Aluminum ion that remains at equilibrium is mathematically given as
x=9.1*10^{-10}M
Aluminum ion that remains at equilibrium
Question Parameters:
A solution contains 3.8x10-2 M in Al3+ and 0.29 M in NaF.
If the Kf for AlF63- is 7x1019,
Generally the equation for the Chemical reaction is is mathematically given as
AL^3-+6F--------{AiF6}
Therefore
Kf=\fraac{AiF6}{(AL^3-)(F-)}
7*10^19=\frac{x}{.8x10-2*(0.29-6x)^6}
x=9.1*10^{-10}M
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