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

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

B) write kf equilibrium expression for the reaction a

Explanation / Answer

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

B) Kc = ([AlF6]3-) / [(Al3+)(F-)6]

Kc is equilibrium expression, Kf is forward reaction expression. Rate = Kf x [AlF6]3-

Kc = Kf / Kr

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.

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

Kc = 5.86 x 108

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