Above what Fe2+ concentration will Fe(OH), precipitate from a buffer solution that has a pH of 8.75? The Ksp of Fe(OH), is 4.87×10-17. [Fe2+] = N
Above what Fe2+ concentration will Fe(OH), precipitate from a buffer solution that has a pH of 8.75? The Ksp of Fe(OH), is 4.87×10-17. [Fe2+] = N
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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![**Determining the Concentration of Fe²⁺ for Precipitation of Fe(OH)₂ from a Buffer Solution**
**Problem Statement:**
Above what Fe²⁺ concentration will Fe(OH)₂ precipitate from a buffer solution that has a pH of 8.75?
The \( K_{sp} \) of Fe(OH)₂ is \( 4.87 \times 10^{-17} \).
\[ \text{[Fe}^{2+}\text{]} = \underline{\hspace{12cm}} \text{M}
\]
**Solution Guide:**
To solve this problem, we need to apply the solubility product constant (\( K_{sp} \)) for Fe(OH)₂ and the provided pH value of the buffer solution.
1. **Write the expression for the solubility product (\( K_{sp} \))**:
\[ K_{sp} = [Fe^{2+}][OH^-]^2 \]
2. **Determine the concentration of \( OH^- \) ions from the pH value**:
\[ \text{pH} + \text{pOH} = 14 \]
\[ \text{pOH} = 14 - 8.75 = 5.25 \]
\[ [OH^-] = 10^{-\text{pOH}} = 10^{-5.25} \approx 5.62 \times 10^{-6} \text{M} \]
3. **Rearrange the solubility product expression to solve for \([Fe^{2+}]\)**:
\[ K_{sp} = [Fe^{2+}][OH^-]^2 \]
\[ [Fe^{2+}] = \frac{K_{sp}}{[OH^-]^2} = \frac{4.87 \times 10^{-17}}{(5.62 \times 10^{-6})^2} \]
\[ [Fe^{2+}] \approx \frac{4.87 \times 10^{-17}}{31.57 \times 10^{-12}} \]
\[ [Fe^{2+}] \approx 1.54 \times 10^{-6} \text{M} \]
Therefore, Fe(OH)₂ will precipitate from the buffer solution at Fe²⁺ concentrations **above \( 1.54 \times 10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbd6e2cac-a13e-4af0-8d42-5fb7a66b94d3%2F21652f9a-a599-496a-83e3-66e8215f38f3%2F68o8e6z_processed.png&w=3840&q=75)
Transcribed Image Text:**Determining the Concentration of Fe²⁺ for Precipitation of Fe(OH)₂ from a Buffer Solution**
**Problem Statement:**
Above what Fe²⁺ concentration will Fe(OH)₂ precipitate from a buffer solution that has a pH of 8.75?
The \( K_{sp} \) of Fe(OH)₂ is \( 4.87 \times 10^{-17} \).
\[ \text{[Fe}^{2+}\text{]} = \underline{\hspace{12cm}} \text{M}
\]
**Solution Guide:**
To solve this problem, we need to apply the solubility product constant (\( K_{sp} \)) for Fe(OH)₂ and the provided pH value of the buffer solution.
1. **Write the expression for the solubility product (\( K_{sp} \))**:
\[ K_{sp} = [Fe^{2+}][OH^-]^2 \]
2. **Determine the concentration of \( OH^- \) ions from the pH value**:
\[ \text{pH} + \text{pOH} = 14 \]
\[ \text{pOH} = 14 - 8.75 = 5.25 \]
\[ [OH^-] = 10^{-\text{pOH}} = 10^{-5.25} \approx 5.62 \times 10^{-6} \text{M} \]
3. **Rearrange the solubility product expression to solve for \([Fe^{2+}]\)**:
\[ K_{sp} = [Fe^{2+}][OH^-]^2 \]
\[ [Fe^{2+}] = \frac{K_{sp}}{[OH^-]^2} = \frac{4.87 \times 10^{-17}}{(5.62 \times 10^{-6})^2} \]
\[ [Fe^{2+}] \approx \frac{4.87 \times 10^{-17}}{31.57 \times 10^{-12}} \]
\[ [Fe^{2+}] \approx 1.54 \times 10^{-6} \text{M} \]
Therefore, Fe(OH)₂ will precipitate from the buffer solution at Fe²⁺ concentrations **above \( 1.54 \times 10
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