Question

\(K_{\mathrm{sp}}\) for \(\mathrm{CaSO}_{4}\) at \(100^{\circ} \mathrm{C}\) is estimated to be \(1.6 \times 10^{-5}\). At \(25^{\circ} \mathrm{C}, 0.915 \mathrm{~g}\) of \(\mathrm{CaSO}_{4}\) is added to one liter of water. (a) Will all the \(\mathrm{CaSO}_{4}\) dissolve? (b) If the solution is heated to \(100^{\circ} \mathrm{C}\), will all the \(\mathrm{CaSO}_{4}\) dissolve?

Short answer

Answer: (a) We cannot determine if all the CaSO4 will dissolve at 25°C as no Ksp value is given for that temperature. (b) At 100°C, not all CaSO4 will dissolve as the ion product (Q) is greater than the Ksp value given.

Step-by-step solution

Calculate the molar mass of CaSO4

First, we need to find the molar mass of CaSO4. The molar mass of CaSO4 can be found using the molar masses of each element in the compound: Ca: 40.08 g/mol, S: 32.07 g/mol, and O: 16.00 g/mol. So, the molar mass of CaSO4 = 40.08 + 32.07 + (4 × 16.00) = 136.15 g/mol.

Calculate moles of CaSO4 at 25°C

Now, we will calculate the number of moles of CaSO4 in 0.915g by dividing the mass by its molar mass: moles of CaSO4 = 0.915g / 136.15 g/mol = 0.00672 mol.

Calculate the molarity of CaSO4 at 25°C

Next, we need to find the molarity of CaSO4 in a 1L solution. The molarity is calculated by dividing the number of moles by the volume of the solution in liters: Molarity of CaSO4 = 0.00672 mol / 1L = 0.00672 M. Since the stoichiometry of the dissolution of CaSO4 is 1:1 for Ca²⁺ and SO₄²⁻, the molarities of both ions will be equal to that of CaSO4. The ion product (Q) is the product of the concentrations of the dissolved ions: Q = [Ca²⁺][SO₄²⁻] = (0.00672 M)(0.00672 M) = 4.52 × 10⁻².

Compare Q to Ksp at 25°C

No information about the Ksp of CaSO4 at 25°C is given, so we can't determine if all CaSO4 will dissolve completely or not. However, we can determine if it will dissolve or precipitate if the solution is heated by comparing the ion product Q (which we already calculated) with the given Ksp at 100°C.

Compare Q to Ksp at 100°C

At 100°C, the Ksp of CaSO4 is given as 1.6 × 10⁻⁵. Comparing Q and Ksp: Q = 4.52 × 10⁻² > 1.6 × 10⁻⁵ = Ksp Since Q is greater than the Ksp, the solution is supersaturated, and the CaSO4 will not dissolve completely even at 100°C. In conclusion, (a) We cannot determine if all the CaSO4 will dissolve at 25°C as no Ksp value is given for that temperature. (b) At 100°C, not all CaSO4 will dissolve as the ion product (Q) is greater than the Ksp value given.

Key concepts

Ksp Calculation

Understanding the solubility product constant, commonly known as Ksp, is integral to predicting the solubility of ionic compounds in water. The Ksp value is specific to each compound and varies with temperature, representing the extent to which a compound will dissociate to form its constituent ions in solution. For example, the Ksp for calcium sulfate (CaSO4) at a high temperature of 100 degrees Celsius is given as 1.6 x 10^-5.

When tackling exercises involving Ksp, you first calculate the ion product (Q), which is similar to Ksp but for a solution not at equilibrium. You compare Q to Ksp to determine the level of saturation of the solution; saturated when Q equals Ksp, unsaturated when Q is less than Ksp, and supersaturated when Q is greater than Ksp. In the given exercise, Q calculated for CaSO4 at room temperature exceeds the Ksp at 100°C, telling us that even at higher temperatures, CaSO4 will not fully dissolve.

Molarity and Dilution

Molarity, a measure of concentration, represents moles of solute per liter of solution and plays a pivotal role in solutions chemistry. Calculating molarity is a straightforward process: divide the number of moles of the solute by the volume of the solution in liters. In our example involving calcium sulfate, the number of moles of CaSO4 was derived by dividing its mass by its molar mass. From there, determining its molarity in a 1-liter solution was simple.

Dilution involves adding more solvent to a solution in order to decrease its molarity. The core principle that underlies dilutions is that the number of moles of solute remains constant. So if you know the initial and final volumes of the solution, you can determine the molarity after dilution using the formula: \(M_1 \times V_1 = M_2 \times V_2\) where M is molarity and V is volume. This exercise, however, did not require a dilution step as the volume of the solution remained unchanged.

Thermochemistry

Thermochemistry is the study of heat changes accompanying chemical reactions, including dissolution. Temperature plays a vital role in solubility; an increase in temperature often increases the solubility of solids in liquids, but there are exceptions.

In the context of our exercise, understanding thermochemistry can help explain why the Ksp of CaSO4 changes with temperature. Generally, for endothermic dissolving processes, solubility increases as the temperature rises. However, increasing temperature will also shift the equilibrium, illustrated by the difference in solubility product constants at room temperature and at 100 degrees Celsius.

Thermochemical considerations add complexity to solubility predictions because temperature changes alter the equilibrium consistently defined by Le Chatelier’s principle. Thus, a compound may become more soluble at higher temperatures, but it is always crucial to refer to empirical data, such as the provided Ksp values, to make accurate and precise predictions.