Question

What are the individual ion concentrations and the total ion concentration in \(1.04 \mathrm{M}\) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} ?\)

Short answer

[Al^{3+}] = 2.08 \, M, [SO_4^{2-}] = 3.12 \, M; Total concentration = 5.20 \, M.

Step-by-step solution

Identify the Dissociation Reaction

Aluminum sulfate \(\mathrm{Al}_2(\mathrm{SO}_4)_3\) dissociates in water to form aluminum ions \(\mathrm{Al}^{3+}\) and sulfate ions \(\mathrm{SO}_4^{2-}\). The dissociation reaction is: \[\mathrm{Al}_2(\mathrm{SO}_4)_3 \rightarrow 2 \, \mathrm{Al}^{3+} + 3 \, \mathrm{SO}_4^{2-}\]

Calculate the Aluminum Ion Concentration

Since \(1\, \text{mole of} \, \mathrm{Al}_2(\mathrm{SO}_4)_3\) produces \(2\, \text{moles of} \, \mathrm{Al}^{3+}\), the concentration of \(\mathrm{Al}^{3+}\) ions is:\[ [\mathrm{Al}^{3+}] = 2 \times 1.04 \, \mathrm{M} = 2.08 \, \mathrm{M} \]

Calculate the Sulfate Ion Concentration

Since \(1\, \text{mole of} \, \mathrm{Al}_2(\mathrm{SO}_4)_3\) produces \(3\, \text{moles of} \, \mathrm{SO}_4^{2-}\), the concentration of \(\mathrm{SO}_4^{2-}\) ions is:\[ [\mathrm{SO}_4^{2-}] = 3 \times 1.04 \, \mathrm{M} = 3.12 \, \mathrm{M} \]

Determine the Total Ion Concentration

The total ion concentration in the solution is the sum of the concentrations of \(\mathrm{Al}^{3+}\) and \(\mathrm{SO}_4^{2-}\) ions. Since there are \(2 \, \mathrm{Al}^{3+}\) ions and \(3 \, \mathrm{SO}_4^{2-}\) ions per formula unit of \(\mathrm{Al}_2(\mathrm{SO}_4)_3\), the total ion concentration in the solution is:\[\text{Total ion concentration} = 2 \times 1.04 \, \mathrm{M} + 3 \times 1.04 \, \mathrm{M} = 5.20 \, \mathrm{M} \]

Key concepts

Dissociation Reaction

When a chemical compound is placed in water, it can undergo a process known as dissociation. This is when the compound splits into its individual ions. For aluminum sulfate, \(\mathrm{Al}_2(\mathrm{SO}_4)_3\), this involves separating into aluminum ions \(\mathrm{Al}^{3+}\) and sulfate ions \(\mathrm{SO}_4^{2-}\).
Understanding the dissociation reaction is crucial for predicting the types and amounts of ions present in a solution. The equation for aluminum sulfate's dissociation is as follows:

\[\mathrm{Al}_2(\mathrm{SO}_4)_3 \rightarrow 2 \, \mathrm{Al}^{3+} + 3 \, \mathrm{SO}_4^{2-}\]

This means each formula unit of \(\mathrm{Al}_2(\mathrm{SO}_4)_3\) produces two aluminum ions and three sulfate ions.

• Recognizing ionic compounds' dissociation helps in calculating ion concentrations.
• The coefficients in the dissociation equation directly affect the resulting ion concentrations.

Understanding this step is foundational to solving problems involving ions in solution.

Molarity

Molarity is a measure of concentration that tells us how many moles of a solute are present in one liter of solution. It's often expressed in moles per liter (M).
When we say a solution is \(1.04\, \mathrm{M}\) in \(\mathrm{Al}_2(\mathrm{SO}_4)_3\), it means there is \(1.04\) moles of aluminum sulfate in every liter of the solution.

• Molarity provides a way to quantify the concentration of ions after dissociation.
• This measurement is critical when calculating how much reactant is needed or produced in a reaction.

For solutions containing dissolved ionic compounds, molarity can help in determining the concentration of the individual ions that result from dissociation. It is an essential tool in many areas of chemistry.

Aluminum Sulfate

Aluminum sulfate is an ionic compound commonly used in water purification and the paper industry. Its chemical formula is \(\mathrm{Al}_2(\mathrm{SO}_4)_3\). When dissolved in water, aluminum sulfate dissociates into its constituent aluminum and sulfate ions.
This compound is known for the following properties:

• Dissociation: As illustrated by its dissociation reaction, it releases multiple ions.
• Applications: It's used in processes requiring coagulation and precipitation of particles, like in water treatment.

The understanding of aluminum sulfate’s properties and reactions is important not only in chemical calculations but also for its practical applications.

Ions in Solution

In a solution of an ionic compound like aluminum sulfate, the compound itself no longer exists as a whole. Instead, its ions, in this case, \(\mathrm{Al}^{3+}\) and \(\mathrm{SO}_4^{2-}\), are free to move within the liquid.
The concentration of each ion is calculated from the molarity of the original compound multiplied by the ratio of ions it produces, derived from its dissociation equation.
For \(\mathrm{Al}_2(\mathrm{SO}_4)_3\):

• The concentration of \(\mathrm{Al}^{3+}\): \([\mathrm{Al}^{3+}] = 2 \times 1.04 \, \mathrm{M} = 2.08 \, \mathrm{M}\)
• The concentration of \(\mathrm{SO}_4^{2-}\): \([\mathrm{SO}_4^{2-}] = 3 \times 1.04 \, \mathrm{M} = 3.12 \, \mathrm{M}\)
• Total ion concentration is the sum of individual ion concentrations: \(5.20 \, \mathrm{M}\)

Understanding ion concentrations in a solution aids in predicting the solution's behavior in chemical reactions and processes.