Consider the following reaction: Al(OH)3 Al(OH)3(s) The Ksp for aluminum hydroxide is 3.01 x 10-34. Calculate the equilibrium concentrations for this reaction if 0.500 moles of aluminum hydroxide was added to 1.00 L of water. Calculate your answer to 3 SIGNIFICANT FIGURES.. 13+ Al³+ (aq) + 3OH-(aq)
Consider the following reaction: Al(OH)3 Al(OH)3(s) The Ksp for aluminum hydroxide is 3.01 x 10-34. Calculate the equilibrium concentrations for this reaction if 0.500 moles of aluminum hydroxide was added to 1.00 L of water. Calculate your answer to 3 SIGNIFICANT FIGURES.. 13+ Al³+ (aq) + 3OH-(aq)
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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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![### Equilibrium Concentration Calculation of Aluminum Hydroxide
#### Reaction:
Consider the following reaction:
\[ \text{Al(OH)}_3(\text{s}) \rightleftharpoons \text{Al}^{3+}(\text{aq}) + 3\text{OH}^{-}(\text{aq}) \]
#### Given Data:
- The solubility product constant (Ksp) for aluminum hydroxide (\(\text{Al(OH)}_3\)) is \(3.01 \times 10^{-34}\).
- \(0.500\) moles of aluminum hydroxide was added to \(1.00\) L of water.
#### Problem Statement:
Calculate the equilibrium concentrations for this reaction. Provide your answer to 3 significant figures.
##### Step-by-Step Solution:
1. **Establish the Initial Concentrations:**
Since aluminum hydroxide is a solid, its concentration remains constant and does not appear in the equilibrium expression. Initially, there are 0 moles of both \(\text{Al}^{3+}\) and \(\text{OH}^{-}\).
2. **Define the Changes at Equilibrium:**
Let \(s\) be the solubility of \(\text{Al(OH)}_3\) in mol/L. At equilibrium:
- The concentration of \(\text{Al}^{3+}\) will be \(s\)
- The concentration of \(\text{OH}^{-}\) will be \(3s\)
3. **Write the Expression for \(Ksp\):**
\[
\text{Ksp} = [\text{Al}^{3+}][\text{OH}^{-}]^3
\]
Substituting the equilibrium concentrations:
\[
3.01 \times 10^{-34} = (s)(3s)^3 = 27s^4
\]
4. **Solve for \(s\):**
\[
27s^4 = 3.01 \times 10^{-34}
\]
\[
s^4 = \frac{3.01 \times 10^{-34}}{27}
\]
\[
s^4 = 1.115 \times 10^{-35}
\]
\[
s = (1.115 \times 10^{-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3ba71b9f-6645-4b82-be5d-cda1c9d5ec57%2F05f98cfb-60bd-4f56-82e9-e2f83a16b0d3%2Fjvsu80r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Equilibrium Concentration Calculation of Aluminum Hydroxide
#### Reaction:
Consider the following reaction:
\[ \text{Al(OH)}_3(\text{s}) \rightleftharpoons \text{Al}^{3+}(\text{aq}) + 3\text{OH}^{-}(\text{aq}) \]
#### Given Data:
- The solubility product constant (Ksp) for aluminum hydroxide (\(\text{Al(OH)}_3\)) is \(3.01 \times 10^{-34}\).
- \(0.500\) moles of aluminum hydroxide was added to \(1.00\) L of water.
#### Problem Statement:
Calculate the equilibrium concentrations for this reaction. Provide your answer to 3 significant figures.
##### Step-by-Step Solution:
1. **Establish the Initial Concentrations:**
Since aluminum hydroxide is a solid, its concentration remains constant and does not appear in the equilibrium expression. Initially, there are 0 moles of both \(\text{Al}^{3+}\) and \(\text{OH}^{-}\).
2. **Define the Changes at Equilibrium:**
Let \(s\) be the solubility of \(\text{Al(OH)}_3\) in mol/L. At equilibrium:
- The concentration of \(\text{Al}^{3+}\) will be \(s\)
- The concentration of \(\text{OH}^{-}\) will be \(3s\)
3. **Write the Expression for \(Ksp\):**
\[
\text{Ksp} = [\text{Al}^{3+}][\text{OH}^{-}]^3
\]
Substituting the equilibrium concentrations:
\[
3.01 \times 10^{-34} = (s)(3s)^3 = 27s^4
\]
4. **Solve for \(s\):**
\[
27s^4 = 3.01 \times 10^{-34}
\]
\[
s^4 = \frac{3.01 \times 10^{-34}}{27}
\]
\[
s^4 = 1.115 \times 10^{-35}
\]
\[
s = (1.115 \times 10^{-
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