The solubility products of #Fe(OH)_2# and #Fe(OH)_3# are #10^(-17)# and #10^(-38)# respectively. If the concentrations of #Fe^(2+)# and #Fe^(3+)# are each #10^(-5)#, what is the #[OH^-]# when each just begin to precipitate?

1 Answer
Andrea B.
Mar 19, 2018

#[OH^-]=1 xx 10^-6#
and#[OH^-]=1 xx 10^-11#

Explanation:

for the reactions
#Fe(OH)_2 = Fe^(2+) + 2 OH^-# and
#Fe(OH)_3 = Fe^(3+) + 3 OH^-# we have the espressions of Kps that are
#Kps = [Fe^(2+)] xx [OH^-]^2 = 1 xx 10^-17# and
#Kps = [Fe^(3+)] xx [OH^-]^3 = 1 xx 10^-38#
from wich you have
#[OH^-]= sqrt ((Kps)/[Fe^(2+)])= sqrt ((1 xx 10^-17)/(1 xx 10^-5))=1 xx 10^-6 #
pOH = 6; pH = 8
and
#[OH^-]= root(3)((Kps)/[Fe^(3+)])= root(3) ((1 xx 10^-38)/(1 xx 10^-5))=1 xx 10^-11#
pOH = 11; pH = 3

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