pOH = 14 - pH
pOH = 14 - 8.42 = 5.58
pOH = -log[OH-]
- log[OH-] = 5.58
[OH-] = 2.63× 10-6M
solubility equilibrium of Fe(OH)2 is
Fe(OH)2(s) <---------> Fe2+(aq) + 2OH-(aq)
Ksp = [Fe2+][OH-]2
[Fe2+][OH-]2 = 4.87 ×10-17M3
[Fe2+] × (2.63 ×10-6M)2 = 4.87 ×10-17M3
[Fe2+] = 7.04 × 10-6M
Therefore,
concentration of Fe2+ above which Fe(OH)2 will precipitate = 7.04×10-6M
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