Answer
The molar solubility (concentration) at this temperature is 4.25 x 10⁻⁶ mol/L
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
[tex]K_{sp}=[Zn^{2+}][OH^-]=(x)(x)^2[/tex]Putting Ksp = 7.7 x 10⁻¹⁷, we have
[tex]\begin{gathered} 7.7\times10⁻¹⁷=(x)(x)^2 \\ \\ 7.7\times10⁻¹⁷=x^3 \\ \\ x=\sqrt[3]{7.7\times10⁻¹⁷} \\ \\ x=4.25\times10^{-6}\text{ }mol\text{/}L \end{gathered}[/tex]Therefore, the molar solubility (concentration) at this temperature is 4.25 x 10⁻⁶ mol/L
