# Question 14d2b

Dec 15, 2016

From this site we learn that for zinc hydroxide, ${K}_{s p} = 3 \times {10}^{-} 16$.

#### Explanation:

How did they know this? How else but by measurement? The ${K}_{s p}$ value really should have been included with the problem.

So we write out the solubility equilibrium:

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

So, here, ${K}_{s p} = \left[Z {n}^{2 +}\right] {\left[H {O}^{-}\right]}^{2} = 3 \times {10}^{-} 16$

## Given the equilibrium expression, we call the solubility of $Z n {\left(O H\right)}_{2}$, $S$.

Thus ${K}_{s p} = \left[Z {n}^{2 +}\right] {\left[H {O}^{-}\right]}^{2} = S \times {\left(2 S\right)}^{2} = 3 \times {10}^{-} 16$

So we have a cubic to solve:

$4 {S}^{3} = 3 \times {10}^{-} 16$

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

To get a gram solubility, we simply multiply this by the molecular mass, $4.22 \times {10}^{-} 6 \cdot m o l \cdot {L}^{-} 1 \times 99.42 \cdot g \cdot m o {l}^{-} 1$

$= 0.42 \cdot m g \cdot {L}^{-} 1$, i.e. less than $\text{1 ppm} .$

But we are not finished yet. We were asked to find the $p H$ of a saturated solution.

Clearly, $\left[H {O}^{-}\right] = 8.44 \times {10}^{-} 6 \cdot m o l \cdot {L}^{-} 1$; why clearly?

And thus $p O H = - {\log}_{10} \left(8.44 \times {10}^{-} 6\right) = - \left(- 5.07\right) = 5.07$.

And $p H = 14 - p O H = 8.93$

Does this value of $p H$ make sense, given that we dissolve an hydroxide? Why?