Above what fe2 concentration will fe(oh)2 precipitate from a buffer solution that has a ph of 8.59? the ksp of fe(oh)2 is 4.87×10-17.

First, we need to get POH value & we can get it from this formula:

PH + POH = 14 

when we have PH = 8.59 

∴ POH = 14 - 8.59 = 5.41

then, we need to get [OH-] & we can get it using POH:

POH = -㏒[OH]

5.41  = -㏒[OH]

∴ [OH] = 3.89 x 10^-6

when Ksp = [Fe2-][OH-]^2

and when we have Ksp = 4.87 x 10^-17

so, by substitution:
 
 4.87 x 10^-17 = [Fe2-] * (3.89 x 10 ^-6)^2

∴[Fe2-] = 3.22 x 10^-6 M

∴ above 3.22 x 10^-6 M of Fe2+ , Fe(OH)2 will precipitate

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