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A solution contains 3.8x10-2 M in Al3+ and 0.29 M in NaF. If the Kf for AlF63- is 7x1019, how much aluminum ion remains at equilibrium?

Sagot :

okpalawalter8

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

For more information on Chemical Reaction

https://brainly.com/question/11231920

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