Problem 33 - Chemistry: Principles and Reactions

Question

A 65-mL solution of \(0.40 \mathrm{M} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) is mixed with \(125 \mathrm{~mL}\) of \(0.17 \mathrm{M}\) iron(II) nitrate. Solid sodium hydroxide is then added without a change in volume. (a) Which will precipitate first, \(\mathrm{Al}(\mathrm{OH})_{3}\) or \(\mathrm{Fe}(\mathrm{OH})_{2} ?\) (b) What is \(\left[\mathrm{OH}^{-}\right]\) when the first compound begins to precipitate?

Step-by-Step Solution

Answer

Answer: (a) Aluminum hydroxide (\(\mathrm{Al}(\mathrm{OH})_{3}\)) will precipitate first. (b) The concentration of hydroxide ions when the first compound begins to precipitate is \(6.07 \times 10^{-12} \ \mathrm{M}\).

Step 1: Calculate initial ion concentrations

First, we need to calculate the initial concentrations of Al³⁺ and Fe²⁺ ions in the mixture. The initial concentration of aluminum ions can be found by: \([\mathrm{Al^{3+}}]_{initial}=\frac{(\mathrm{Molarity \ of \ Al(NO_3)_3})(\mathrm{Volume \ of \ Al(NO_3)_3})}{\mathrm{Total \ volume}}\) The initial concentration of iron(II) ions can be found by: \([\mathrm{Fe^{2+}}]_{initial}=\frac{(\mathrm{Molarity \ of \ Fe(NO_3)_2})(\mathrm{Volume \ of \ Fe(NO_3)_2})}{\mathrm{Total \ volume}}\) Plug in the given values to find the concentrations: \([\mathrm{Al^{3+}}]_{initial}=\frac{(0.40 \ \mathrm{M})(65 \ \mathrm{mL})}{65 \ \mathrm{mL} + 125 \ \mathrm{mL}} = 0.146 \ \mathrm{M}\) \([\mathrm{Fe^{2+}}]_{initial}=\frac{(0.17 \ \mathrm{M})(125 \ \mathrm{mL})}{65 \ \mathrm{mL} + 125 \ \mathrm{mL}} = 0.0925 \ \mathrm{M}\)

Step 2: Calculate solubility product constants

In this step, we need to find the solubility product constants (\(K_{sp}\)) for \(\mathrm{Al}(\mathrm{OH})_{3}\) and \(\mathrm{Fe}(\mathrm{OH})_{2}\). These constants can be found in a reference source or given by the problem. For this example, we will use the following values: \(K_{sp}(\mathrm{Al}(\mathrm{OH})_{3}) = 3.0 \times 10^{-34}\) \(K_{sp}(\mathrm{Fe}(\mathrm{OH})_{2}) = 8.0 \times 10^{-16}\)

Step 3: Compare concentrations of hydroxide ions needed to precipitate Al³⁺ and Fe²⁺

In this step, we will calculate the concentrations of hydroxide ions needed for respective metal ions to precipitate as their hydroxides. For \(\mathrm{Al}(\mathrm{OH})_{3}\) and \(\mathrm{Fe}(\mathrm{OH})_{2}\), we have: \(K_{sp} = [\mathrm{Al^{3+}}][\mathrm{OH^-}]^3\) \(3.0 \times 10^{-34} = (0.146)[\mathrm{OH^-}]^3\) \(K_{sp} = [\mathrm{Fe^{2+}}][\mathrm{OH^-}]^2\) \(8.0 \times 10^{-16} = (0.0925)[\mathrm{OH^-}]^2\) Now, we need to solve for the values of \([\mathrm{OH^-}]\) in each case: \([\mathrm{OH^-}]_{Al} = \sqrt[3]{\frac{3.0 \times 10^{-34}}{0.146}} = 6.07 \times 10^{-12}\) \([\mathrm{OH^-}]_{Fe} = \sqrt{\frac{8.0 \times 10^{-16}}{0.0925}} = 9.17 \times 10^{-8}\) Comparing these two values, we see that \([\mathrm{OH^-}]_{Al} < [\mathrm{OH^-}]_{Fe}\).

