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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### 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^{-
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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