Question

For the aqueous [AlF₆]^3- complex K_f= 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.

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