Question

A student dilutes \(75.0 \mathrm{~mL}\) of a \(2.00 \mathrm{M}\) solution of iron(III) nitrate with sufficient water to prepare \(2.00 \mathrm{~L}\) of solution. (a) What is the molar concentration of iron(III) nitrate in the diluted solution? Once in solution, the iron(III) nitrate exists not intact but rather as dissociated ions. What are the molar concentrations (b) of \(\mathrm{Fe}^{3+}(a q)\) in the diluted solution and (c) of \(\mathrm{NO}^{3-}(a q)\) in the diluted solution?

Short answer

The molar concentrations in the diluted solution are: (a) Iron(III) nitrate: 0.075 M, (b) Fe³⁺: 0.075 M, and (c) NO₃⁻: 0.225 M.

Step-by-step solution

Use the dilution formula to find the molar concentration of the diluted solution.

The dilution formula is given by: \[ M_1V_1 = M_2V_2 \] Where: - \(M_1\) is the initial concentration of the solution. - \(V_1\) is the initial volume of the solution. - \(M_2\) is the final concentration of the solution after dilution. - \(V_2\) is the final volume of the solution after dilution. Given that the initial concentration is 2.00 M and the initial volume is 75.0 mL, we can plug these values into the dilution formula, keeping in mind that the final volume is 2.00 L. We want to find \(M_2\), the molar concentration of iron(III) nitrate after dilution, so let's rearrange the equation: \[ M_2 = \frac{M_1V_1}{V_2} \] Then plug in the known values and solve for \(M_2\).

Calculate the molar concentration of iron(III) nitrate after dilution.

Substituting the values: \[ M_2 = \frac{(2.00 \mathrm{~M})(75.0 \mathrm{~mL})}{2.00 \mathrm{~L}} \] Convert 75.0 mL to L: \[ 75.0 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}} = 0.075 \mathrm{~L} \] Now, calculate \(M_2\): \[ M_2 = \frac{(2.00 \mathrm{~M})(0.075 \mathrm{~L})}{2.00 \mathrm{~L}} = 0.075 \mathrm{~M} \] The molar concentration of iron(III) nitrate in the diluted solution is 0.075 M.

Write the dissociation equation for iron(III) nitrate in water.

To find the molar concentrations of Fe³⁺ and NO₃⁻, we must first understand how iron(III) nitrate dissociates in water: \[ \mathrm{Fe(NO_3)_3} (aq) \rightarrow \mathrm{Fe^{3+}}(aq) + 3\mathrm{NO_3^-}(aq) \] From the stoichiometry of the equation, for every 1 mole of iron(III) nitrate that dissociates, 1 mole of Fe³⁺ and 3 moles of NO₃⁻ are produced.

Calculate the molar concentration of Fe³⁺ in the diluted solution.

Since the dissociation produces 1 mole of Fe³⁺ for every 1 mole of iron(III) nitrate, the molar concentration of Fe³⁺ in the diluted solution is equal to the molar concentration of iron(III) nitrate: \[ [\mathrm{Fe^{3+}}] = 0.075 \mathrm{~M} \]

Calculate the molar concentration of NO₃⁻ in the diluted solution.

Since the dissociation produces 3 moles of NO₃⁻ for every 1 mole of iron(III) nitrate, the molar concentration of NO₃⁻ in the diluted solution is three times the molar concentration of iron(III) nitrate: \[ [\mathrm{NO_3^-}] = 3 \times 0.075 \mathrm{~M} = 0.225 \mathrm{~M} \] So, the molar concentrations in the diluted solution are: a) Iron(III) nitrate: 0.075 M b) Fe³⁺: 0.075 M c) NO₃⁻: 0.225 M

Key concepts

Molar Concentration

Molar concentration, often referred to as molarity, measures the concentration of a substance in a solution. It is defined as the number of moles of a solute per liter of solution. The formula is given by:

• \( M = \frac{n}{V} \)

Where:

• \( M \) is the molarity in moles per liter (M).
• \( n \) is the number of moles of solute.
• \( V \) is the volume of the solution in liters.

To determine the molarity of a diluted solution, we use the dilution formula:

• \( M_1V_1 = M_2V_2 \)

Here, \( M_1 \) and \( V_1 \) are the initial concentration and volume, and \( M_2 \) and \( V_2 \) are the final concentration and volume, respectively.
In our exercise, iron(III) nitrate's initial molarity is 2.00 M, and its initial volume is converted to 0.075 L. After dilution to 2.00 L, we calculate the final molarity as 0.075 M.

Dissociation Equation

A dissociation equation represents how a compound separates into its ions in an aqueous solution. Understanding this process is key to calculating the concentrations of individual ions.
For iron(III) nitrate, the dissociation equation in water is:

• \( \mathrm{Fe(NO_3)_3} (aq) \rightarrow \mathrm{Fe^{3+}}(aq) + 3\mathrm{NO_3^-}(aq) \)

This equation shows that one mole of iron(III) nitrate produces one mole of \( \mathrm{Fe^{3+}} \) and three moles of \( \mathrm{NO_3^-} \).
By interpreting this reaction, we can deduce that the molarity of \( \mathrm{Fe^{3+}} \) ions will be equal to the molarity of the original iron(III) nitrate solution, while the \( \mathrm{NO_3^-} \) ions will be three times more concentrated.

Ions in Solution

In a solution, compounds like salts disassociate into their respective ions. Knowing how ions behave and interact in a solution is fundamental to chemistry.
For iron(III) nitrate, once it is dissolved, it separates into individual \( \mathrm{Fe^{3+}} \) and \( \mathrm{NO_3^-} \) ions.
After calculating the molar concentration of the diluted solution:

• The concentration of \( \mathrm{Fe^{3+}} \) ions is the same as the dissolved iron(III) nitrate, which is 0.075 M.
• The concentration of \( \mathrm{NO_3^-} \) ions is three times that of \( \mathrm{Fe^{3+}} \), resulting in 0.225 M.

Understanding these ion concentrations helps predict how solutions will react chemically and physically, which is crucial for experiments and industrial applications.