Answered

Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

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

Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.