Ksp of Fe(OH)2 = 4.87*10-17
now, for Fe(OH)2
Fe(OH)2 Fe2+ (aq) + 2 OH- (aq)
Ksp = [Fe2+] [OH-]2 (1)
given , pH = 8.16
then , pOH = 14 - 8.16 = 5.84
now,
[OH-] = 10-5.84 = 1.45 *10-6 M
putting in Eq . 1
4.87*10-17 = [Fe2+] * ( 1.45 *10-6)2
or, 4.87*10-17 = [Fe2+] * 2.089*10-12
or, [Fe2+] = (4.87*10-17/ 2.089*10-12)
or, [Fe2+] = 2.33*10-5 M
the Fe(OH)2 will precipitate when Ksp < [Fe2+] [OH-]2 .
therefore above 2.33*10-5 M Fe2+ concentration Fe(OH)2 will precipitate.
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