Step 4: Determine the concentration of hydroxide ions when the first compound begins to precipitate

Since \([\mathrm{OH^-}]_{Al} < [\mathrm{OH^-}]_{Fe}\), aluminum hydroxide will precipitate first. Therefore, the concentration of hydroxide ions needed for the first compound to precipitate is: \([\mathrm{OH^-}] = 6.07 \times 10^{-12} \ \mathrm{M}\) (a) \(\mathrm{Al}(\mathrm{OH})_{3}\) will precipitate first (b) The concentration of hydroxide ions when the first compound begins to precipitate is \(6.07 \times 10^{-12} \ \mathrm{M}\).

Key Concepts

Solute Concentration

In any chemical solution, solute concentration describes the amount of a solute dissolved in a given quantity of solvent. It's often expressed in molarity (M), which is moles of solute per liter of solution.
Understanding solute concentration is vital in predicting how substances interact in a solution. For example, when mixing solutions like aluminum nitrate and iron(II) nitrate, calculating the initial concentrations of ions is essential. It helps us predict which ions will precipitate first when a reagent like sodium hydroxide is added.

• To find the concentration of Al³⁺ ions: Use the formula \( [\text{Al}^{3+}] = \frac{(0.40 \, \text{M})(65 \, \text{mL})}{65\, \text{mL} + 125 \, \text{mL}} \)
• Similarly, for Fe²⁺ ions: Use \( [\text{Fe}^{2+}] = \frac{(0.17 \, \text{M})(125 \, \text{mL})}{65\, \text{mL} + 125 \, \text{mL}} \)

This knowledge allows us to assess how these ions will behave in reactions.

Solubility Product Constant

The solubility product constant (K_{sp}) is a measure of a compound's solubility in solution. It tells us how much of a substance can dissolve before it starts forming a solid, or precipitating, in the solution.
For compounds like \(Al(OH)_3\) and \(Fe(OH)_2\), knowing their respective \(K_{sp}\) values enables us to predict when a precipitate will form when the hydroxide ions are added. These constants are often provided in tables or reference materials:

• \(K_{sp}(\text{Al(OH)}_3) = 3.0 \times 10^{-34}\)
• \(K_{sp}(\text{Fe(OH)}_2) = 8.0 \times 10^{-16}\)

The smaller the \(K_{sp}\), the less soluble the compound is, hence it precipitates more readily at lower concentrations of the ion product.

Hydroxide Ion Concentration

Hydroxide ion concentration \([\text{OH}^-]\) is crucial in determining when a compound will begin to precipitate from a solution.
When calculating for compounds like \(Al(OH)_3\) and \(Fe(OH)_2\), we use the solubility product expression:

• For \(\text{Al(OH)}_3\): \(K_{sp} = [\text{Al}^{3+}][\text{OH}^-]^3\)
• For \(\text{Fe(OH)}_2\): \(K_{sp} = [\text{Fe}^{2+}][\text{OH}^-]^2\)

Solving the equations gives us the required \([\text{OH}^-]\) for each compound to precipitate:

• \([\text{OH}^-]_{\text{Al}} = \sqrt[3]{ \frac{3.0 \times 10^{-34}}{0.146} } = 6.07 \times 10^{-12}\,\text{M}\)
• \([\text{OH}^-]_{\text{Fe}} = \sqrt{ \frac{8.0 \times 10^{-16}}{0.0925} } = 9.17 \times 10^{-8}\,\text{M}\)

This shows that a lower \([\text{OH}^-]\) concentration is needed for \(\text{Al(OH)}_3\) to precipitate first.

Precipitation Reaction

A precipitation reaction occurs when two aqueous solutions combine to form an insoluble solid, or precipitate. This happens often because the ion product exceeds the solubility product constant of the compound.
In our scenario, adding sodium hydroxide to solutions of aluminum and iron nitrates triggers precipitation reactions as hydroxide ions interact with metal ions:

• \(\text{Al}(\text{NO}_3)_3\) forms \(\text{Al(OH)}_3\) due to lower \([\text{OH}^-]\) required
• \(\text{Fe}(\text{NO}_3)_2\) forms \(\text{Fe(OH)}_2\) only at a higher \([\text{OH}^-]\) concentration

The critical factor in determining which precipitates first is the hydroxide ion concentration needed to surpass each compound's \(K_{sp}\). Hence, in this case, \(\text{Al(OH)}_3\) forms sooner as its \(K_{sp}\) is extremely low, indicating it precipitates at a lower \([\text{OH}^-]\).