Problem 123 - 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 \(4.458 \times 10^{-25} \mathrm{M}\). Answer: (a)

Step 1: Write the Solubility Product Expression

The molar solubility of \( \mathrm{Fe(OH)}_3 \) is given by:\( \mathrm{Fe(OH)}_3(s) \rightleftharpoons \mathrm{Fe}^{3+}(aq) + 3 \mathrm{OH}^-(aq) \)The solubility product, \( K_{sp} \), is:\[ K_{sp} = [\mathrm{Fe}^{3+}][\mathrm{OH}^-]^3 = 2.6 \times 10^{-39} \]

Step 2: Use Henderson-Hasselbalch Equation for pH

First, calculate the pH of the buffer solution.The relevant equation is the Henderson-Hasselbalch equation: \[ \text{pH} = \text{p}K_a + \log{\left( \frac{[\mathrm{NH}_3]}{[\mathrm{NH}_4^+]} \right)} \]Since \( K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10} \). Therefore, \( \text{p}K_a = -\log{(5.56 \times 10^{-10})} \approx 9.26 \).So, the pH is:\[ \text{pH} = 9.26 + \log{\left( \frac{0.10}{0.10} \right)} = 9.26 \]

Step 3: Calculate pOH from pH

Since \( \text{pH} + \text{pOH} = 14 \), the \( \text{pOH} \) is:\[ \text{pOH} = 14 - 9.26 = 4.74 \]Therefore, \([\mathrm{OH}^-]\) is given by:\[ [\mathrm{OH}^-] = 10^{-4.74} \approx 1.82 \times 10^{-5} \mathrm{M} \]

Step 4: Plug into the Solubility Product

Using the \( K_{sp} \) expression:\[ 2.6 \times 10^{-39} = [\mathrm{Fe}^{3+}][1.82 \times 10^{-5}]^3 \]Solving for \([\mathrm{Fe}^{3+}]\):\[ [\mathrm{Fe}^{3+}] = \frac{2.6 \times 10^{-39}}{(1.82 \times 10^{-5})^3} \]Calculate the value:\[ [\mathrm{Fe}^{3+}] \approx 4.458 \times 10^{-25} \mathrm{M} \]

Step 5: Verify and Conclude

The calculations give us a molar solubility of \([\mathrm{Fe}^{3+}] \approx 4.458 \times 10^{-25} \mathrm{M}\). The closest answer choice is \( a) 4.458 \times 10^{-25} \mathrm{M} \).

Key Concepts

Solubility Product Constant (Ksp)

The Solubility Product Constant, often abbreviated as \( K_{sp} \), is a key concept when discussing the solubility of sparingly soluble salts in water. It describes the product of the concentrations of the ions in a saturated solution of a compound. For example, in the case of \( \mathrm{Fe(OH)}_3 \), the dissociation in water gives \( \mathrm{Fe}^{3+} \) and \( \mathrm{OH}^- \). Therefore, the expression is given by:\[ K_{sp} = [\mathrm{Fe}^{3+}][\mathrm{OH}^-]^3 \]- The numerical value of \( K_{sp} \) tells us how much of the substance can dissolve in a solution before the ions start to precipitate out as a solid.- A small \( K_{sp} \) value indicates low solubility, meaning the compound does not dissolve well.
In the case of \( \mathrm{Fe(OH)}_3 \), the \( K_{sp} \) is given as \( 2.6 \times 10^{-39} \), a very low value, suggesting that \( \mathrm{Fe(OH)}_3 \) is sparingly soluble. The concept of \( K_{sp} \) is crucial in many areas of chemistry, especially in predicting whether a precipitate will form in reactions.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch Equation is a powerful tool used to estimate the pH of a buffer solution. Buffer solutions are crucial because they can resist changes in pH upon the addition of small amounts of acids or bases. Here's the equation:\[ \text{pH} = \text{p}K_a + \log{\left( \frac{[\mathrm{Base}]}{[\mathrm{Acid}]} \right)} \]In our exercise, the buffer solution contains ammonia \( (\mathrm{NH}_3) \) and ammonium chloride \( (\mathrm{NH}_4Cl) \). Ammonia acts as the base, while the ammonium ion \( (\mathrm{NH}_4^+) \) acts as the acid.- The \( \text{p}K_a \) value provides insights into the strength of the weak acid part of the buffer.- For ammonium, \( K_a \) is derived from the relationship \( K_a = \frac{K_w}{K_b} \), leading to a \( \text{p}K_a \) of approximately 9.26.
This equation helped us calculate the pH of the buffer solution as approximately 9.26, providing a base for further calculations in determining molar solubility.

Buffer Solution Chemistry

Buffer solutions play a critical role in maintaining the pH of a solution within a narrow range, even when an external acid or base is added. They consist of a weak acid and its conjugate base or a weak base and its conjugate acid. For the scenario examined, a buffer system of \( \mathrm{NH}_3 \) and \( \mathrm{NH}_4Cl \) is utilized.- Buffers rely on the weak acid-base pair balancing each other's effects, preventing sharp changes in pH.- This is essential in many biological and chemical processes where pH must remain stable.The chemistry of buffer solutions ensures that the concentration of hydrogen ions \( (\mathrm{H}^+) \) or hydroxide ions \( (\mathrm{OH}^-) \) do not swing extensively, giving way to stable environments. This quality of buffer solutions was crucial in calculating the pH and ultimately the solubility of \( \mathrm{Fe(OH)}_3 \) in our exercise.
Understanding buffer solution chemistry allows for manipulating the pH to either increase or decrease the solubility of certain compounds in solutions, as highlighted by the problem set presented.