Really hard chem question...solubility and acids..

Question

Fe(OH) can be dissolved in an aqueous solution of CN-...It forms Fe(CN)6^-4

Kf= 4 x 10^45

Calculate the molar solubility in a 1M CN- solution?

Can someone please explain this in steps?

Thanks~

I really don't understand this...How do you calculate Molar solubility??

Answer 1

First FeOH does not exist.  It would have to be Fe(OH)2 or Fe(OH)3.  So I'm going to assume Fe(OH)2, based on the final product.

Fe(OH)2 (s)  +  6CN-  =  Fe(CN)6(-4) + 2(OH-)

Kf =  [Fe(CN)6(-4)] [OH-]^2 / [CN-]^6

4 x 10^45  = (x)(2x)^2 / (1-6x)^6

x is the molar solubility