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 of \(\mathrm{Fe(OH)_3}\) is approximately \(4.458 \times 10^{-25} \mathrm{M}\).

Step 1: Identify pH of the buffer solution

The buffer solution consists of 0.10 M NH\(_4\)Cl and 0.10 M NH\(_3\). NH\(_3\) is a weak base, and its conjugate acid is NH\(_4^+\), thus this is a buffer solution. To find the pH, we first calculate the pOH using the ratio of the concentrations and the K\(_b\) of NH\(_3\).

Step 2: Calculate pOH of the buffer

Use the Henderson-Hasselbalch equation for a base: \[ \text{pOH} = \text{pK}_b + \log \left( \frac{[\text{NH}_4^+]}{[\text{NH}_3]} \right) \]Substitute the values:\[ \text{pOH} = -\log(1.8 \times 10^{-5}) + \log \left( \frac{0.10}{0.10} \right) = 4.74 + 0 = 4.74 \]

Step 3: Convert pOH to pH

Convert pOH to pH using the relation:\[ \text{pH} = 14 - \text{pOH} \]Therefore:\[ \text{pH} = 14 - 4.74 = 9.26 \]

Step 4: Determine [OH-] in the buffer

Use the pOH to find the hydroxide concentration:\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-4.74} \approx 1.82 \times 10^{-5} \text{ M} \]

Step 5: Set up expression using the Ksp of Fe(OH)3

The solubility product (K\(_{sp}\)) of Fe(OH)\(_3\) is:\[ \text{K}_{sp} = [\text{Fe}^{3+}][\text{OH}^-]^3 \]And given \(\text{Ksp} = 2.6 \times 10^{-39}\), we know\[ [\text{Fe}^{3+}][\text{OH}^-]^3 = 2.6 \times 10^{-39} \]

Step 6: Solve for molar solubility of Fe(OH)3

Substitute [OH\(^{-}\)] from Step 4 into the K\(_{sp}\) expression:\[ [\text{Fe}^{3+}] = \frac{2.6 \times 10^{-39}}{(1.82 \times 10^{-5})^3} \]Calculate [Fe\(^{3+}\)]:\[ [\text{Fe}^{3+}] \approx \frac{2.6 \times 10^{-39}}{6.03 \times 10^{-15}} \approx 4.31 \times 10^{-25} \text{ M} \]

Step 7: Choose the closest answer

Given the calculated molar solubility of \(4.31 \times 10^{-25}\) M, the closest choice from the provided options is (a) \(4.458 \times 10^{-25}\) M.

Key Concepts

Buffer Solutions

Buffer solutions are aqueous systems that can resist changes in pH when small amounts of acids or bases are added to them. This stability in pH is crucial in various chemical and biological processes. The buffer solution in the exercise is composed of 0.10 M NH extsubscript{4}Cl and 0.10 M NH extsubscript{3}. These components work together because NH extsubscript{3} is a weak base, and NH extsubscript{4} extsuperscript{+} is its conjugate acid.
In a buffer, the presence of significant amounts of both the weak base and its conjugate acid allows the solution to neutralize added H extsuperscript{+} or OH extsuperscript{-} ions. This means that even if an acid or base is introduced, the pH will not change drastically, maintaining the environment within a desired range.
This property of buffers is extensively used in experiments and manufacturing where maintaining a consistent pH is crucial. The buffer solution in this exercise is used to determine the influence of pH on the solubility of iron(III) hydroxide.

Ksp Calculation

Ksp, or the solubility product constant, is a measure of the solubility of a compound under specific conditions. It applies to sparingly soluble salts like Fe(OH) extsubscript{3}. In the dissolution process, one molecule of Fe(OH) extsubscript{3} dissolves to yield one Fe extsuperscript{3+} ion and three OH extsuperscript{-} ions.
The formula for K extsubscript{sp} is derived from the equilibrium concentrations of these ions:

• For Fe(OH) extsubscript{3}: \[ \text{K\_sp} = [\text{Fe}^{3+}][\text{OH}^-]^3 \]

The goal in this exercise is to use the given K extsubscript{sp} value of Fe(OH) extsubscript{3} to determine the molar solubility in a buffer solution. By substituting the concentration of OH extsuperscript{-} found from the pH calculations, we can calculate the concentration of Fe extsuperscript{3+}, or the molar solubility.
Solving this equation involves algebraic manipulation and substitution to find the amount of Fe extsuperscript{3+}, which gives an insight into how much of the solid compound can dissolve in a given solution.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a key formula in acid-base chemistry which provides a quick way to calculate the pH of a buffer solution. For a basic buffer, the equation is expressed as:

• \[ \text{pOH} = \text{pK}_b + \log \left( \frac{[\text{Conjugate Acid}]}{[\text{Base}]} \right) \]

This equation was utilized in the exercise to calculate the pOH of the buffer containing NH extsubscript{3} and NH extsubscript{4} extsuperscript{+}. Here, pK extsubscript{b} represents the base dissociation constant of NH extsubscript{3}, while the concentrations ([NH extsubscript{4} extsuperscript{+}] and [NH extsubscript{3}]) were known as equal.
This makes the logarithmic part of the equation effectively zero, maintaining simplicity in calculation. The resulting pOH can then be converted to pH, illustrating how buffer compositions impact pH regulation. Understanding the Henderson-Hasselbalch equation is essential for managing the balance of weak acids or bases and their conjugate partners in solutions.

pH and pOH Relationship

In aqueous solutions, the relationship between pH and pOH is always interconnected, represented through the equation:

• \[ \text{pH} + \text{pOH} = 14 \]

This relationship is significant because it underscores the complementary nature of hydrogen ions (H extsuperscript{+}) and hydroxide ions (OH extsuperscript{-}) in water. If you know either the pH or the pOH, you can effortlessly calculate the other using this relationship.
In the context of the given exercise, after finding the pOH of the buffer, the pH could be swiftly determined. This is pivotal when you're dealing with reactions sensitive to hydrogen ion concentration, like those involving solubility of salts.
Remembering this straightforward relationship simplifies the handling of acidic and basic solutions both practically and theoretically, enabling effective design and interpretation of experiments revolving around equilibrium chemistry.