# How does zinc form an insoluble zinc hydroxide in water?

Jul 10, 2017

It does so in moderately basic solution, modeled by the reaction seen here. Note that in overly basic solution, "Zn"("OH")_2(s) will dissolve and "Zn"("OH")_4^(2-)(aq) will form instead.

Consider the following Pourbaix Diagram:

This graphs the EMF of the redox process against the $\text{pH}$ of the solution.

The higher up you go in the diagram, the more oxidized the species are, and the farther right you go, the more base-ic the species become.

You can see "Zn"("OH")_2(s) forms in $\text{pH}$ $9 - 11$ solutions. In acidic conditions, the redox half-reactions for this process would be:

${\text{Zn"(s) -> "Zn}}^{2 +} \left(a q\right) + \cancel{2 {e}^{-}}$
$2 {\text{H"^(+)(aq) + cancel(2e^(-)) -> "H}}_{2} \left(g\right)$
$\text{--------------------------------------------------}$
${\text{Zn"(s) + 2"H"^(+)(aq) -> "Zn"^(2+)(aq) + "H}}_{2} \left(g\right)$

In basic conditions, we just have to add $2 \times {\text{OH}}^{-}$ to both sides to represent such conditions by neutralizing all the strong acid.

${\text{Zn"(s) + 2"H"^(+)(aq) + 2"OH"^(-)(aq) -> "Zn"^(2+)(aq) + 2"OH"^(-)(aq) + "H}}_{2} \left(g\right)$

This gives:

$\textcolor{b l u e}{{\text{Zn"(s) + 2"H"_2"O"(l) -> "Zn"("OH")_2(s) + "H}}_{2} \left(g\right)}$