Question

Problem 4 (30 pts) Ignore activity corrections. Fe3+ + OH = Fe(OH)2+ Fe3+ + 2OH = Fe(OH)2+ Fe3+ + 3OH = Fe(OH)3(aq) Fe3+ + 40H = Fe(OH)4 Fe(OH)3°(s) = Fe3+ + 3OH log K10H = log B20H = log B3OH = log B4OH = log Ksp = 11.81 22.4 30.2 34.4 -38.8 a. (10 pts) Will Fe(OH)3'(s) precipitate in a solution prepared in water and with Fer = 1.00 mM and buffered to pH 7.0? Hint: There are several ways to do this, but one way involves calculating an "a Fe3+", or the fraction of Fer that is Fe3+. This alpha is analogous to, but not the same as, the alphas for weak acids/bases. Yes or No b. (4 pts) What is the concentration of Fe3+ in a solution prepared in water and with Fer = 1.00 mM and buffered to pH 7.0? mm C. (2 pts) Calculate QFe3+ from your answer in b.

Answer 1

Fe_T = [Fe³⁺] + [Fe(OH)]²⁺ +[Fe(OH)₂]⁺+ [Fe(OH)₃] +[Fe(OH)₄]^-

The  fraction, $\alpha$_Fe³+ present as uncomplexed Fe³⁺ is :

$\alpha$_Fe³+ = [Fe³⁺] /Fe_T

(a) Fe_T  = 1.00 mM in water pH = 7.0 (OH-) = 10^-7 M

$\alpha$_Fe³+ = [Fe³⁺] /Fe_T = 1/ (1 + K1 (OH-) + $\beta$2 (OH-)²+ $\beta$3 (OH-)³ +$\beta$4 (OH-)⁴ )

$\alpha$_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) $\rightleftharpoons$ 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) $\alpha$_Fe³+ = [Fe³⁺] /Fe_T

$\alpha$_Fe³+ = 5.44 *10^-10  mM / 1.00 mM = 5.44 *10^-10