Problem 20
Question
If \(100.0 \mathrm{mL}\) of \(0.0025 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is saturated with \(\mathrm{CaSO}_{4},\) how many grams of \(\mathrm{CaSO}_{4}\) would be present in the solution? [Hint: Does the usual simplifying assumption hold?]
Step-by-Step Solution
Verified Answer
The solution would contain 0.034 grams of \(\mathrm{CaSO}_{4}\).
1Step 1: Calculate the molar mass of \(\mathrm{CaSO}_{4}\)
Add the atomic masses of calcium, sulfur, and four oxygen atoms to get the molar mass of \(\mathrm{CaSO}_{4}\): \(\mathrm{Molar \: mass\: of\: CaSO}_{4}=40.08+32.06+4*16.00=136.14 \: \mathrm{g/mol}\)
2Step 2: Determine the moles of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) present in the solution
Use the molarity and volume of the solution to find the moles of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\):\(\mathrm{Moles \: of\: Na}_{2}\mathrm{SO}_{4}=\mathrm{Molarity}\times\mathrm{Volume}=0.0025\: M\times 0.100\: L=0.00025\: \mathrm{moles}\)Given the stoichiometry of the reaction, the moles of \(\mathrm{Na}_{2}\mathrm{SO}_{4}\) equal the moles of \(\mathrm{CaSO}_{4}\). So, you have 0.00025 moles of \(\mathrm{CaSO}_{4}\).
3Step 3: Convert moles of \(\mathrm{CaSO}_{4}\) to grams
Use the molar mass of \(\mathrm{CaSO}_{4}\) calculated in Step 1 to convert the moles of \(\mathrm{CaSO}_{4}\) to grams:\(\mathrm{Grams \: of\: CaSO}_{4}=\mathrm{Moles\: of\: CaSO}_{4}\times \mathrm{molar \: mass\: of\: CaSO}_{4}=0.00025\: \mathrm{moles}\times 136.14\: \mathrm{g/mol}=0.034\: \mathrm{g}\)
Key Concepts
Molar Mass CalculationStoichiometryMolarity and Solutions
Molar Mass Calculation
Understanding how to calculate the molar mass of a compound is crucial in chemistry. The molar mass is the weight of one mole of a substance and usually expressed in g/mol. To calculate the molar mass of calcium sulfate (\(\mathrm{CaSO}_4\)), you need to know the atomic masses of its constituent elements:
- Calcium (Ca): 40.08 g/mol
- Sulfur (S): 32.06 g/mol
- Oxygen (O): 16.00 g/mol (but keep in mind there are four oxygen atoms in the compound)
Stoichiometry
Stoichiometry is the section of chemistry dealing with the quantitative relationships between reactants and products in a chemical reaction. In our exercise, stoichiometry helps us determine how the substances interact in solution.Given \(100.0 \mathrm{mL} \) of \(0.0025 \mathrm{M} \mathrm{Na}_2 \mathrm{SO}_4\) is saturated with \(\mathrm{CaSO}_4\), the stoichiometry of the reaction dictates that one mole of \(\mathrm{Na}_2 \mathrm{SO}_4\) generates one mole of \(\mathrm{CaSO}_4\). To determine the moles of \(\mathrm{Na}_2 \mathrm{SO}_4\) used, apply the formula:\[\mathrm{Moles \: of \: Na}_2\mathrm{SO}_4 = \text{molarity} \times \text{volume in liters}\]Substitute the given values:\[0.0025 \: \mathrm{M} \times 0.100 \: \mathrm{L} = 0.00025 \: \mathrm{moles}\]Since the moles of \(\mathrm{Na}_2 \mathrm{SO}_4\) equal the moles of \(\mathrm{CaSO}_4\), equate them directly. This implies you have 0.00025 moles of \(\mathrm{CaSO}_4\), demonstrating the stoichiometric relationship and enabling conversion to grams. Stoichiometry ensures you use quantities effectively by applying direct mole relationships.
Molarity and Solutions
Molarity is a measure of the concentration of a solute in a solution. It’s defined as moles of solute per liter of solution and indicated as "M" (molar). Getting this concept right is key to solving the given exercise.Consider our problem setup with \(0.0025 \mathrm{M} \mathrm{Na}_2 \mathrm{SO}_4\). Molarity becomes useful as you determine the number of moles present. Molarity can also predict changes in concentrations during chemical reactions, providing a ratio for reaction stoichiometry.For example, in the exercise:
- The volume of solution: 100.0 mL or 0.100 L
- Molarity: 0.0025 M
Other exercises in this chapter
Problem 17
Describe the effects of the salts \(\mathrm{KI}\) and \(\mathrm{AgNO}_{3}\) on the solubility of AgI in water.
View solution Problem 19
A \(0.150 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) solution that is saturated with \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) has \(\left[\mathrm{Ag}^{+}\right]=9.
View solution Problem 23
Can the solubility of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) be lowered to \(5.0 \times 10^{-8} \mathrm{mol} \mathrm{Ag}_{2} \mathrm{CrO}_{4} / \mathrm{L}\) by us
View solution Problem 24
A handbook lists the \(K_{\mathrm{sp}}\) values \(1.1 \times 10^{-10}\) for \(\mathrm{BaSO}_{4}\) and \(5.1 \times 10^{-9}\) for \(\mathrm{BaCO}_{3} .\) When sa
View solution