Ferric chloride (FeCl_3) is added to pure water at a dose of 10 mg/L. Ferric chl

Question

Ferric chloride (FeCl_3) is added to pure water at a dose of 10 mg/L. Ferric chloride completely, but the following hydrolysis form with Fe^3+ The system is closed, and you can assume that no solids precipitate. Remember that the activity of water in water is 1. Fe^3+ + H _2 O = FeOH^2 + H^+ logK = -2.19 Fe^3+ +2 H _2 O = Fe(OH) _2 + 2H^+ logK = -5.67 Fe^3+ + 3H _2 O = Fe(OH) _3(qa) + 3H^+ logK = -13.6 Fe^3+ + 4H _2 O = Fe(OH) _4 + 4H^+ logK = -21.6 Write equations for all of the species in the system as functions of [H^+] Prepare a Log C-pH diagram for the system over the pH range 1-13. Determine the pH of the solution to within 0.2 pH units.

Explanation / Answer

[Fe+3] = 10mg / L = 10 X 10^-3 g / L = 10 X 10^-3 / 162.2 = 6.17 X 10-5

1) Fe+3 + H2O = FeOH2+ + H+                            logK1 = -2.19 K1 = 0.0065

K1 = [ FeOH2+] [H+ ] / [Fe+3]

0.0065 X 6.17 X 10-5 = [H+]^2

[H+] = 0.633 X 10^-3

pH = -log [H+] = 3.19

2) Fe+3 + 2H2O = Fe(OH)2+ + 2H+                      logK2 = -5.67   K2 = 2.14 X 10^-6

K2 = [Fe(OH)2+ ] [H+ ]2 / [Fe+3]

[H+] = 2 X [Fe(OH)2+]

2.14 X 10-6 X 6.17 X 10-5 = [H+]^2 [0.5 H+] = 0.5 [H+]3

[H+] = 6.415 X 10^-4

In other two equilibrium the H+ concentration will be very low due to less equilibrium constant value

3) Fe+3 + 3H2O = Fe(OH)3 + 3H+                        logK3 = -13.6   K3 = 2.511 X 10^-14

K3 = [Fe(OH)3][H+]3 / [Fe+3]

4) Fe+3 + 4H2O = Fe(OH)4- + 4H+                       logK4 = -21.6    K4 = 2.51 X 10^-22

K4 = [Fe(OH)4-][H+]4 / [Fe+3]

pH will be due to H+ produced in the first two equilibrium

[H+] = 6.415 X 10^-4 + 0.633 X 10^-3 = 2.89

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