Calculate the solubility of zn(oh)2(s) in 2.0 m naoh solution. (hint: you must take into account the formation of zn(oh)2−4, which has a kf=2×1015.)

When we have:

Zn(OH)2 → Zn2+ 2OH-  with Ksp = 3 x 10 ^-16

and:

Zn2+ + 4OH- → Zn(OH)4 2-  with Kf = 2 x 10^15
 
by mixing those equations together:

Zn(OH)2 + 2OH- → Zn(OH)4 2- with K = Kf *Ksp = 3 x 10^-16 * 2x10^15 =0.6

by using ICE table:

         Zn(OH)2 + 2OH- → Zn(OH)4 2-

initial                     2m              0

change                  -2X                +X     

Equ                       2-2X                 X

when we assume that the solubility is X

and when K = [Zn(OH)4 2-] / [OH-]^2

        0.6 = X / (2-2X)^2    by solving this equation for X

∴ X = 0.53 m

∴ the solubility of Zn(OH)2 = 0.53 M

It would be (at) + cu (s)