Consider the compounds shown below along with their solubility product constants: Co(OH)2 Ksp= 1.60e-15 CdCO3 Ksp= 5.20e-12 Cd(OH)2 Ksp= 7.20e-15 ?

In order to compare relative solubility and put these compounds in rank order from least to most soluble, you first need to calculate the molar solubility of each. The molar solubility of Co(OH)2 is _ moles/L. The molar solubility of CdCO3 is __ moles/L. The molar solubility of Cd(OH)2 is _ moles/L. Based on these, the correct order of solubility, from least to most is:
(a) Cd(OH)2,CdCO3,Co(OH)2
(b) CdCO3,Co(OH)2,Cd(OH)2
(c) Cd(OH)2,Co(OH)2,CdCO3

1 Answer
Mar 30, 2018

A lot of work for a multiple choice question....

Explanation:

We gots THREE solubility equilibria to consider...

#Co(OH)_2(s) rightleftharpoonsCo^(2+) + 2HO^-#

#K_"sp"=[Co^(2+)][HO^-]^2=1.60xx10^-15#...

And if we put #"Solubility of cobalt(II) hydroxide"=S#, then...

#1.60xx10^-15=Sxx(2S)^2=4S^3#

#S=""^(3)sqrt{(1.60xx10^-15)/4}=7.4xx10^-6*mol*L^-1#...

#7.4xx10^-6*mol*L^-1xx92.95*g*mol^-1=6.88xx10^-4*g*L^-1<1*"ppm"#

And...

#CdCO_3(s) rightleftharpoonsCd^(2+) + CO_3^(2-)#

#K_"sp"=[Cd^(2+)][CO_3]=5.20xx10^-12#...

And if we put #"Solubility of cadmium(II) carbonate"=S#, then...

#5.20xx10^-12=SxxS=S^2#

#S=sqrt{5.20xx10^-12}=2.28xx10^-6*mol*L^-1#

#-=2.28xx10^-6*mol*L^-1xx172.42*g*mol^-1-=3.93xx10^-4*g*L^-1#

And...

#Cd(OH)_2(s) rightleftharpoonsCd^(2+) + 2HO^(-)#

#K_"sp"=[Cd^(2+)][HO^-]^2=5.20xx10^-12#...

And if we put #"Solubility of cadmium(II) hydroxide"=S#, then...

#5.20xx10^-12=Sxx2S=2S^3#

#S=""^(3)sqrt((5.20xx10^-12)/2)=1.38xx10^-4*mol*L^-1#

#-=1.38xx10^-4*mol*L^-1xx146.43*g*mol^-1-=0.0201*g*L^-1#

And so in terms of solubility #g*L^-1#...we write the order...

#"cadmium carbonate"<"cobalt hydroxide"<"cadmium hydroxide"#...which is which option....I've forgotten them...?