Question

The solubility-product constant for Zn(OH)2 is Ksp=3.00×10−16.

The formation constant for the hydroxo complex, Zn(OH)42−, is Kf=4.60×1017.

When Zn(OH)2(s) was added to 1.00 L of a basic solution, 1.10×10−2 mol of the solid dissolved. What is the concentration of OH− in the final solution?

Answer 1

You need K for the reaction: Zn(OH)2(s) + 2OH- ----> [Zn(OH)4{-2}]
Remember that solid Zn(OH)2(s) will not appear in the equilibrium constant expression, so:
Zn+2 + 4OH- ==⇒ [Zn(OH)4{-2}];

Kf = [Zn(OH)4{-2}]/ [Zn+2][OH-]^4
and
Ksp = [Zn+2][OH-]^2

Multiply Kf by Ksp
[Zn(OH)4{-2}]/ [Zn+2][OH-]^4 x [Zn+2][OH-]^2 yields: [Zn(OH)4{-2}]/[OH-]^2 ;
The K for this equilibrium constant is:
4.60 x 10^17 (3.00 x 10^-16) = 13.8 x 10^1

Since Kf is extremely large, assume that the 1.24 x 10^-2 moles of [Zn(OH)4{-2}]
are formed in the 1.0 liter
138 = 1.10 x 10^-2/[OH]^2
[OH]^2= 1.24 x 10^-2/138 = 7.97 x 10^-5
[OH-] = 8.928x10^-3 M