Why don't we multiply Ksp and Kf?

My understanding is that for a process with two reactions having the equilibrium constants K1 and K2, K for the full process is K = K1 x K2. However, in this problem in my book where complexation increases solubility of CuS in water, they just argue that Kf >> Ksp and drives the reaction forward. Why is it not K = Kf x Ksp?
Thank you.


The problem:
A 0.1 mol sample of CuS is added to 1.00 L of 1.00 M NH3. What is the final concentration of the complex ion, tetraaminecopper(II)?


$$\ce{CuS(s) <=> Cu^2+ (aq) + S^2- (aq)}\ \ \ K\_\mathrm{sp} = \pu{8E-37}$$


$$\ce{Cu^2+ (aq) + 4NH3 (aq) <=> [Cu(NH3)4]^2+ (aq)}\ \ \ K\_f = \pu{1.1E13}$$


The answer key: $K\_f$ will drive the dissociation of $\ce{CuS}$ forward such that the amount of $\ce{Cu^2+}$ available to react is [Cu2+] = .1 mol/1.00 L = 0.1 M, and due to the 1:1 relationship between $\ce{Cu^2+}$ and the complex ion, the concentration of the complex ion will also be 0.1 M. I believe this implies the overall K >> 1, rather than ~10^-24, which it would be if we multiplied Ksp and Kf.


