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 \(\mathrm{CaSO}_{4}(s)\) 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 \(3.0 \mathrm{~L}\), what is the minimum mass of \(\mathrm{CaSO}_{4}(s)\) needed to achieve equilibrium?

Short answer

The equilibrium concentrations of Ca²⁺ and SO₄²⁻ ions in a saturated solution of CaSO₄ are both 0.0049 mol/L. The minimum mass of CaSO₄ needed to achieve equilibrium in a 3.0 L solution is approximately 2.00 g.

Step-by-step solution

(Part a) Calculate equilibrium concentrations of Ca²⁺ and SO4²⁻ ions

To find the equilibrium concentrations, first, set up an ICE table for the reaction. Initial concentrations (in mol/L): [Ca²⁺] = 0 [SO4²⁻] = 0 Change in concentrations (in mol/L): - Reaction proceeds, x mol/L of Ca²⁺ ions and x mol/L of SO4²⁻ ions are formed, so [Ca²⁺] = x [SO4²⁻] = x Equilibrium concentrations (in mol/L): [Ca²⁺] = x [SO4²⁻] = x Now, use the equilibrium constant expression for the reaction: \[K_{c}=\frac{[\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}]}{[\mathrm{CaSO}_{4}]}\] Since CaSO4 is a solid, its concentration does not affect the equilibrium, so we can ignore it in the expression. \[K_{c} = [\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}] = x^2\] Now, plug in the value of Kc and solve for x: \(2.4 \times 10^{-5} = x^2\) To find x, take the square root of both sides: \[x = \sqrt{2.4 \times 10^{-5}}\] After calculating x, \[x = 0.0049\] Now, we have the equilibrium concentrations for Ca²⁺ and SO4²⁻ ions: [Ca²⁺] = 0.0049 mol/L [SO4²⁻] = 0.0049 mol/L

(Part b) Calculate the minimum mass of CaSO4 needed to achieve equilibrium

In a 3.0 L solution, with the equilibrium concentration of Ca²⁺ ions calculated earlier, we can find the amount of CaSO4 needed in moles: Amount of Ca²⁺ ions in the solution = [Ca²⁺] × volume of solution Amount of Ca²⁺ ions = (0.0049 mol/L) × (3.0 L) Amount of Ca²⁺ ions = 0.0147 mol Since the balanced reaction involves one mole of CaSO4 producing one mole of Ca²⁺ ions, the amount of CaSO4 required is equal to the amount of Ca²⁺ ions. Amount of CaSO4 = 0.0147 mol Now, use the molar mass of CaSO4 to find the mass of CaSO4 required: Molar mass of CaSO4 = 40.08 (for Ca) + 32.07 (for S) + 4 × 16 (for O) = 136.14 g/mol Mass of CaSO4 = (0.0147 mol) × (136.14 g/mol) Mass of CaSO4 = 2.0006 g The minimum mass of CaSO4 needed to achieve equilibrium in a 3.0 L solution is approximately 2.00 g.

Key concepts

Equilibrium Constant

The equilibrium constant, represented as \( K_c \), is a crucial concept in chemical equilibrium that offers a quantitative measure of the position of equilibrium for a reversible reaction in a closed system. It is defined for a general reaction, such as \( aA + bB \rightleftharpoons cC + dD \), by the expression:

\[ K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]
Here, the square brackets indicate concentrations in mol/L, and the letters a, b, c, and d, represent the stoichiometric coefficients of the reactants and products. It's important to note that in the expression for \( K_c \), only gaseous and aqueous species are included, while pure solids and liquids are excluded since their concentrations are constants.

In our exercise, the solubility of calcium sulfate (\(\mathrm{CaSO}_{4}\)) is being examined. The equilibrium constant expression for its dissolution in water is simplified to \( K_c = [\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}] \) because calcium sulfate is a solid and thus not included in the expression. By knowing \( K_c \), we can predict whether a reaction favours the formation of reactants or products at equilibrium. A small \( K_c \) value, like in our case (\(2.4 \times 10^{-5}\)), suggests the reaction lies far to the left, indicating low solubility of the compound.

Solubility Product

The concept of solubility product, represented as \( K_{sp} \), is closely related to the equilibrium constant. It applies specifically to the dissolution of ionic solids in water. For the dissolution of a sparingly soluble salt, such as \(\mathrm{CaSO}_{4}\), the solubility product is the constant for the equilibrium that exists between the dissolved ions and the undissolved solid.

In mathematical terms, for our exercise, the solubility product is given as: \[ K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{SO}_{4}^{2-}] \]
This expression is identical to that of the equilibrium constant but is often used specifically when discussing solubility equilibria. \( K_{sp} \) helps us understand the maximum concentration of ions that can exist in a solution before the solid precipitates. The lower the value of \( K_{sp} \) is, the less soluble the compound. It's important to differentiate between \( K_c \) and \( K_{sp} \) - while both describe equilibrium positions, \( K_{sp} \) is used when the equilibrium involves a solid dissolving into its constituent ions.

ICE Table Method

The ICE table method stands for Initial, Change, and Equilibrium; it is a systematic approach to handle chemical equilibrium problems, especially when calculating unknown concentrations. It involves setting up a table where you document the initial concentrations, the change that occurs as the system reaches equilibrium, and the final equilibrium concentrations.

To use the ICE table method:

• Start by writing the balanced reaction
• List the initial concentrations of reactants and products under the 'I' column
• Under the 'C' column, indicate the changes in concentrations as the reaction proceeds towards equilibrium
• Under the 'E' column, sum the initial concentration and the change to find the equilibrium concentrations

In our exercise, we started with no ions in solution (zero concentrations). As calcium sulfate dissolves, we denote the increase in ion concentration with \( x \). Upon reaching equilibrium, both \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) have concentrations of \( x \). This method simplifies solving for \( x \) using the \( K_c \) or \( K_{sp} \) expression, eventually helping to determine the equilibrium concentrations and the quantities required for predictive purposes, as done in Step 1 and Step 2 of our solution. Clearly, ICE tables are an indispensable tool in equilibrium calculations.