rate of the reverse reaction (Ca\(^{2+}\) and SO\(_4^{2-}\) recombining to form CaSO\(_4\)). This balance represents a dynamic but stable condition, characteristic of chemical equilibrium. The given equilibrium constant \( K_c \) allows us to calculate these maximal concentrations in a saturated solution.

Dissociation Reaction

A dissociation reaction involves the breakdown of a compound into its component ions when dissolved in water. It's a fundamental aspect of certain reactions, especially in ionic compounds. In our example, calcium sulfate (CaSO\(_4\)) undergoes dissociation when mixed with water, forming calcium ions (Ca\(^{2+}\)) and sulfate ions (SO\(_4^{2-}\)).
This reaction is represented as:

• \( \text{CaSO}_4 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + \text{SO}_4^{2-} (aq) \)

As the reaction proceeds, CaSO\(_4\) reaches a point where it can no longer dissolve further, achieving equilibrium. For each unit of CaSO\(_4\) dissolving, the solution gains equal molar amounts of Ca\(^{2+}\) and SO\(_4^{2-}\), a key aspect of dissociation reactions.Understanding how dissociation balances with reformation of the solid is critical, as it defines the concentrations present in the solution at equilibrium, directly linked to \( K_c \). Hence, the study of dissociation provides insight into why the equilibrium concentrations calculated are the values reached.