Problem 110 - General Chemistry

Question

An aqueous solution contains \(3.45 \mathrm{~g}\) of iron(III) sulfate, \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), per liter. What is the molarity of \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) ? When the compound dissolves in water, the \(\mathrm{Fe}^{3+}\) ions and \(\mathrm{SO}_{4}{\underline{\phantom{xx}}}^{2-}\) ions in the crystal go into the solution. What is the molar concentration of each ion in the solution?

Step-by-Step Solution

Answer

The molarity of \(\mathrm{Fe}_{2}(\mathrm{SO}_{4})_{3}\) is 0.00863 M. The molar concentrations are 0.0173 M for \(\mathrm{Fe}^{3+}\) ions and 0.0259 M for \(\mathrm{SO}_{4}^{2-}\) ions.

Step 1: Calculate Molar Mass

First, we need to find the molar mass of the compound \(\mathrm{Fe}_{2}(\mathrm{SO}_{4})_{3}\). The molar masses of the elements are approximately: \(\mathrm{Fe} = 55.85 \, \mathrm{g/mol}\), \(\mathrm{S} = 32.07 \, \mathrm{g/mol}\), and \(\mathrm{O} = 16.00 \, \mathrm{g/mol}\). Calculate the molar mass: \[ 2 \times 55.85 + 3 \times (32.07 + 4 \times 16.00) = 2 \times 55.85 + 3 \times (32.07 + 64) = 399.88 \, \mathrm{g/mol}. \]

Step 2: Determine Moles of Compound

Using the molar mass, calculate the number of moles in \(3.45 \, \mathrm{g}\) of \(\mathrm{Fe}_{2}(\mathrm{SO}_{4})_{3}\): \[\text{moles} = \frac{3.45 \, \mathrm{g}}{399.88 \, \mathrm{g/mol}} \approx 0.00863 \, \mathrm{mol}. \]

Step 3: Calculate Molarity of the Compound

Molarity is defined as the moles of solute per liter of solution. Given that the solution volume is \(1 \, \mathrm{L}\), the molarity is \[0.00863 \, \mathrm{mol/L}.\]

Step 4: Determine Ionization in Solution

When \(\mathrm{Fe}_{2}(\mathrm{SO}_{4})_{3}\) dissolves, one formula unit dissociates into \(2 \mathrm{Fe}^{3+}\) ions and \(3 \mathrm{SO}_{4}^{2-}\) ions. Therefore, for every mole of compound, there are 2 moles of \(\mathrm{Fe}^{3+}\) ions and 3 moles of \(\mathrm{SO}_{4}^{2-}\) ions.

Step 5: Calculate Molar Concentration of Each Ion

Using the molarity of \(\mathrm{Fe}_{2}(\mathrm{SO}_{4})_{3}\) (0.00863 \, \mathrm{mol/L}) and the dissociation stoichiometry, calculate the molarities: \[\text{For } \mathrm{Fe}^{3+}, \: 2 \times 0.00863 = 0.0173 \, \mathrm{mol/L}.\] \[\text{For } \mathrm{SO}_{4}^{2-}, \: 3 \times 0.00863 = 0.0259 \, \mathrm{mol/L}.\]

Key Concepts

Iron(III) sulfate

Iron(III) sulfate, represented by the chemical formula \(\text{Fe}_2(\text{SO}_4)_3\), is an important compound in chemistry, often encountered in basic and applied fields. It is a crystalline solid made up of iron and sulfate ions. This compound is typically used in various applications such as water purification, and as a coagulant. Understanding its chemical composition involves recognizing that it comprises two iron ions \((\text{Fe}^{3+})\) and three sulfate ions \((\text{SO}_4^{2-})\) in each formula unit. Hence, the name Iron(III) comes from the oxidation state of iron, which is +3 in this compound. Chemistry students often need to calculate its molar mass for various applications, as we do here for determining molarity in solution.

Dissociation in solution

When Iron(III) sulfate dissolves in water, it undergoes a process known as dissociation. Dissociation is when an ionic compound separates into its respective ions when it dissolves. In the case of Iron(III) sulfate, the dissolution results in the release of \(2 \text{Fe}^{3+}\) ions and \(3 \text{SO}_4^{2-}\) ions. This means that for each mole of \(\text{Fe}_2(\text{SO}_4)_3\), you end up with two moles of iron ions and three moles of sulfate ions.

Understanding dissociation helps chemists predict how a compound will behave when added to water. It is essential for calculating concentrations of ions in solutions, which can affect everything from chemical reaction rates to conductivity and more.

Molar concentration

Molarity, or molar concentration, is a way to express the concentration of a solution. It is calculated as the number of moles of solute divided by the volume of the solution in liters. For our example, calculating the molarity of \(\text{Fe}_2(\text{SO}_4)_3\) requires us to first determine the number of moles of the compound in the given mass.

We use the molar mass of Iron(III) sulfate, which we calculated as \(399.88 \, \text{g/mol}\). Dividing the given mass \(3.45 \, \text{g}\) by this molar mass yields approximately \(0.00863\, \text{mol}\). Since the solution's volume is \(1\, \text{L}\), this directly translates to a molarity of \(0.00863 \, \text{mol/L}\). This concept is pivotal in chemistry as it helps in conducting reactions with precise stoichiometry.

Ion concentration calculation

Calculating the ion concentration in a solution of \(\text{Fe}_2(\text{SO}_4)_3\) involves understanding its dissociation into ions. Given the molarity of the compound \(0.00863 \, \text{mol/L}\), we use the dissociation stoichiometry to find the concentration of each ion.

For the \(\text{Fe}^{3+}\) ions, with a dissociation factor of 2, the concentration becomes \(2 \times 0.00863 = 0.0173 \, \text{mol/L}\). Similarly, for the \(\text{SO}_4^{2-}\) ions, which have a dissociation factor of 3, the concentration is \(3 \times 0.00863 = 0.0259 \, \text{mol/L}\).

This precise calculation is crucial for accurately analyzing the reactivity and the physical properties of ionic solutions in a laboratory or industrial setting.