Calculate the solubility of Zn(OH)2 in water at 25°C.

The solubility of Zn(OH)₂ at 25ºC is 1.96x 10⁻⁶,

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.

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