A solution is 0.017 M in Al3+ and 0.136 M in NaF. If the Kf for...

A solution is 0.017 M in Al3+ and 0.136 M in NaF. If the Kf for AlF63- is 7.0 × 10^19, what is the aluminum ion concentration at equilibrium?

Expert Solution

NaF --------> Na⁺ + F^-

0.136M 0.136M

Al⁺³ + 6F^- ---------> AlF₆^3-

I 0.017 0.136 0

C -0.017 -0.017 0.017

E 0 0.119 0.017

Kf = [AlF₆^3-]/[Al⁺³][F^-]⁶

7*10¹⁹ = 0.017/x*(0.119)⁶

x = 0.017/7*10¹⁹ * (0.119)⁶

x = 8.55*10^-17

[Al⁺³] = x = 8.55*10^-17 M

queen_honey_blossom answered 1 year ago

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