The generic metal a forms an insoluble salt ab(s) and a complex ac5(aq). the equilibrium concentrations in a solution of ac5 were found to be [a] = 0.100 m, [c] = 0.0320 m, and [ac5] = 0.100 m. determine the formation constant, kf, of ac5.

AB (s)plus 5C(aq)  <-> AC5 (aq)plus B(aq)
,KF  is equal to (AC5)/(A) (C)^5  which  is  0.100/ (0.001m) (0.032)^5  which  is equal  to 2.9x10^7

AB  sokubility in  1.000 solution of    C   is   0.121M.
 find  the  ksp  of  AB

from ICE  table the  change  is  equivalent  to  solubility  which  is  0.121m.  therefore  k for (AC)  is 0.121and  since  ratio of AC  to B  is  1:1 that  of
(B)is also 0.121
(C) is  1.00- 5(0.121)  which is  0.605
k  is  equal  to (0.121) / (0.0121)(0.605)^5 which is  1.5237
1.5237 is  then  divided  by  2.9x10^7  which  is  5.3x10^-8

Do u have the n? or maybe the m?

Adozen is 1221/12 = 1.751.75 x 12 = 21therefore, 1.75 is the answer.