Problem 54 - Chemistry The Molecular Science

Question

At \(20 .{ }^{\circ} \mathrm{C}, 2.03 \mathrm{~g} \mathrm{CaSO}_{4}\) dissolves per liter of water. From these data calculate the \(K_{\mathrm{sp}}\) of calcium sulfate at 20\. \({ }^{\circ} \mathrm{C}\). Assume that there are no other reactions but the \(K_{\mathrm{sp}}\) reaction.

Step-by-Step Solution

Answer

The \( K_{sp} \) of calcium sulfate at 20\(\,^{\circ}\mathrm{C}\) is \(2.22 \times 10^{-4}\).

Step 1: Identify Dissolution Reaction

Determine the dissolution reaction from the given information: \( \text{CaSO}_4 (s) \leftrightarrow \text{Ca}^{2+} (aq) + \text{SO}_4^{2-} (aq) \). This reaction shows that one mole of \( \text{CaSO}_4 \) produces one mole of \( \text{Ca}^{2+} \) and one mole of \( \text{SO}_4^{2-} \).

Step 2: Determine Molar Mass of Calcium Sulfate

To find the solubility in moles per liter, calculate the molar mass of calcium sulfate (\( \text{CaSO}_4 \)). The molar masses are approximately: \( \text{Ca} = 40.08 \text{ g/mol} \), \( \text{S} = 32.07 \text{ g/mol} \), \( \text{O} = 16.00 \text{ g/mol} \). Therefore, \( \text{CaSO}_4 = 40.08 + 32.07 + (16.00 \times 4) = 136.14 \text{ g/mol} \).

Step 3: Calculate Solubility in Moles/Liter

Using the mass-to-mole conversion: \[ \text{Moles of } \text{CaSO}_4 = \frac{2.03 \text{ g}}{136.14 \text{ g/mol}} = 0.01491 \text{ mol} \]This is the concentration of both \( \text{Ca}^{2+} \) and \( \text{SO}_4^{2-} \), as they dissolve in a 1:1 ratio from \( \text{CaSO}_4 \).

Step 4: Write Expression for \( K_{sp} \)

The solubility product constant expression for the dissolution of \( \text{CaSO}_4 \) is \( K_{sp} = [\text{Ca}^{2+}][\text{SO}_4^{2-}] \). Given the dissolution stoichiometry, \([\text{Ca}^{2+}] = [\text{SO}_4^{2-}] = 0.01491\text{ mol/L}\).

Step 5: Calculate \( K_{sp} \)

Substitute the concentration values into the \( K_{sp} \) expression:\[ K_{sp} = (0.01491)(0.01491) = 2.22 \times 10^{-4} \]

Key Concepts

Calcium Sulfate

Calcium sulfate, denoted as \( \text{CaSO}_4 \), is a common chemical compound that you might encounter in various chemistry problems. It's a white, crystalline solid often found as a mineral in nature. One of its well-known forms is gypsum, which is used in plaster and wallboard.

In aqueous solutions, calcium sulfate demonstrates limited solubility. This means it doesn't dissolve very well in water, making it an interesting subject for studying solubility products. When calcium sulfate does dissolve, it breaks apart into calcium ions \( \text{Ca}^{2+} \) and sulfate ions \( \text{SO}_4^{2-} \). This dissociation behavior is crucial to understanding chemical reactions and equilibria involving calcium sulfate.

Understanding the solubility properties of calcium sulfate helps chemists predict how it behaves in different environments, which is significant in fields like environmental science and industrial chemistry.

Dissolution Reaction

The dissolution process of calcium sulfate can be described with a chemical equation that represents its breakdown into ions. For calcium sulfate, the dissolution reaction is:\[ \text{CaSO}_4 (s) \leftrightarrow \text{Ca}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]. This reaction indicates that solid calcium sulfate separates into calcium ions and sulfate ions when it dissolves in water.

This reaction is reversible, meaning that precipitated calcium sulfate can also form by the combination of \( \text{Ca}^{2+} \) ions with \( \text{SO}_4^{2-} \) ions in a solution. The point at which no more calcium sulfate can dissolve in the water, resulting in a balance between dissolved ions and solid particles, is called equilibrium.

Understanding the dissolution reaction is crucial because it sets the stage for considering the equilibrium condition and the parameters that define the extent of solubility, namely, the solubility product constant \( K_{sp} \).

Molar Mass Calculation

To interpret the solubility of calcium sulfate, one necessary step involves calculating its molar mass. This calculation helps convert from grams to moles, which is essential for chemistry problems.

The molar mass of calcium sulfate is computed by adding the atomic masses of its constituent elements. Calcium (\( \text{Ca} \)) has a mass of approximately 40.08 g/mol, sulfur (\( \text{S} \)) is about 32.07 g/mol, and each oxygen (\( \text{O} \)) atom is 16.00 g/mol. Therefore, the molar mass of \( \text{CaSO}_4 \) is the sum: \( 40.08 + 32.07 + (16.00 \times 4) = 136.14 \text{ g/mol} \).

Knowing the molar mass allows you to convert a known mass of calcium sulfate in grams (such as the 2.03 g specified in our problem) to moles, thus giving the molarity of the dissolution in the solution.

Chemical Equilibrium

Chemical equilibrium in the context of solubility involves the balance between the undissolved solid and its dissolved ions. When calcium sulfate reaches equilibrium in water, the rates of dissolution and precipitation equalize. The concentration of dissolved ions remains constant, which defines the equilibrium state.

At equilibrium, we use the solubility product constant \( K_{sp} \) to quantify the solubility level. The \( K_{sp} \) expression for calcium sulfate is \( [\text{Ca}^{2+}][\text{SO}_4^{2-}] \). Each term in this expression represents the molarity of the ions at equilibrium; since each mole of \( \text{CaSO}_4 \) yields one mole of each ion, their concentrations are equal.

Substituting into the \( K_{sp} \) expression with the known concentrations gives the measure of how much calcium sulfate can dissolve before the solution becomes saturated. Hence, \( K_{sp} \) serves as a critical factor in predicting solubility under specific conditions.