Question

Above what Fe2+ concentration will Fe(OH), precipitate from a buffer solution that has a pH of 9.56? ut The Ksp of Fe(OH), is 4.87x10-17 ЬВ Fe2]= ЬВ Ьвr

Answer 1

The expression of solubility product constant can be obtained as : Fe(OH)2 D Fe 2+ + 2OH" Ksp = [Fe2+] [OH^"]2 Rearrange the above expression to calculate [Fe] as Nowe qö pH + DOH = 14 on, DOH = 14 - PH => pOH = 14 – 9.56 7 pOH = 4.44 Ako, LOH = 10 por 10-4,44 = 3.63 x 10-5 M

Substituting the above value of [on") and the given values in equation (1) , we get 2 + 1 = 4.87 x 10 lt (3.63 x 10- 52 3.69 x 10-8 M