Problem 114 - Chemistry The Molecular Science

Question

The salt calcium sulfate is sparingly soluble in water with a solubility of \(0.209 \mathrm{~g} / 100 \mathrm{~mL}\) water at \(30^{\circ} \mathrm{C}\). If you stirred \(0.550 \mathrm{~g} \mathrm{CaSO}_{4}\) into \(100.0 \mathrm{~mL}\) water at \(30^{\circ} \mathrm{C}\) calculate the molarity of the resulting solution. Calculate the mass of \(\mathrm{CaSO}_{4}\) that would remain undissolved.

Step-by-Step Solution

Answer

Molarity is 0.0154 M; undissolved CaSO_4 is 0.341 g.

Step 1: Calculate Moles of Dissolved Calcium Sulfate

First, determine how many grams of calcium sulfate can dissolve in 100 mL of water at 30°C. Given solubility is 0.209 g/100 mL.The molar mass of calcium sulfate (CaSO_4) is approximately 136.14 g/mol. To find the number of moles that dissolve:\[\text{Moles of } \text{CaSO}_4 = \frac{0.209 \text{ g}}{136.14 \text{ g/mol}} \approx 0.00154 \text{ moles}\]

Step 2: Calculate the Volume in Liters

Convert the volume of water from milliliters to liters since molarity is expressed in terms of moles per liter. \[100 \text{ mL} = 0.100 \text{ L}\]

Step 3: Calculate Molarity of the Solution

Molarity \((M)\) is calculated as the number of moles of solute per liter of solution. Use the moles from Step 1 and the volume from Step 2:\[M = \frac{0.00154 \text{ moles}}{0.100 \text{ L}} = 0.0154 \text{ M}\]

Step 4: Calculate Mass of Undissolved Calcium Sulfate

Determine the amount of calcium sulfate initially added:\[0.550 \text{ g} \text{ total}\]Subtract the amount that can dissolve:\[\text{Undissolved } \text{CaSO}_4 = 0.550 \text{ g} - 0.209 \text{ g} = 0.341 \text{ g}\]

Key Concepts

Solubility

Solubility refers to the maximum amount of a substance (solute) that can dissolve in a specific amount of solvent at a stated temperature. This concept is important when learning about solutions because it affects how much of a substance can be present in the liquid. For calcium sulfate, its solubility is given as 0.209 g/100 mL of water at 30°C. This means at this temperature, only this amount can dissolve completely in water without leaving any undissolved residue.

In real-life scenarios, different factors like temperature and pressure can impact solubility. For instance, increasing temperature generally enhances the solubility of solids in liquids, though this isn't always the case for every substance.

• Generally, solubility decreases as temperature decreases for solids.
• The nature of the solute and solvent also plays a role.

Understanding solubility helps in calculating how much of a substance like calcium sulfate can be dissolved in water, influencing how balanced solutions are prepared.

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate, resulting in no net change in the concentration of reactants and products. In the context of dissolving calcium sulfate in water, equilibrium is reached when the rate of calcium sulfate dissolving equals the rate of it precipitating out of the solution.

For sparingly soluble salts like calcium sulfate, reaching equilibrium means that the solution is saturated, and any additional substance added will not dissolve further.

• This is why 0.341 g of calcium sulfate remains undissolved in the exercise.
• The dissolved calcium sulfate reaches its solubility limit at equilibrium.

The concept of equilibrium is crucial when calculating the molarity of solutions, as it allows us to determine the point at which further dissolution does not occur.

Calcium Sulfate

Calcium sulfate, represented chemically as CaSO₄, is a white, crystalline salt that is slightly soluble in water. Its molar mass is approximately 136.14 g/mol, which is essential for calculating the number of moles present in a solution for determining molarity.

Being sparingly soluble means that calcium sulfate does not dissolve completely in water, leading to the presence of undissolved solid when excess is added. This property is utilized in many practical applications such as in construction (plaster), and in medical settings (plaster casts).

• It forms hemihydrate plaster (gypsum) used in drywall and orthopedic casts.
• Its low solubility makes it suitable for applications involving slow release of calcium and sulfate ions.

Understanding its characteristics is key to effectively working with it in both academic and practical contexts.