Question

For the aqueous [AlF6]3- complex Kf= 4.0 x 1019 at 25 degrees celcius. Suppose equal volumes...

For the aqueous [AlF6]3- complex Kf= 4.0 x 1019 at 25 degrees celcius.

Suppose equal volumes of 0.0064 M Al (NO3)3 solution and 0.50 M KF solution are mixed. Calculate the equilibrium molarity of aqueous Al3+ ion.

Round your answer to 2 signifcant digits.

Homework Answers

Answer #1

concentration of Al(NO3)3 = 0.0064 / 2 = 0.0032 M

concentration of KF = 0.50 / 2 = 0.25 M

Al3+ (aq) +   6 F- (aq)   -----------------> AlF63-

1                    6                                          1

0.0032         0.25

here limiting reagent is Al3+.

concentration of F- remaining = 0.25 - 6 x 0.0032 = 0.2308 M

Kf = [AlF6]3- / [Al3+] [F-]^6

4.0 x 1019 = 0.0032 / [Al3+] (0.2308)^6

[Al3+] = 5.3 x 10^-19 M

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