What is the molar solubility of Fe(OH)2 when buffered at pH of 9.00? [Fe(OH)2 : Ksp = 7.9 x 10^-16]
Fe(OH)2 ---------> Fe2+ + 2OH-
pH = 9
pOH = 14- pH = 14 - 9
pOH = 5
[OH] = 10-pOH = 10-5
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Ksp = [Fe2+].[OH-]2
[Fe2+] = (Ksp) / ([OH-]2)
Therefore, Molar solubility is given by:
[Fe2+] = (7.9 x 10-16) / (10-5)2
[Fe2+] = 7.9 x 10-6 M -------------------(ANSWER)
What is the molar solubility of Fe(OH)2 when buffered at pH of 9.00? [Fe(OH)2 : Ksp...
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