Problem 128 - General Chemistry

Question

An aluminum nitrate solution is labeled \(0.256 \mathrm{M}\). If \(31.6\) \(\mathrm{mL}\) of this solution is diluted to a total of \(63.7 \mathrm{~mL}\), calculate the molarity of nitrate ion in the resulting solution.

Step-by-Step Solution

Answer

The molarity of nitrate ions in the diluted solution is \(0.3807 \, \mathrm{M}\).

Step 1: Understanding Initial and Final Volumes

We start with a solution of aluminum nitrate with an initial concentration of \(0.256 \, \mathrm{M}\). The initial volume of the solution is \(31.6 \, \mathrm{mL}\). The solution is then diluted to a final volume of \(63.7 \, \mathrm{mL}\).

Step 2: Calculate the Moles of Aluminum Nitrate

First, we need to find the moles of aluminum nitrate present in the initial solution. Use the formula \(\text{moles} = \text{molarity} \times \text{volume in liters}\): \[\text{moles of } \text{Al}(\text{NO}_3)_3 = 0.256 \, \mathrm{M} \times 0.0316 \, \mathrm{L} = 0.0080896 \, \mathrm{moles}\]

Step 3: Consider the Contribution to Nitrate Ion

Aluminum nitrate \(\text{Al}(\text{NO}_3)_3\) dissociates into \(\text{Al}^{3+}\) and \(3\, \text{NO}_3^-\) ions upon dissolving. Therefore, each mole of \(\text{Al}(\text{NO}_3)_3\) yields three moles of \(\text{NO}_3^-\) ions. Thus: \[\text{moles of } \text{NO}_3^- = 3 \times 0.0080896 = 0.0242688 \, \mathrm{moles}\]

Step 4: Calculate the Molarity of Nitrate Ion After Dilution

Now find the molarity of the nitrate ion in the final, diluted solution by dividing the moles of nitrate by the final volume of the solution in liters: \[\text{Molarity of } \text{NO}_3^- = \frac{0.0242688 \, \mathrm{moles}}{0.0637 \, \mathrm{L}} = 0.3807 \, \mathrm{M}\]

Key Concepts

Dilution

In chemistry, dilution is a process of reducing the concentration of a solute in a solution, often by adding more solvent. For this exercise, it is key to understand how dilution affects the concentration of ions in a solution. The initial solution of aluminum nitrate undergoes dilution when the initial volume, 31.6 mL, is expanded to a final volume of 63.7 mL.

When a solution is diluted, the number of solute particles remains constant; only the total volume of the solution changes. This principle allows us to use the formula:

• Initially calculate the number of moles using: \( ext{moles} = ext{initial molarity} \times ext{initial volume in liters} \)
• After dilution, use the same number of moles to find the new molarity: \( rac{ ext{moles}}{ ext{final volume in liters}} \)

This is how we can determine the concentration of nitrate ions after the dilution process has been completed.

Aluminum Nitrate

Aluminum nitrate, represented by the chemical formula \( ext{Al(NO}_3)_3 \), is a type of salt composed of the metal aluminum and nitrate ions. In an aqueous solution, it dissociates completely into aluminum ions \( ext{Al}^{3+} \) and nitrate ions \( ext{NO}_3^- \).

The dissociation can be expressed by the equation:
\[ ext{Al(NO}_3)_3 (s) \rightarrow ext{Al}^{3+} (aq) + 3 ext{NO}_3^- (aq) \]

This dissociation means that each unit of aluminum nitrate contributes one aluminum ion and three nitrate ions when dissolved. The calculation of how much solute is present and how it contributes to the ions in solution is essential to understanding this problem.

• The molar ratio of aluminum nitrate to nitrate ion is 1:3.
• This means each mole of aluminum nitrate will produce three moles of nitrate ions when dissolved in water.

Nitrate Ion

The nitrate ion, \( ext{NO}_3^- \), carries a negative charge and is a significant component of many inorganic compounds and fertilizers. In this problem, understanding the number of nitrate ions produced is crucial to calculating their concentration in the solution.

• In the initial solution, we determined there are \( 0.0080896 \) moles of aluminum nitrate.
• Since each mole of aluminum nitrate yields three moles of nitrate ions, we multiply to find we have \( 0.0242688 \) moles of nitrate ions.

When the solution is diluted, the total volume increases but the amount of nitrate ion remains the same as determined before dilution. The molarity after dilution is given by:
\[ ext{Molarity of } ext{NO}_3^- = \frac{0.0242688 \, ext{moles}}{0.0637 \, ext{L}} = 0.3807 \, ext{M} \]
By understanding how dilution affects molarity, students can compute the change in concentration and predict the behavior of ions in solutions.