calculate the molar solubility of fe(OH)2 at a ph 8 ksp of fe(oh)2=1.6X1o 2 d power -14
We Know,
p[OH] + p[H] = 14
Given : pH = 8
pOH = 14 - 8= 6
[OH-] = 1.0 x 10-6 M
Reaction involved :
Fe(OH)2 → Fe2+ + 2OH-
Consider ‘s‘ be the concentration of Fe(OH)2 of dissolved form and result into the formation of ‘s’ M Fe2+ & 2’s’ M OH-
[Fe2+]= s
[OH-] = 2s + 1.0 x 10-6 M
=> Ksp = 1.6 x 10-14 = [Fe2+][OH-]2
=> (s)(2s + 1.0 x 10-6)2 = 1.6 x 10-14
=> ‘s’= 0.016 M
p[OH] + p[H] = 14
Given : pH = 8
pOH = 14 - 8= 6
[OH-] = 1.0 x 10-6 M
Reaction involved :
Fe(OH)2 → Fe2+ + 2OH-
Consider ‘s‘ be the concentration of Fe(OH)2 of dissolved form and result into the formation of ‘s’ M Fe2+ & 2’s’ M OH-
[Fe2+]= s
[OH-] = 2s + 1.0 x 10-6 M
=> Ksp = 1.6 x 10-14 = [Fe2+][OH-]2
=> (s)(2s + 1.0 x 10-6)2 = 1.6 x 10-14
=> ‘s’= 0.016 M