As in the other parts the problem stays the same in each example. It is:
Calculate the molar solubility (in mol/L) of a saturated solution of the substance.
I'm going to assume you've gone through the other files first, so the presentations here will be a bit abbreviated. I'm hoping that you have begun to see a pattern developing.
If not, the pattern is that the coefficient winds up doing two things:
1) it puts a power on the 'x' which represents that particular ion and
2) it puts a coefficient in front of the 'x'
The two example substances are:
Fe(OH)3 Ksp = 2.64 x 10¯39 Ag3PO4 Ksp = 8.88 x 10¯17
Iron(III) hydroxide
Fe(OH)3 <===> Fe3+ + 3 OH¯Ksp = [Fe3+] [OH¯]3
2.64 x 10¯39 = (x) (3x)3
27x4 = 2.64 x 10¯39
x = 9.94 x 10¯11 M
How's that for abbreviated? The 3x comes from the fact that there are three hydroxides produced for every Fe3+ ion, but you knew that, didn't you?
Silver phosphate
Ag3PO4 <===> 3 Ag+ + PO43¯Ksp = [Ag+]3 [PO43¯]
8.88 x 10¯17 = (3x)3 (x)
27x4 = 8.88 x 10¯17
x = 4.26 x 10¯5 M
Here are some more problems of this type.
Go to Solving Ksp Problems - Part One
Go to Solving Ksp Problems - Part Two
Go to Solving Ksp Problems - Part Four