What is the approximate concentration of free Al³+ ion at equilibrium when 1.06x10-2 mol aluminum nitrate is added to 1.00 L of solution that is 1.390 M in OH-. For [Al(OH)4], Kf = 1.1×10³3. [A1³+] = M

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Chapter1: Chemical Foundations
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**Problem Statement:**

Determine the approximate concentration of free \( \text{Al}^{3+} \) ion at equilibrium under the given conditions. 

- **Initial Conditions:**
  - \( 1.06 \times 10^{-2} \) mol of aluminum nitrate is added to \( 1.00 \) L of solution.
  - The solution has a concentration of \( 1.390 \) M in hydroxide ions (\( \text{OH}^- \)).

- **Equilibrium Expression:**
  - \([\text{Al(OH)}_4]^-\) 
  - Formation constant (\( K_f \)) is given as \( 1.1 \times 10^{33} \).

**Required:**

Calculate the concentration of \( \text{Al}^{3+} \) ions:
\[ [\text{Al}^{3+}] = \text{_____ M} \]
Transcribed Image Text:**Problem Statement:** Determine the approximate concentration of free \( \text{Al}^{3+} \) ion at equilibrium under the given conditions. - **Initial Conditions:** - \( 1.06 \times 10^{-2} \) mol of aluminum nitrate is added to \( 1.00 \) L of solution. - The solution has a concentration of \( 1.390 \) M in hydroxide ions (\( \text{OH}^- \)). - **Equilibrium Expression:** - \([\text{Al(OH)}_4]^-\) - Formation constant (\( K_f \)) is given as \( 1.1 \times 10^{33} \). **Required:** Calculate the concentration of \( \text{Al}^{3+} \) ions: \[ [\text{Al}^{3+}] = \text{_____ M} \]
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