# What is the solubility of zinc hydroxide in aqueous solution if pH=9?

##### 1 Answer
Sep 14, 2015

You find the ${K}_{\text{sp}}$ expression, ${K}_{\text{sp}} = \left[Z {n}^{2 +}\right] {\left[H {O}^{-}\right]}^{2}$, i.e. ${K}_{\text{sp}} = 3 \times {10}^{-} 16$...

#### Explanation:

Here we know the $\left[O {H}^{-}\right]$ value (how; what is the $p H$?). Therefore you have the tools to solve for $\left[Z {n}^{2 +}\right]$, whose concentration represents the solubility of zinc hydroxide. Remember that hydroxide ion concentration is squared in the ${K}_{s p}$ expression.

As an extension, it may be speculated that we could reduce the $\left[Z {n}^{2 +}\right]$ value to any degree, simply by increasing $\left[O {H}^{-}\right]$. Note that at very high levels of $\left[O {H}^{-}\right]$, we are likely to form complex ions such as, $Z n {\left(O H\right)}_{4}^{2 -}$ (I assure you this species is real). This is a competing equilibrium that can occur.

And so, we address the equilibrium reaction...

$Z n {\left(O H\right)}_{2} \left(s\right) r i g h t \le f t h a r p \infty n s Z {n}^{2 +} + 2 H {O}^{-}$...

For which ${K}_{\text{sp}} = \left[Z {n}^{2 +}\right] {\left[H {O}^{-}\right]}^{2}$...

And if we set $S = \text{solubility of zinc hydroxide}$...IN PURE WATER...

Then...${K}_{\text{sp}} = S \times {\left(2 S\right)}^{2} = 4 {S}^{3} = 3 \times {10}^{-} 16$

And so finally, S=""^(3)sqrt((3xx10^-16)/4)=4.22xx10^-6*mol*L^-1...

And so under the given conditions, we can find the mass of zinc hydroxide in solution... $99.42 \cdot g \cdot m o {l}^{-} 1 \times 4.22 \times {10}^{-} 6 \cdot m o l \cdot {L}^{-} 1 = 0.42 \cdot m g \cdot {L}^{-} 1$, i.e. less than $\text{1 ppm}$...

But in the given problem, $\left[H {O}^{-}\right] = {10}^{- 5} \cdot m o l \cdot {L}^{-} 1$...(why?..because $p H = 9$, and ${K}_{w} = {10}^{-} 14$)...and so...

${K}_{\text{sp}} = S \times {\left(2 S + 2 \times {10}^{-} 5\right)}^{2}$ And so $S \cong {K}_{\text{sp}} / {\left(2 \times {10}^{-} 5\right)}^{2} = 7.5 \times {10}^{-} 7 \cdot m o l \cdot {l}^{-} 1$...and so the solubility of the salt is reduced tenfold...