Question

The Ksp of zinc hydroxide, Zn(OH)2 (s) is 7.7 x 10-17 at 25℃. What is its molar solubility(concentration) at this temperature?

Answer 1

The molar solubility (concentration) at this temperature is 4.25 x 10⁻⁶ mol/L

Step-by-step explanation

Given:

Ksp of Zn(OH)₂ = 7.7 x 10⁻¹⁷

What to find:

The molar solubility (concentration) at this temperature.

Step-by-step solution:

Zn(OH)₂ ⇄ Zn²⁺ + 2OH⁻

Initial concentration 1 0 0

Change (1 - x) +x +2x

Equilibrium x x 2x

The Ksp of the solubility of Zn(OH)₂ is

`K_(sp)=[Zn^(2+)][OH^-]=(x)(x)^2`

Putting Ksp = 7.7 x 10⁻¹⁷, we have

`\begin{gathered} 7.7*10⁻¹⁷=(x)(x)^2 \\ \\ 7.7*10⁻¹⁷=x^3 \\ \\ x=\sqrt[3]{7.7*10⁻¹⁷} \\ \\ x=4.25*10^(-6)\text{ }mol\text{/}L \end{gathered}`

Therefore, the molar solubility (concentration) at this temperature is 4.25 x 10⁻⁶ mol/L