Question

A solution is prepared by mixing 50 mL of 0.047 M Al(NO3)3 with 50 mL of 0.50 M NaF. For the complex [AlF6]3-, Kf = 6.9 x 1019. In the final solution, {[AlF6]3-}= ___ M.

Answer 1

concentration of Al3+ = 50 x 0.047 / 50 + 50 = 0.0235 M

concentration of F- = 50 x 0.50 / 50 + 50 = 0.25 M

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

1                     6                                  1

0.0235          0.25                     

here limiting reagent is Al3+.

so concentration of [AlF6]3- = 0.0235 M