Problem 90 - Chemistry and Chemical Reactivity

Question

You will often work with salts of \(\mathrm{Fe}^{3+}, \mathrm{Pb}^{2+},\) and \(\mathrm{Al}^{3+}\) in the laboratory. (All are found in nature, and all are important economically.) If you have a solution containing these three ions, each at a concentration of \(0.10 \mathrm{M},\) what is the order in which their hydroxides precipitate as aqueous NaOH is slowly added to the solution?

Step-by-Step Solution

Answer

Fe(OH)_3 precipitates first, followed by Al(OH)_3, and then Pb(OH)_2.

Step 1: Write the equilibrium reaction for each ion

For precipitation to occur, hydroxide ions must react with the metal ions. The reactions are as follows: \( \text{Fe}^{3+} + 3 \text{OH}^- \rightarrow \text{Fe(OH)}_3 \), \( \text{Pb}^{2+} + 2 \text{OH}^- \rightarrow \text{Pb(OH)}_2 \), and \( \text{Al}^{3+} + 3 \text{OH}^- \rightarrow \text{Al(OH)}_3 \).

Step 2: Determine the solubility product (Ksp) for each hydroxide

The solubility product constants \( K_{sp} \) for each hydroxide are: \( K_{sp}(\text{Fe(OH)}_3) = 4 \times 10^{-38} \), \( K_{sp}(\text{Pb(OH)}_2) = 1.2 \times 10^{-15} \), \( K_{sp}(\text{Al(OH)}_3) = 3.0 \times 10^{-34} \).

Step 3: Calculate the hydroxide ion concentration needed for initial precipitation

To calculate the concentration of \([\text{OH}^-]\) needed to begin precipitation, use the formula \([\text{OH}^-]^n = \frac{K_{sp}}{[\text{Ion}]}\). For \(\text{Fe}^{3+}\), \([\text{OH}^-]^3 = \frac{4 \times 10^{-38}}{0.10}\). For \(\text{Pb}^{2+}\), \([\text{OH}^-]^2 = \frac{1.2 \times 10^{-15}}{0.10}\). For \(\text{Al}^{3+}\), \([\text{OH}^-]^3 = \frac{3.0 \times 10^{-34}}{0.10}\).

Step 4: Solve for \([\text{OH}^-]\) for each ion

Calculate \([\text{OH}^-]\) required for initial precipitation: - \(\text{Fe(OH)}_3: [\text{OH}^-] = \sqrt[3]{4 \times 10^{-37}} \approx 1.58 \times 10^{-13}\)- \(\text{Pb(OH)}_2: [\text{OH}^-] = \sqrt[]{1.2 \times 10^{-14}} \approx 1.1 \times 10^{-7}\)- \(\text{Al(OH)}_3: [\text{OH}^-] = \sqrt[3]{3.0 \times 10^{-33}} \approx 3.1 \times 10^{-11}\)

Step 5: Determine the order of precipitation

The ion requiring the lowest \([\text{OH}^-]\) will precipitate first. Therefore, the order of precipitation from first to last is \(\text{Fe(OH)}_3, \text{Al(OH)}_3, \text{Pb(OH)}_2\).

Key Concepts

Solubility Product (Ksp)

A solubility product, often abbreviated as Ksp, is a constant that provides insight into the solubility of a compound. It's particularly useful for sparingly soluble salts, where it reveals how much of the solid can dissolve to form a saturated solution. The value of Ksp is unique to each compound and is determined under specific conditions of temperature and pressure.

Ksp helps predict if a precipitate will form when two solutions are mixed. For metal hydroxides in precipitation reactions, like those with \[\mathrm{Fe^{3+}}, \mathrm{Pb^{2+}},\text{ and } \mathrm{Al^{3+}}\], their precipitation is driven by their respective solubility products:

• \( K_{sp}(\text{Fe(OH)}_3) = 4 \times 10^{-38} \)
• \( K_{sp}(\text{Pb(OH)}_2) = 1.2 \times 10^{-15} \)
• \( K_{sp}(\text{Al(OH)}_3) = 3.0 \times 10^{-34} \)

A smaller Ksp implies less solubility, meaning the compound will more readily precipitate out from the solution. This relationship is crucial in determining the sequence in which metals precipitate from a solution. Understanding Ksp is vital to predicting and controlling chemical reactions, especially in laboratory and industrial applications.

Hydroxide Ion Concentration

Hydroxide ion concentration is a key factor in precipitation reactions involving metal ions. The concentration of hydroxide ions ([OH-]) essentially determines when a metal starts and finishes precipitating from a solution.

To calculate the necessary [OH-] for precipitation, we use the equation \[[\text{OH}^-]^n = \frac{K_{sp}}{[\text{Ion}]}\], where \( n \) is the number of hydroxide ions involved in the formation of the precipitate. Here's how it applies to our case:

• For \( \text{Fe}^{3+} \): \([\text{OH}^-]^3 = \frac{4 \times 10^{-38}}{0.10} \) giving \([\text{OH}^-] \approx 1.58 \times 10^{-13}\)
• For \( \text{Pb}^{2+} \): \([\text{OH}^-]^2 = \frac{1.2 \times 10^{-15}}{0.10} \) giving \([\text{OH}^-] \approx 1.1 \times 10^{-7}\)
• For \( \text{Al}^{3+} \): \([\text{OH}^-]^3 = \frac{3.0 \times 10^{-34}}{0.10} \) giving \([\text{OH}^-] \approx 3.1 \times 10^{-11}\)

These calculations help identify the exact moment a complex ion like a metal hydroxide will begin to fall out of solution as a solid. This step is crucial in guiding whether the addition of a reagent will indeed result in the desired reaction occurring under specified conditions.

Metal Ion Precipitation Order

The order in which metal ions precipitate from solution is dictated by their differing requirements for hydroxide ions, determined through calculations considering Ksp. In simpler terms, it depends on which ion reaches its threshold concentration of hydroxide ions first.

In the case of the salts of \( \mathrm{Fe}^{3+} \), \( \mathrm{Pb}^{2+} \), and \( \mathrm{Al}^{3+} \) studied here, the sequence of initial precipitation can be mapped out as follows:

• First, \( \text{Fe}^{3+} \) ion forms \( \text{Fe(OH)}_3 \) because it requires the lowest \([\text{OH}^-]\).
• Second, \( \text{Al}^{3+} \) ion forms \( \text{Al(OH)}_3 \) as it comes next in line with its need for \([\text{OH}^-]\).
• Finally, \( \text{Pb}^{2+} \) ion forms \( \text{Pb(OH)}_2 \) with its requirement for the highest \([\text{OH}^-]\).

This process shows how the solubility principles govern the order of precipitation. By understanding this concept, scientists and chemists can efficiently manipulate procedures to selectively separate or recover specific metals from mixtures, vital for both research and industry practices.