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

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Chemistry

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

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4.0 Ratings (492 Votes)

    NaF --------> Na+ + F-

0.136M 0.136M

            Al+3 + 6F-    ---------> AlF63-

I         0.017   0.136                 0

C      -0.017   -0.017              0.017

E        0           0.119             0.017

Kf   = [AlF63-]/[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
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