T Part B How does this compare to the solubility of Mg(OH)2 in pure water? Express your answer as a ratio to two significant figures. View Available Hint(s) |2| ΑΣΦΑ -6 $1.63 107 Submit Previous Answers ?
T Part B How does this compare to the solubility of Mg(OH)2 in pure water? Express your answer as a ratio to two significant figures. View Available Hint(s) |2| ΑΣΦΑ -6 $1.63 107 Submit Previous Answers ?
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
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Please use pictures as reference. Use part A information to find part B answer.
Question: How does this compare to the solubility of Mg(OH)2 in pure water? Find Part B answer
![**Educational Website Transcription:**
---
**Part B: Solubility Comparison**
How does this compare to the solubility of Mg(OH)₂ in pure water? Express your answer as a ratio to two significant figures.
- **Hint Option:** View Available Hint(s)
- **Input Provided:**
\[
\frac{S_1}{S} = 1.63 \times 10^{-6}
\]
- **Submission Button:** Submit
- **Attempt Status:** Incorrect; Try Again; 5 attempts remaining
---
*Note: The image contains a mathematical expression input field and a submit button typical for an educational quiz or exercise platform, along with a notification indicating an incorrect submission and the remaining attempts.*](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4ef2acf2-70fd-40cc-bcbe-08eec48c6b31%2Fa9fa3d76-42e4-4f45-b8f4-3d24d2e8ab36%2Fcw2jvsx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Educational Website Transcription:**
---
**Part B: Solubility Comparison**
How does this compare to the solubility of Mg(OH)₂ in pure water? Express your answer as a ratio to two significant figures.
- **Hint Option:** View Available Hint(s)
- **Input Provided:**
\[
\frac{S_1}{S} = 1.63 \times 10^{-6}
\]
- **Submission Button:** Submit
- **Attempt Status:** Incorrect; Try Again; 5 attempts remaining
---
*Note: The image contains a mathematical expression input field and a submit button typical for an educational quiz or exercise platform, along with a notification indicating an incorrect submission and the remaining attempts.*
![**Part A**
Calculate the solubility (in grams per \(1.00 \times 10^2\) mL of solution) of magnesium hydroxide in a solution buffered at pH = 12.
Express your answer to two significant figures.
**\[ S_1 = 1.2 \times 10^{-8}\ \text{g} / (1.00 \times 10^2 \ \text{mL}) \]**
**Solution**
Begin by writing the reaction by which solid Mg(OH)₂ dissolves into its constituent aqueous ions:
\[ \text{Mg(OH)}_2(s) \rightarrow \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq) \]
Write the equilibrium expression for \(K_{sp}\):
\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \]
The solubility of Mg(OH)₂ depends on the concentration of OH⁻ ions. The calculation of the OH⁻ ion concentration is summarized in the hint for this part (\([\text{OH}^-] = 1.0 \times 10^{-2} \text{M}\)). In a buffered solution, \([\text{OH}^-]\) remains constant, and therefore, the molar solubility of Mg(OH)₂ equals \([\text{Mg}^{2+}]\):
\[ [\text{Mg}^{2+}] = \frac{K_{sp}}{[\text{OH}^-]^2} = \frac{2.6 \times 10^{-13}}{(1.0 \times 10^{-2})^2} = 2.1 \times 10^{-9}\ \text{M} \]
Then convert the molar solubility to g per 100 mL:
\[ S_1 = 100\ \mu L \times \frac{1.2 \times 10^{-9}\ \text{mol Mg(OH)}_2}{1\ L} \times \frac{58.33\ \text{g Mg(OH)}_2}{\text{1 mol Mg(OH)}_2} \]
\[ S_1 = 1.2 \times 10^{-8}\ \text{g Mg(OH)}_2/100\ \text](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4ef2acf2-70fd-40cc-bcbe-08eec48c6b31%2Fa9fa3d76-42e4-4f45-b8f4-3d24d2e8ab36%2Fnqpg93c_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Part A**
Calculate the solubility (in grams per \(1.00 \times 10^2\) mL of solution) of magnesium hydroxide in a solution buffered at pH = 12.
Express your answer to two significant figures.
**\[ S_1 = 1.2 \times 10^{-8}\ \text{g} / (1.00 \times 10^2 \ \text{mL}) \]**
**Solution**
Begin by writing the reaction by which solid Mg(OH)₂ dissolves into its constituent aqueous ions:
\[ \text{Mg(OH)}_2(s) \rightarrow \text{Mg}^{2+}(aq) + 2\text{OH}^-(aq) \]
Write the equilibrium expression for \(K_{sp}\):
\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \]
The solubility of Mg(OH)₂ depends on the concentration of OH⁻ ions. The calculation of the OH⁻ ion concentration is summarized in the hint for this part (\([\text{OH}^-] = 1.0 \times 10^{-2} \text{M}\)). In a buffered solution, \([\text{OH}^-]\) remains constant, and therefore, the molar solubility of Mg(OH)₂ equals \([\text{Mg}^{2+}]\):
\[ [\text{Mg}^{2+}] = \frac{K_{sp}}{[\text{OH}^-]^2} = \frac{2.6 \times 10^{-13}}{(1.0 \times 10^{-2})^2} = 2.1 \times 10^{-9}\ \text{M} \]
Then convert the molar solubility to g per 100 mL:
\[ S_1 = 100\ \mu L \times \frac{1.2 \times 10^{-9}\ \text{mol Mg(OH)}_2}{1\ L} \times \frac{58.33\ \text{g Mg(OH)}_2}{\text{1 mol Mg(OH)}_2} \]
\[ S_1 = 1.2 \times 10^{-8}\ \text{g Mg(OH)}_2/100\ \text
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