Question

a 200.0 ml volume of solution contains 3.8*10^-2 M in Al^3+ and 0.29 m in NaF

A) writeout a chemical equation fir Al^3 ion and F ion forming AlF6^-3

B) write kf equilibrium expression for the reaction a

Answer 1

A) Al³⁺ (aq) + 6F^-(aq) --> [AlF₆]^3-(aq)

B) K_c = ([AlF₆]^3-) / [(Al³⁺)(F^-)⁶]

K_c is equilibrium expression, K_f is forward reaction expression. Rate = K_f x [AlF₆]^3-

K_c = K_f / K_r

moles Al³⁺ = 200 mL x 10^-3 L x 3.8 x 10^-2 M = 0.0076 moles

moles F^- = 200 mL x 10^-3 L x 0.29 M = 0.058 moles

0.0076 moles Al³⁺ reacts with ( 6 x 0.0078 = 0.0468 moles) of F^- and gives 0.0468 moles of [AlF₆]^3- only.

K_c = (0.0468) / (0.0076)(0.0468)⁶ = 5.86 x 10⁸

K_c = 5.86 x 10⁸

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