Problem 127 - General Chemistry

Question

If \(38.2 \mathrm{~mL}\) of \(0.248 \mathrm{M}\) aluminum sulfate solution is diluted with deionized water to a total volume of \(0.639 \mathrm{~L}\), how many grams of aluminum ion are present in the diluted solution?

Step-by-Step Solution

Answer

0.511 grams of aluminum ions are present in the diluted solution.

Step 1: Calculate Moles of Aluminum Sulfate

First, we need to calculate the number of moles of aluminum sulfate in the original solution. Use the formula for moles: \( \text{moles} = \text{concentration} \times \text{volume} \). Here, concentration \( = 0.248 \) M and volume \( = 0.0382 \) L (since 38.2 mL equals 0.0382 L). Calculate the moles: \[\text{moles of } \text{Al}_2(\text{SO}_4)_3 = 0.248 \times 0.0382 = 0.0094736\] mole.

Step 2: Determine Moles of Aluminum Ions

Each formula unit of aluminum sulfate \((\text{Al}_2(\text{SO}_4)_3)\) produces two aluminum ions \((\text{Al}^{3+})\) in solution due to its dissociation. Hence, the moles of \(\text{Al}^{3+}\) equals twice the moles of aluminum sulfate.Calculate the moles of \(\text{Al}^{3+}\):\[\text{moles of } \text{Al}^{3+} = 2 \times 0.0094736 = 0.0189472\] mole.

Step 3: Calculate Mass of Aluminum Ions

To find the grams of \(\text{Al}^{3+}\), use the molar mass of aluminum \(\approx 26.98 \text{ g/mol}\). Calculate the mass:\[\text{mass of } \text{Al}^{3+} = 0.0189472 \times 26.98 = 0.511\] grams.

Key Concepts

Understanding Aluminum Sulfate

Aluminum sulfate, represented by the chemical formula \(\text{Al}_2(\text{SO}_4)_3\), is a compound commonly used in industries for water purification, paper manufacturing, and as a mordant in dyeing. It is composed of two aluminum ions \((\text{Al}^{3+})\) and three sulfate ions \((\text{SO}_4^{2-})\). When dissolved in water, it dissociates into these ions, making aluminum sulfate solutions conductive. This behavior is important because it determines how the compound interacts in solution, crucial for calculations involving dilution and concentration adjustments.

Mole Calculation Demystified

Moles are a way to quantify the amount of substance using Avogadro's number \((6.022 \times 10^{23})\). For any mole calculation, the relationship between moles, concentration, and volume is key.

• Concentration (Molarity): Given in moles per liter \((\text{mol/L})\), it tells you how much of the substance is present in a solution.
• Volume: Typically converted from milliliters \(\text{(mL)}\) to liters \((\text{L})\) for accuracy in calculations.

Using these, the formula \( \text{moles} = \text{concentration} \times \text{volume} \) allows for the determination of the number of moles of a compound in a given volume.

The Dissociation Process

The term 'dissociation' refers to the process of a compound breaking apart into its constituent ions in solution. For aluminum sulfate in water, this reaction can be represented as follows: \[\text{Al}_2(\text{SO}_4)_3 \rightarrow 2\text{Al}^{3+} + 3\text{SO}_4^{2-}\]This equation means that when aluminum sulfate dissolves, each formula unit produces two aluminum ions and three sulfate ions. Understanding dissociation is essential for mole calculations because it dictates the number of ions produced from a single formula unit, directly affecting the concentration of those ions in solution.

The Role of Molar Mass

Molar mass is the mass of one mole of a given substance, expressed in grams per mole \((\text{g/mol})\). For any element, molar mass corresponds to its relative atomic mass found on the periodic table. For instance, aluminum has a molar mass of approximately 26.98 \(\text{g/mol}\). Calculations involving mass require converting moles to grams using this molar mass:\[\text{mass} = \text{moles} \times \text{molar mass}\]Thus, knowing the molar mass allows you to find out how much a specific number of moles weighs or how much mass corresponds to the moles you are working with in a solution adjustment.