# What is the dissociation of "Zn"("OH")_2?

Jun 27, 2017

${\text{Zn"("OH")_ (2(s)) rightleftharpoons "Zn"_ ((aq))^(2+) + 2"OH}}_{\left(a q\right)}^{-}$

#### Explanation:

Zinc hydroxide is considered insoluble in water, so right from the start, you know that you're dealing with a solubility equilibrium between the undissociated solid and the solvated ions.

Now, each mole of zinc hydroxide contains

• one mole of zinc cations, $1 \times {\text{Zn}}^{2 +}$
• two moles of hydroxide anions, $2 \times {\text{OH}}^{-}$

This means that when you dissolve solid zinc hydroxide in water, the following equilibrium will be established

${\text{Zn"("OH")_ (2(s)) rightleftharpoons "Zn"_ ((aq))^(2+) + 2"OH}}_{\left(a q\right)}^{-}$

This equilibrium will lie to the left, meaning that only very small quantities of solid zinc hydroxide will dissociate in aqueous solution to produce zinc cations and hydroxide anions.

The solubility product constant, ${K}_{s p}$, which is essentially a measure of the degree of dissociation of zinc hydroxide in water, is equal to $3.0 \cdot {10}^{- 16}$ at room temperature.

https://en.wikipedia.org/wiki/Zinc_hydroxide

This basically tells you that the electrostatic attraction that exists between the zinc cations and the hydroxide anions far exceeds the attraction between the polar water molecules and the two ions $\to$ zinc hydroxide is considered insoluble in aqueous solution.