Problem 104 - 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 M, what is the order in which their hydroxides precipitate as aqueous NaOH is slowly added to the solution?

Step-by-Step Solution

Answer

Order of precipitation: \(\mathrm{Fe(OH)_3}\), \(\mathrm{Al(OH)_3}\), \(\mathrm{Pb(OH)_2}\).

Step 1: Identify the relevant solubility concept

To determine the order of precipitation, we need to consider the solubility products (Ksp) of the metal hydroxides formed by each ion. The ion with the lowest solubility product will precipitate first as the concentration of hydroxide ions increases.

Step 2: Find the solubility products

The Ksp values for the hydroxides of these ions are as follows: \(\mathrm{Fe(OH)_3}\) has a \(K_{sp}\) of \(4 \times 10^{-38}\), \(\mathrm{Pb(OH)_2}\) has a \(K_{sp}\) of \(1.2 \times 10^{-15}\), and \(\mathrm{Al(OH)_3}\) has a \(K_{sp}\) of \(3 \times 10^{-34}\).

Step 3: Compare the Ksp values

The metal hydroxide with the lowest \(K_{sp}\) will precipitate first. Comparing the \(K_{sp}\) values:- \(\mathrm{Fe(OH)_3}\) (\(4 \times 10^{-38}\)) is the least soluble.- \(\mathrm{Al(OH)_3}\) (\(3 \times 10^{-34}\)) is next.- \(\mathrm{Pb(OH)_2}\) (\(1.2 \times 10^{-15}\)) is the most soluble.

Step 4: Determine the order of precipitation

Based on \(K_{sp}\) values, \(\mathrm{Fe(OH)_3}\) with the smallest \(K_{sp}\) will precipitate first, followed by \(\mathrm{Al(OH)_3}\), and finally \(\mathrm{Pb(OH)_2}\).

Key Concepts

Solubility Product (Ksp)

The solubility product, denoted as \(K_{sp}\), helps us understand how substances dissolve and form precipitates. It’s an equilibrium constant specific to the dissociation of a compound into its constituent ions in a solution. For hydroxide ions, the equation goes like this: \[ \text{M(OH)_n(s) \rightleftharpoons M^{n+}(aq) + nOH^-(aq)}\]The \(K_{sp}\) expression for this would be:\[\K_{sp} = [M^{n+}][OH^-]^n\]

• Where \([M^{n+}]\) is the concentration of the metal ion.
• And \([OH^-]^n\) is the concentration of hydroxide ions raised to the power of n.

The smaller the \(K_{sp}\) value, the less soluble the compound, meaning it's more likely to form a precipitate. This is crucial for understanding the order in which substances will precipitate as the concentration of a particular ion, like hydroxide, increases in solution. In the case of metal hydroxides like \(\text{Fe(OH)_3}\), \(K_{sp}\) values help predict precipitation order.

Hydroxide Precipitation

When it comes to precipitation reactions, particularly with hydroxide ions, the key is the addition of a base. As aqueous NaOH is added, the concentration of \(OH^-\) ions in the solution increases.This increase leads to various reactions, each wanting to form a solid precipitate. The precipitation order follows the \(K_{sp}\) values: a lower \(K_{sp}\) means the compound will precipitate first. Consider these ions:

• Iron Hydroxide \(\text{Fe(OH)_3}\): With a \(K_{sp}\) of \(4 \times 10^{-38}\), it precipitates first.
• Aluminum Hydroxide \(\text{Al(OH)_3}\): Next, with a slightly higher \(K_{sp}\) of \(3 \times 10^{-34}\).
• Lead Hydroxide \(\text{Pb(OH)_2}\): Precipitates last, due to a much larger \(K_{sp}\) of \(1.2 \times 10^{-15}\).

This sequence depends solely on the increasing \(OH^-\) concentration reaching critical levels for each ion, reaching saturation according to their specific \(K_{sp}\) values.

Concentration of Ions

Understanding ion concentration is vital in managing chemical reaction outcomes. Initially, each ion (\(\mathrm{Fe}^{3+}\), \(\mathrm{Pb}^{2+}\), \(\mathrm{Al}^{3+}\)) is at a concentration of 0.10 M. As sodium hydroxide is added, the \(OH^-\) concentration starts to dominate. This affects the system's equilibrium.

• Each metal ion will continue to exist in its free form until the \(OH^-\) concentration reaches a point where it exceeds the solubility product for the given metal hydroxide.
• When this happens, the ions begin to combine with the \(OH^-\) ions to form a precipitate as per reaction equations illustrated in \(K_{sp}\) analysis.

By controlling the concentration of added \(OH^-\) ions, one can selectively precipitate different metal ions based on their respective \(K_{sp}\) values. This controlled addition helps predict and manipulate the outcome of typical laboratory and industrial processes.