The 𝐾sp of zinc hydroxide, Zn(OH)2, is 3.00×10−17 . Calculate the molar solubility, 𝑠 , of this compound.

Answer 1

Zn(OH)₂  <.............>        Zn²⁺    +    2OH^-

Initial              :       s                             0                 0  

At equilibrium:       s                                 s                2s   

Ksp = [Zn²⁺] x [OH^-]²

Substituting the given Ksp value in the above formula and concentrations of Zn²⁺, OH^-

3.00 * 10^-17 = [s] x [2s]²

3.00 x 10^-17 = 4s³

(3.00 x 10^-17 ) / 4 =  s³

7.5 x 10^-18 = s³

 S = (7.5 x 10^-18)^1/3 = 1.96 x 10^-6. Molar solubility of Zn(OH)₂ is 1.96 x 10^-6 M

Answer 2

Zn(OH)₂ ↔ Zn²⁺ + 2OH^- K_SP = 3.00 x 10^-17

Given the salt, what is x, the molar solubilty or the molarity of the salt that ionizes at equilibrium (= molarity of Zn²⁺ put into solution):

K_SP expression is [Zn²⁺][OH^-]²

At equilibrium 3 x 10^-17 = (x)(2x)² = 4x² (note x molar of salt ionized produces x M Zn²⁺ and 2x M OH^-)

Now, solve for x

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