What is the solubility of zinc hydroxide in aqueous solution if #pH=9#?

1 Answer
Sep 14, 2015

Answer:

You find the #K_"sp"# expression, #K_"sp"=[Zn^(2+)][HO^-]^2#, i.e. #K_"sp"=3xx10^-16#...

Explanation:

Here we know the #[OH^-]# value (how; what is the #pH#?). Therefore you have the tools to solve for #[Zn^(2+)]#, whose concentration represents the solubility of zinc hydroxide. Remember that hydroxide ion concentration is squared in the #K_(sp)# expression.

As an extension, it may be speculated that we could reduce the #[Zn^(2+)]# value to any degree, simply by increasing #[OH^-]#. Note that at very high levels of #[OH^-]#, we are likely to form complex ions such as, #Zn(OH)_4^(2-)# (I assure you this species is real). This is a competing equilibrium that can occur.

And so, we address the equilibrium reaction...

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

For which #K_"sp"=[Zn^(2+)][HO^-]^2#...

And if we set #S="solubility of zinc hydroxide"#...IN PURE WATER...

Then...#K_"sp"=Sxx(2S)^2=4S^3=3xx10^-16#

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

And so under the given conditions, we can find the mass of zinc hydroxide in solution... #99.42*g*mol^-1xx4.22xx10^-6*mol*L^-1=0.42*mg*L^-1#, i.e. less than #"1 ppm"#...

But in the given problem, #[HO^-]=10^(-5)*mol*L^-1#...(why?..because #pH=9#, and #K_w=10^-14#)...and so...

#K_"sp"=Sxx(2S+2xx10^-5)^2# And so #S~=K_"sp"/(2xx10^-5)^2=7.5xx10^-7*mol*l^-1#...and so the solubility of the salt is reduced tenfold...