FeT = [Fe3+] + [Fe(OH)]2+ +[Fe(OH)2]++ [Fe(OH)3] +[Fe(OH)4]-
The fraction, Fe3+ present as uncomplexed Fe3+ is :
Fe3+ = [Fe3+] /FeT
(a) FeT = 1.00 mM in water pH = 7.0 (OH-) = 10-7 M
Fe3+ = [Fe3+] /FeT = 1/ (1 + K1 (OH-) + 2 (OH-)2+ 3 (OH-)3 +4 (OH-)4 )
Fe3+ = 5.44 *10-10
so, [Fe3+] = 5.44 *10-10 * 10-3 M = 5.44 *10-13 M
we have following equilibrium,
Fe3+(aq) + 3 OH-(aq) Fe(OH)3(s)
Ksp = [Fe3+] [ OH-]3 = 10-38.8
now for ppt. to form reaction Q should be more than Ksp :
[Fe3+] [ OH-]3 = [5.44 *10-13 ] [ 10-7]3 = 5.44 *10-34
since. Q> Ksp , so Fe(OH)3 will precipitate in this solution. (Yes)
(b) as from calculation above,
[Fe3+] = 5.44 *10-13 M = 5.44 *10-10 mM
(c) Fe3+ = [Fe3+] /FeT
Fe3+ = 5.44 *10-10 mM / 1.00 mM = 5.44 *10-10
Problem 4 (30 pts) Ignore activity corrections. Fe3+ + OH = Fe(OH)2+ Fe3+ + 2OH =...