Question

Above what Fe2+ concentration will Fe(OH), precipitate from a buffer solution that has a pH of 8.36? The Kop of Fe(OH), is 4.87x10-17 [Fe2+1 =

Answer 1

pH = 8.36 POH = 14.00 -8.36 = 5.64 [OH-] =10-564 = 2.29x106M - - - - - - - - - - - - - - - - - - - - - - - - - - - Solubility equilibrium: Fe(OH), (s) 2 Fe3+ (aq) + 2OH- (aq) Solubility product, K, =[Fe2+ [OH-] = 4.87x10-2* Ionic product of Fe(OH),, e = [Fe2+ ]mkin [OH-] Concentration of OH” = 2.29x10M When the ionic product exceeds the solubility product, precipitation of the salt takes place. 02 Kg [Fe²+ 1.. [OH-], 24.87x10-17 [Fe* Id. (2.29x109)* 2 4.87x10-4* 4.87x10-17 (2.29x106) [Fe** Id. 2 9.29x10M Fe(OH), will start precipitate when [Fe2+ ]ntas > 9.29x106M [Fe2+7 Lotillon