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) \(4.458 \times 10^{-22} \mathrm{M}\)

Short answer

The molar solubility of \( \mathrm{Fe}( ext{OH})_3 \) is approximately \( 4.458 \times 10^{-25} \text{ M} \). This corresponds to option (a).

Step-by-step solution

Calculate the Hydroxide Ion Concentration from The Buffer Solution

The solution contains a buffer with equal concentrations of \( \mathrm{NH}_4\mathrm{Cl} \) and \( \mathrm{NH}_3 \). We can use the Henderson-Hasselbalch equation to find the \( \mathrm{pH} \) and then calculate the \( \mathrm{pOH} \).Since it's a base, we calculate the \( \mathrm{pOH} \) directly:\[\mathrm{pOH} = \mathrm{pK}_b + \log \left( \frac{\left[ \mathrm{NH}_4^+ \right]}{\left[ \mathrm{NH}_3 \right]} \right)\]\[ \mathrm{pK}_b = -\log (1.8 \times 10^{-5}) \approx 4.74 \]Because the concentrations are equal:\[\mathrm{pOH} = 4.74 + \log(1) = 4.74\]

Calculate Hydroxide Ion Concentration

Convert \( \mathrm{pOH} \) to \( \left[\mathrm{OH}^-\right] \):\[\left[\mathrm{OH}^-\right] = 10^{-\mathrm{pOH}} = 10^{-4.74} \approx 1.82 \times 10^{-5} \ \text{M}\]

Write the Solubility Product Expression for \( \mathrm{Fe}( ext{OH})_3 \)

The solubility product expression for \( \mathrm{Fe}( ext{OH})_3 \) is:\[\mathrm{K}_{\mathrm{sp}} = \left[\mathrm{Fe}^{3+}\right]\left[\mathrm{OH}^-\right]^3\]Given, \( \mathrm{K}_{\mathrm{sp}} = 2.6 \times 10^{-39} \).

Calculate the Molar Solubility of \( \mathrm{Fe}( ext{OH})_3 \)

Let \( s \) be the molar solubility of \( \mathrm{Fe}( ext{OH})_3 \) in mol/L. Then,\[\mathrm{K}_{\mathrm{sp}} = s \times \left[ 1.82 \times 10^{-5} \right]^3 = 2.6 \times 10^{-39}\]Rearranging to solve for \( s \):\[s = \frac{2.6 \times 10^{-39}}{\left( 1.82 \times 10^{-5} \right)^3}\]\[s = \frac{2.6 \times 10^{-39}}{6.03 \times 10^{-15}}\]\[s \approx 4.31 \times 10^{-25} \text{ M}\]

Round and Compare with Given Options

The calculated value of \( s \) is approximately \( 4.31 \times 10^{-25} \text{ M} \). Comparing this with the provided choices, it is closest to option (a), \( 4.458 \times 10^{-25} \text{ M} \).

Key concepts

Buffer Solution

A buffer solution consists of a weak acid and its conjugate base, or a weak base and its conjugate acid, in comparable concentrations. In our exercise, we have a buffer solution made with 0.10 M of both ammonium chloride ( \(\text{NH}_4\text{Cl}\)) and ammonia ( \(\text{NH}_3\)). This composition ensures that the solution can resist changes in \(\text{pH}\) when small amounts of acid or base are added.

The usefulness of a buffer lies in its ability to maintain a relatively constant \(\text{pH}\), which is crucial in many chemical and biological systems. This balance is achieved because the weak acid/base pair in the buffer can "absorb" the excess hydrogen ions (H+) or hydroxide ions (OH-), thereby preventing significant \(\text{pH}\) changes. In the laboratory, buffer solutions are invaluable for controlling \(\text{pH}\) and stabilizing reactions where hydrogen ion concentration might otherwise vary greatly.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a formula that provides quick insights into the relationship between the \(\text{pH}\), \(\text{pK}_a\) (or \(\text{pK}_b\)), and the concentration of the acid/base pair in a buffer solution. It is given by: \[\text{pH} = \text{pK}_a + \log \left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\]for acidic buffers, and a similar equation involving \(\text{pK}_b\) for basic buffers like ammonia ( \(\text{NH}_3\)): \[\text{pOH} = \text{pK}_b + \log \left(\frac{[\text{NH}_4^+]}{[\text{NH}_3]}\right)\]

In the context of our exercise, the concentration of \(\text{NH}_4^+\) and \(\text{NH}_3\) are equal. This simplifies the equation to: \(\text{pOH} = \text{pK}_b = 4.74\), which indicates a stable hydroxide ion concentration at equilibrium. The Henderson-Hasselbalch equation is especially useful as it allows for the prediction of buffer capacity and facilitates the design of solutions that resist \(\text{pH}\) changes.

Solubility Product

The solubility product ( \(\text{K}_{\text{sp}}\)) is a critical concept when dealing with the solubility of sparingly soluble compounds like \(\text{Fe(OH)}_3\). It represents the maximum product of ion concentrations that can exist in solution before precipitation occurs. For \(\text{Fe(OH)}_3\), we write the solubility product expression as: \[\text{K}_{\text{sp}} = [\text{Fe}^{3+}][\text{OH}^-]^3\]

This exercise involves calculating the molar solubility of \(\text{Fe(OH)}_3\) in a buffer solution where the hydroxide ion concentration is affected by the buffering action. By knowing the \(\text{K}_{\text{sp}}\), and the concentration of \([\text{OH}^-]\) from the buffer, we can compute the molar solubility ( \(s\)) as required. The solubility product helps us understand how much \(\text{Fe(OH)}_3\) can dissolve under given conditions before the solution becomes saturated and precipitation begins.

Hydroxide Ion Concentration

Understanding hydroxide ion concentration ( \([\text{OH}^-]\)) is vital when studying the solubility of metal hydroxides like \(\text{Fe(OH)}_3\). In an aqueous system, \([\text{OH}^-]\) determines the basicity of the solution, which affects the solubility of various compounds. The concentration of hydroxide ions directly interacts with metal ions, impacting their ability to remain dissolved.

In buffered solutions, such as our system with \(\text{NH}_4\text{Cl}\) and \(\text{NH}_3\), the \([\text{OH}^-]\) concentration is intentionally kept at a predictable level. By calculating the \(\text{pOH}\), we derived \([\text{OH}^-] = 1.82 \times 10^{-5} \text{ M}\). Knowing this concentration allows us to utilize the solubility product to determine the conditions under which \(\text{Fe(OH)}_3\) will dissolve or precipitate. This concentration gives insight into the chemistry of solutions, tailoring conditions to achieve desired solubility outcomes.