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 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 \(\mathrm{SO}_{4}^{2-}\) ? (b) If the resulting solution has a volume of \(1.4 \mathrm{L},\) what is the minimum mass of \(\mathrm{CaSO}_{4}(s)\) needed to achieve equilibrium?

Short answer

(a) The equilibrium concentrations of Ca²⁺ and SO₄²⁻ are \(4.9 \times 10^{-3}\) mol/L. (b) The minimum mass of CaSO₄ needed to achieve equilibrium in 1.4 L of solution is 0.933 g.

Step-by-step solution

Write the expression for the equilibrium constant

For the given reaction: \[\mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\] The equilibrium constant expression is given by: \[K_c = \frac{[\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}]}{[\mathrm{CaSO}_{4}]}\] Since the concentration of CaSO₄(s) is considered constant, we can write it as: \[K_c = [\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}]\] Given, \(K_c = 2.4 \times 10^{-5}\)

Calculate the equilibrium concentrations

Let [Ca²⁺]=[SO₄²⁻]=x mol/L, at equilibrium since in the balanced reaction, for every one mole of CaSO₄ that dissolves, one mole of Ca²⁺ and one mole of SO₄²⁻ ions are produced. So we can write: \[K_c = x^2\] Now, we can solve for x: \[2.4 \times 10^{-5} = x^2\] \[x = \sqrt{2.4 \times 10^{-5}}\] \[x = 4.9 \times 10^{-3}\] Hence, the equilibrium concentrations of Ca²⁺ and SO₄²⁻ are: \[[\mathrm{Ca}^{2+}] = [\mathrm{SO}_{4}^{2-}] = 4.9 \times 10^{-3}\,\text{mol/L}\]

Determine the minimum mass of CaSO₄

Given the volume of the solution is 1.4 L, we can find the moles of Ca²⁺ and SO₄²⁻ ions in the solution: Moles of Ca²⁺ = Moles of SO₄²⁻ = \([\mathrm{Ca}^{2+}] \times \text{Volume}\) \[= (4.9 \times 10^{-3}\,\text{mol/L}) \times 1.4\,\text{L}\] \[= 6.86 \times 10^{-3}\,\text{mol}\] Now, we need to convert the moles of Ca²⁺ into grams of CaSO₄. The molar mass of CaSO₄: Molar mass = 40.08 (Ca) + 32.07 (S) + 4*16 (O) = 136.14 g/mol Minimum mass of CaSO₄ = moles of Ca²⁺ * molar mass of CaSO₄ \[= 6.86 \times 10^{-3}\,\text{mol} \times 136.14\,\text{g/mol}\] \[= 0.933\,\text{g}\] #Answer# (a) The equilibrium concentrations of Ca²⁺ and SO₄²⁻ are 4.9 x 10⁻³ mol/L. (b) The minimum mass of CaSO₄ needed to achieve equilibrium in 1.4 L of solution is 0.933 g.

Key concepts

Equilibrium Constant

Understanding the equilibrium constant is essential when studying chemical equilibria. It is a numerical value characterized by the ratio of the concentrations of the products to the reactants for a reversible reaction at equilibrium. The equation for the equilibrium constant, denoted as \( K_c \) for reactions in solution, is given by following formula:

• \( K_c = \frac{[products]}{[reactants]} \)

The reaction for calcium sulfate dissolution is:\[ \mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq)+\mathrm{SO}_{4}^{2-}(aq) \]Here, the equilibrium constant is \( K_c = 2.4 \times 10^{-5} \), which suggests that at equilibrium, the product concentrations are relatively low, indicating low solubility. This low value indicates that only a small amount of calcium sulfate dissolves in water. The equilibrium constant is not changed by the initial amount of solid or changes in total pressure, but it will vary with temperature. Knowing \( K_c \) allows us to predict the concentrations of ions at equilibrium, which is incredibly useful for determining how soluble a substance is in a solution.

Saturated Solution

A saturated solution is one where the maximum amount of solute has been dissolved in the solvent, beyond which no more solute can dissolve at a given temperature. When mixing calcium sulfate with water until no more dissolves, you obtain a saturated solution. This is a state where the rate at which the solute dissolves equals the rate at which it precipitates from the solution.

In the context of the given reaction, mixing an excess of calcium sulfate with water results in the establishment of an equilibrium between the dissolved ions and the undissolved solid. At 25°C, once equilibrium is reached, any additional calcium sulfate added will remain undissolved because the solution has achieved its maximum solubility for the temperature. The concentrations of \( \text{Ca}^{2+} \) and \( \text{SO}_{4}^{2-} \) are determined by the equilibrium constant \( K_c \) as demonstrated in the solution steps, reflecting the whole saturated solution.

Ionic Concentration

The concept of ionic concentration describes the amount of ions present in a solution. In a saturated solution of calcium sulfate, the dissolved ions \( \mathrm{Ca}^{2+} \) and \( \mathrm{SO}_{4}^{2-} \) are in equilibrium. The concentration of these ions at equilibrium can be found using the equilibrium constant. For calcium sulfate in water:

• The concentration of \( \text{Ca}^{2+} \) is the same as \( \text{SO}_{4}^{2-} \), due to their 1:1 stoichiometry in the dissociation reaction.
• This concentration is calculated as \( \sqrt{K_c} \) since the dissolved ions contribute equally to the product term \( [\text{Ca}^{2+}][\text{SO}_{4}^{2-}] \).

Upon solving \( K_c = x^2 \) with \( x = 4.9 \times 10^{-3} \text{ mol/L} \), we find that both ion concentrations are equal and reflect the saturation point of the solution. Understanding ionic concentration helps predict the strength and properties of the solution, impacting reactions, solubility, and conductivity.

Solubility Product

The solubility product, often denoted as \( K_{sp} \), is an application of the equilibrium constant specifically for dissolving salts. It's useful in quantifying a salt's solubility in solution. For our reaction:\[ \mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(aq)+\mathrm{SO}_{4}^{2-}(aq) \]The solubility product is the product of the molar concentrations of the ions, each raised to the power of their stoichiometric coefficients, and is given by:

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

For calcium sulfate, \( K_{sp} = 2.4 \times 10^{-5} \), effectively quantifying its solubility in terms of products of ion concentrations at equilibrium. The solubility product is crucial in predicting when a salt will precipitate from a solution or dissolve into ions. If the product of the ion concentrations exceeds \( K_{sp} \), the solution is supersaturated, and precipitation occurs. Conversely, if below \( K_{sp} \), the solution is unsaturated, and additional solid can dissolve. By understanding this concept, chemists can manipulate conditions to control salt solubility and precipitation.