What is the molar solubility of #Cd(OH)_2# given #K_"sp"=5.3xx10^-15#?

1 Answer
Mar 20, 2017

Answer:

#[Cd(OH)_2]=1.10xx10^-5*mol*L^-1#

Explanation:

For the reaction:

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

We write #K_"sp"=[Cd^(2+)][HO^-]^2#

And if we call the solubility of #Cd(OH)_2#, #S#, then.........

#K_"sp"=Sxx(2S)^2=4S^3#

And so #S=""^3sqrt(K_"sp"/4)=""^3sqrt((5.3xx10^-15)/4)=1.10xx10^-5*mol*L^-1#, with respect to #Cd(OH)_2#..........

If the #pH# of the solution were buffered to HIGH or LOW #pH# how would the solubility evolve?