Question #14d2b

1 Answer
Dec 15, 2016

Answer:

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?