Answer:
The solubility of Zn(OH)₂ at 25ºC is 1.96x 10⁻⁶,
Step-by-step explanation:
The solubility is the maximum amount of a solute that can be dissolved in a certain volume of liquid (usually water) at a given temperature.
First, we write the dissolution reaction of Zn(OH)₂ in water, which is:
Zn(OH)₂ (s) ⇄ Zn²⁺ + 2 OH⁻
We know that the Kps, that is, the dissolution equilibrium constant for this compound at 25ºC is 3.0x 10⁻¹⁷.
Therefore, the Kps is:
Kps= [Zn²⁺][OH⁻]² (since Zn(OH)₂ is solid, it is not accounted for).
Given the reaction, we can see that for every single Zn²⁺ ion, two OH⁻ ions are produced. Therefore, if the concentration of Zn²⁺ dissolved is x M, the concentration of OH⁻ dissolved will be 2x M.
Substituting these values in Kps, we have:
[Zn²⁺][OH⁻]² = (x)(2x)² = 3.0x 10⁻¹⁷
4x³ = 3.0x 10⁻¹⁷
x= ∛ (3.0x 10⁻¹⁷ ÷ 4)
x= 1.96x 10⁻⁶
Therefore, the molar solubility of Zn(OH)₂ at 25ºC will be 1.96x 10⁻⁶.