Problem 121 - A Complete Resource Book in Chemistry for JEE Main

Question

Find the molar solubility of \(\mathrm{Fe}(\mathrm{OH})_{3}\) in a buffer solution that \(0.10 \mathrm{M}\) in \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(0.10 \mathrm{M}\) in \(\mathrm{NH}_{3} .\) If \(\mathrm{K}_{\mathrm{b}}\) \(\left(\mathrm{NH}_{3}\right)=1.8 \times 10^{-5}\) and \(\mathrm{Ksp}\left[\mathrm{Fe}(\mathrm{OH})_{3}\right]=2.6 \times 10^{-39}\) (a) \(4.458 \times 10^{-25} \mathrm{M}\) (b) \(3.458 \times 10^{-25} \mathrm{M}\) (c) \(2.229 \times 10^{-24} \mathrm{M}\) (d) \(4.458 \times 10^{-22} \mathrm{M}\)

Step-by-Step Solution

Answer

The molar solubility is approximately \(4.458 \times 10^{-25} \mathrm{M}\), which is option (a).

Step 1: Write the dissociation reaction

The dissociation of iron(III) hydroxide, \( \mathrm{Fe(OH)}_3 \), can be represented by the following equilibrium equation:\[\mathrm{Fe(OH)}_3(s) \rightleftharpoons \mathrm{Fe}^{3+}(aq) + 3\mathrm{OH}^-(aq)\]

Step 2: Calculate the hydroxide ion concentration using Kb

The buffer solution containing \( \mathrm{NH}_4^+ \) and \( \mathrm{NH}_3 \) can be used to calculate the hydroxide ion concentration. First, find the \( \mathrm{K}_\mathrm{w} \) value:\[\mathrm{K}_\mathrm{w} = 1 \times 10^{-14}\]Now, calculate \( \mathrm{K}_a \) for \( \mathrm{NH}_4^+ \):\[\mathrm{K}_a = \frac{\mathrm{K}_\mathrm{w}}{\mathrm{K}_b(\mathrm{NH}_3)} = \frac{1 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10}\]Then, calculate the equilibrium concentration of \([\mathrm{OH}^-]\) using the Henderson-Hasselbalch equation:\[p\text{K}_a = -\log(5.56 \times 10^{-10}) \approx 9.25\]The pH of the buffer is calculated as:\[\text{pH} = \text{p}K_a + \log\left(\frac{[\mathrm{NH}_3]}{[\mathrm{NH}_4^+]}\right) = 9.25 + \log\left(\frac{0.10}{0.10}\right) = 9.25\]\[\text{pOH} = 14 - \text{pH} = 4.75\]Finally, convert pOH to \([\mathrm{OH}^-]\):\[[\mathrm{OH}^-] = 10^{-4.75} \approx 1.78 \times 10^{-5} \mathrm{M}\]

Step 3: Express Ksp in terms of molar solubility

Let \( s \) be the molar solubility of \( \mathrm{Fe(OH)}_3 \). Then the expression for \( \mathrm{K}_{sp} \) is:\[\mathrm{K}_{sp} = [\mathrm{Fe}^{3+}][\mathrm{OH}^-]^3 = s(1.78 \times 10^{-5})^3\]Given \( \mathrm{K}_{sp} = 2.6 \times 10^{-39} \), we have:\[2.6 \times 10^{-39} = s(1.78 \times 10^{-5})^3\]

Step 4: Solve for molar solubility

Calculate \((1.78 \times 10^{-5})^3 \):\[(1.78 \times 10^{-5})^3 = 5.64 \times 10^{-15}\]Now solve for \( s \):\[s = \frac{2.6 \times 10^{-39}}{5.64 \times 10^{-15}} = 4.458 \times 10^{-25} \mathrm{M}\]

Step 5: Match with options

The calculated molar solubility \( s = 4.458 \times 10^{-25} \mathrm{M} \) matches with option (a).

Key Concepts

Dissociation Reaction

The dissociation reaction is a fundamental process that describes how compounds split into their constituent ions in a solution. When dealing with solubility and equilibrium, understanding dissociation reactions is crucial. For the compound iron(III) hydroxide, represented as \( ext{Fe(OH)}_3 \), its dissociation in water can be shown by the equation: \[ ext{Fe(OH)}_3(s) \rightleftharpoons ext{Fe}^{3+}(aq) + 3 ext{OH}^-(aq) \] This equation indicates that for every mole of \( ext{Fe(OH)}_3 \) dissolved, one mole of iron ions and three moles of hydroxide ions are produced. This balance of ions is pivotal when calculating the solubility product and determining the solubility of the compound in various conditions, such as in a buffer solution.

Buffer Solution

A buffer solution is a special solution that maintains its pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base, or a weak base and its conjugate acid. In our exercise, a buffer is formed from \( ext{NH}_4^+ \) and \( ext{NH}_3 \). This buffer is crucial for controlling the pH of the solution.

• The concentration of the two components in the buffer controls its capacity to neutralize added acids or bases.
• In this scenario, the concentrations of \( ext{NH}_4^+ \) and \( ext{NH}_3 \) are both 0.10 M, which helps in determining the pH using the Henderson-Hasselbalch equation.

Understanding buffer solutions is vital as they play a key role in many chemical and biological systems to stabilize pH.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a simplified formula used to calculate the pH of buffer solutions. It relates the pH, the pKa (the negative logarithm of the acid dissociation constant \( K_a \)), and the concentrations of the acid and its conjugate base. The equation is expressed as:
\[ ext{pH} = ext{p}K_a + rac{[ ext{Base}]}{[ ext{Acid}]} \]
In our example, it helps calculate the pH of a buffer made from \( ext{NH}_4^+ \) and \( ext{NH}_3 \) by:

• First calculating \( ext{p}K_a \) using the given \( K_b \) of \( ext{NH}_3 \) and the ion-product constant of water \( K_w \).
• Then, substituting the concentration values into the equation.

This equation simplifies the process of determining the buffer's pH, which is used to find the hydroxide concentration needed for solubility calculations.

Solubility Product (Ksp)

The solubility product, or \( K_{sp} \), is an equilibrium constant used for sparingly soluble salts. It represents the product of the molar concentrations of the ions produced from the salt, each raised to the power of its coefficient in the balanced chemical equation for dissociation. For \( ext{Fe(OH)}_3 \), the formula for the solubility product is:
\[ K_{sp} = [ ext{Fe}^{3+}][ ext{OH}^-]^3 \]
This relation allows us to determine the molar solubility of a compound using known ion concentrations. In the exercise, \( K_{sp} \) is given as \( 2.6 \times 10^{-39} \), and using the hydroxide ion concentration from the buffer's pH, the molar solubility \( s \) is calculated.

• This involves solving the formula \( s imes [ ext{OH}^-]^3 = K_{sp} \).
• By substituting the values, the solubility of \( ext{Fe(OH)}_3 \) in the buffer is determined.

This concept is critical for predicting the behaviors of salts in various chemical environments.