Determine the solubility of Al(OH) 3 (kop-1. (Ksp = 1.8 x10-337 in a solution with pH = 2.00

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### Determining the Solubility of Al(OH)3 in a Solution with pH = 2.00

To determine the solubility of Aluminum Hydroxide \((\text{Al(OH)}_3)\) in a solution with a pH of 2.00, you need to utilize the solubility product constant (Ksp) of \(\text{Al(OH)}_3\). The provided Ksp value for \(\text{Al(OH)}_3\) is:

\[ \text{Ksp} = 1.9 \times 10^{-33} \]

#### Steps to Solve:

1. **Write the Dissociation Equation:**

   \(\text{Al(OH)}_3 \ (s) \rightarrow \text{Al}^{3+} \ (aq) + 3\text{OH}^{-} \ (aq)\)

2. **Establish the Solubility Product Expression:**

   \(\text{Ksp} = [\text{Al}^{3+}][\text{OH}^-]^3\)

3. **Determine the \([\text{H}^+]\) Concentration from the pH:**

   Given the pH is 2.00, the \([\text{H}^+]\) concentration is:

   \[\text{pH} = -\log[\text{H}^+]\]
   \[[\text{H}^+] = 10^{-\text{pH}} = 10^{-2} = 0.01 \ \text{M}\]

4. **Relate \([\text{H}^+]\) and \([\text{OH}^-]\) using the Water Dissociation Constant:**

   \[K_w = [\text{H}^+][\text{OH}^-] = 10^{-14}\]
   \[[\text{OH}^-] = \frac{K_w}{[\text{H}^+]} = \frac{10^{-14}}{0.01} = 10^{-12} \ \text{M}\]

5. **Substitute \([\text{OH}^-]\) into the Ksp Expression and Solve for \([\text{Al}^{3+}]\):**

   \[\text{Ksp} = [\text{Al}^{
Transcribed Image Text:### Determining the Solubility of Al(OH)3 in a Solution with pH = 2.00 To determine the solubility of Aluminum Hydroxide \((\text{Al(OH)}_3)\) in a solution with a pH of 2.00, you need to utilize the solubility product constant (Ksp) of \(\text{Al(OH)}_3\). The provided Ksp value for \(\text{Al(OH)}_3\) is: \[ \text{Ksp} = 1.9 \times 10^{-33} \] #### Steps to Solve: 1. **Write the Dissociation Equation:** \(\text{Al(OH)}_3 \ (s) \rightarrow \text{Al}^{3+} \ (aq) + 3\text{OH}^{-} \ (aq)\) 2. **Establish the Solubility Product Expression:** \(\text{Ksp} = [\text{Al}^{3+}][\text{OH}^-]^3\) 3. **Determine the \([\text{H}^+]\) Concentration from the pH:** Given the pH is 2.00, the \([\text{H}^+]\) concentration is: \[\text{pH} = -\log[\text{H}^+]\] \[[\text{H}^+] = 10^{-\text{pH}} = 10^{-2} = 0.01 \ \text{M}\] 4. **Relate \([\text{H}^+]\) and \([\text{OH}^-]\) using the Water Dissociation Constant:** \[K_w = [\text{H}^+][\text{OH}^-] = 10^{-14}\] \[[\text{OH}^-] = \frac{K_w}{[\text{H}^+]} = \frac{10^{-14}}{0.01} = 10^{-12} \ \text{M}\] 5. **Substitute \([\text{OH}^-]\) into the Ksp Expression and Solve for \([\text{Al}^{3+}]\):** \[\text{Ksp} = [\text{Al}^{
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