What is the approximate concentration of free Al³+ ion at equilibrium when 1.86x10-² mol aluminum nitrate is added to 1.00 L of solution that is 1.440 M in F. For [AIF6]³, Kf = 6.9×10¹⁹. [A1³+] = M
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![**Question:**
What is the approximate concentration of free \( \text{Al}^{3+} \) ion at equilibrium when \( 1.86 \times 10^{-2} \) mol aluminum nitrate is added to \( 1.00 \, \text{L} \) of solution that is \( 1.440 \, \text{M} \) in \( \text{F}^- \)? For \([ \text{AlF}_6 ]^{3-}\), \( K_f = 6.9 \times 10^{19} \).
**Diagram/Equation:**
\[ [ \text{Al}^{3+} ] = \underline{\hspace{2cm}} \, \text{M} \]
**Explanation:**
This problem involves calculating the concentration of free aluminum ions (\( \text{Al}^{3+} \)) in a solution at equilibrium using the provided formation constant (\( K_f \)) for the complex ion \([ \text{AlF}_6 ]^{3-}\). You are given initial amounts of aluminum nitrate and fluoride ions and are expected to use this information in conjunction with the formation constant to find the desired concentration. The equation indicates that the result needs to be filled in after solving the necessary equilibrium expressions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F47386969-f92b-4827-b6d7-02bdac2aa892%2Fb434e8bf-3790-48d9-ae9e-3df2d212ed75%2Fwznabds_processed.jpeg&w=3840&q=75)
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