Problem 197 - Introductory Chemistry Atoms First

Question

A student dilutes \(45.0 \mathrm{~mL}\) of a \(0.500 \mathrm{M}\) solution of aluminum sulfate with sufficient water to prepare \(1.50 \mathrm{~L}\) of solution. (a) What is the molar concentration of aluminum sulfate in the diluted solution? Once in solution, the aluminum sulfate exists not intact but rather as dissociated ions. What are the molar concentrations (b) of \(\mathrm{Al}^{3+}(a q)\) in the diluted solution and (c) of \(\mathrm{SO}_{4}^{2-}(a q)\) in the diluted solution?

Step-by-Step Solution

Answer

(a) The molar concentration of aluminum sulfate in the diluted solution is \(0.0150 \mathrm{~M}\). (b) The molar concentration of \(\mathrm{Al}^{3+}(a q)\) in the diluted solution is \(0.0150 \mathrm{~M}\). (c) The molar concentration of \(\mathrm{SO}_{4}^{2-}(a q)\) in the diluted solution is \(0.0450 \mathrm{~M}\).

Step 1: Calculate moles of aluminum sulfate in the initial solution

Since we have the volume and molarity of the initial solution, we can calculate the moles of aluminum sulfate using the formula: Moles of solute = Molarity × Volume Moles of aluminum sulfate = 0.500 M × \(45.0 \mathrm{~mL}\) × \(\frac{1 \mathrm{~L}}{1000 \mathrm{~mL}}\) = 0.0225 mol

Step 2: Calculate molar concentration of aluminum sulfate in the diluted solution

Since the moles of aluminum sulfate don't change after dilution, we can find the new molarity using the formula: Molarity = Moles of solute / Volume Molar concentration of aluminum sulfate in the diluted solution = 0.0225 mol / 1.50 L = 0.0150 M #answer#(a) The molar concentration of aluminum sulfate in the diluted solution is 0.0150 M.

Step 3: Determine molar concentrations of Al^3+ and SO4^2- ions in the diluted solution

In the dissociation process, aluminum sulfate yields one Al^3+ ion and three SO4^2- ions per molecule. So when we find out the molar concentration of aluminum sulfate, we can get the molar concentrations of the dissociated ions by multiplying by the respective stoichiometry factors. Molar concentration of \(\mathrm{Al}^{3+}(a q)\) = 1 × 0.0150 M = 0.0150 M Molar concentration of \(\mathrm{SO}_{4}^{2-}(a q)\) = 3 × 0.0150 M = 0.0450 M #answer# (b) The molar concentration of \(\mathrm{Al}^{3+}(a q)\) in the diluted solution is 0.0150 M. #answer# (c) The molar concentration of \(\mathrm{SO}_{4}^{2-}(a q)\) in the diluted solution is 0.0450 M.

Key Concepts

Dilution of Solutions

Understanding the concept of dilution is critical for students tackling exercises involving changes in concentration. Dilution occurs when a solvent, usually water, is added to a solution, increasing its total volume, and consequently, lowering the concentration of the solute. This change can be summarized by the formula:

\( C_1V_1 = C_2V_2 \)

where \( C_1 \) and \( C_2 \) are the concentrations before and after dilution, respectively, and \( V_1 \) and \( V_2 \) are the corresponding volumes. Remember, when you add water to dilute a solution, the number of moles of the solute remains constant; only the volume changes, leading to a change in concentration. Dilution is a key process in laboratory work and has practical applications, such as in pharmacology and environmental science.

Molarity

Molarity is one of the most common ways to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. The formula for molarity (\( M \) is:

\( Molarity (M) = \frac{Moles~of~solute}{Volume~of~solution~in~liters} \)

Understanding molarity is crucial when preparing solutions of a specific concentration and when mixing solutions to achieve a desired molar concentration in chemistry labs. It is an essential part of the stoichiometry calculation in reactions and allows scientists to calculate reacting volumes accurately.

Stoichiometry

Stoichiometry is a section of chemistry that involves the calculation of reactants and products in chemical reactions. In stoichiometry, the law of conservation of mass plays a key role – matter is neither created nor destroyed. Stoichiometry allows chemists to predict how much reactant is needed to produce a desired amount of product and to determine the amount of products formed from a given amount of reactants.

For the student's exercise, the stoichiometry of the dissociation of aluminum sulfate is important. Each formula unit of \( \text{Al}_2(\text{SO}_4)_3 \) dissociates into two \( \text{Al}^{3+} \) ions and three \( \text{SO}_4^{2-} \) ions, which then affects the calculation of the individual ionic concentrations in the solution.

Ionic Dissociation in Solutions

Ionic dissociation is the process by which solid ionic compounds, once dissolved in a solvent, separate into their constituent ions. In aqueous solutions, this is often visible when compounds like salts dissolve in water. Understanding ionic dissociation is important because it affects various chemical properties of the solution such as conductivity, boiling point elevation, and freezing point depression.

For the exercise with aluminum sulfate, knowing that it dissociates into \( \text{Al}^{3+} \) and \( \text{SO}_4^{2-} \) ions in solution informs us that for every mole of aluminum sulfate that dissolves, two moles of \( \text{Al}^{3+} \) ions and three moles of \( \text{SO}_4^{2-} \) ions are released into the solution – essential information for correctly calculating their molar concentrations.