What is the concentration of \(\mathrm{Al}^{3+}\) ions in a solution of \(\mathrm{AlBr}_{3}\) having a \(\mathrm{Br}^{-}\)ion concentration of \(0.142 \mathrm{M}\) ?

In this exercise, we need to determine the concentration of aluminum ions \( \text{Al}^{3+} \). Aluminum ions are the positively charged particles that form when aluminum bromide ( \(\text{AlBr}_3 \) ) dissociates in water.
When \(\text{AlBr}_3 \) dissolves in water, it separates (or dissociates) into ions. Specifically:
\[ \text{AlBr}_3 \rightarrow \text{Al}^{3+} + 3 \text{Br}^{-} \]
This equation means each molecule of \(\text{AlBr}_3 \) produces one aluminum ion ( \(\text{Al}^{3+} \) ) and three bromide ions ( \(\text{Br}^{-} \) ).
Thus, understanding the ratio at which the ions are produced (1:3) is fundamental to solving the problem.

Chemical dissociation is a process where a compound breaks apart into smaller chemical species. In our case, \(\text{AlBr}_3 \) dissociates in an aqueous solution, splitting into one \(\text{Al}^{3+} \) ion and three \(\text{Br}^{-} \) ions.
 This dissociation can be summarized by the chemical equation:
\[ \text{AlBr}_3 \rightarrow \text{Al}^{3+} + 3 \text{Br}^{-} \]
To solve the problem, recognize the stoichiometric relationship (the ratio) between the resulting ions. For every mole of \(\text{AlBr}_3 \) that dissolves:

• 1 mole of \(\text{Al}^{3+} \)
• 3 moles of \(\text{Br}^{-} \)

are produced. This key relationship will help you determine the concentration of aluminum ions based on the given bromide ion concentration.

Molar concentration (also known as molarity) is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute dissolved in one liter of solution, with the unit M (moles per liter).
Given the concentration of bromide ions ( \(\text{Br}^{-} \) ) is 0.142 M, we can use the dissociation ratio to find the concentration of aluminum ions ( \(\text{Al}^{3+} \) ). Since each \(\text{AlBr}_3 \) produces one \(\text{Al}^{3+} \) for every three \(\text{Br}^{-} \), the concentration of \(\text{Al}^{3+} \) ions is calculated by:
\[ \text{[Al}^{3+}\text{]} = \frac{\text{[Br}^{-}\text{]}}{3} = \frac{0.142 \text{ M}}{3} = 0.0473 \text{ M} \]
So, the molar concentration of aluminum ions in the solution is 0.0473 M.