Question

Question #14d2b

Answer 1

From this site we learn that for zinc hydroxide, `K_(sp)=3xx10^-16`.

Explanation:

How did they know this? How else but by measurement? The `K_(sp)` value really should have been included with the problem.

So we write out the solubility equilibrium:

`Zn(OH)_2(s) rightleftharpoonsZn^(2+)+2HO^-`

So, here, `K_(sp)=[Zn^(2+)][HO^-]^2=3xx10^-16`

Given the equilibrium expression, we call the solubility of `Zn(OH)_2`, `S`.

Thus `K_(sp)=[Zn^(2+)][HO^-]^2=Sxx(2S)^2=3xx10^-16`

So we have a cubic to solve:

`4S^3=3xx10^-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, `4.22xx10^-6*mol*L^-1xx99.42*g*mol^-1`

`=0.42*mg*L^-1`, i.e. less than `"1 ppm".`

But we are not finished yet. We were asked to find the `pH` of a saturated solution.

Clearly, `[HO^-]=8.44xx10^-6*mol*L^-1`; why clearly?

And thus `pOH=-log_10(8.44xx10^-6)=-(-5.07)=5.07`.

And `pH=14-pOH=8.93`

Does this value of `pH` make sense, given that we dissolve an hydroxide? Why?