Calculate pFe^2+ at each of the following points in the titration of 25.00 mL of

Question

Calculate pFe^2+ at each of the following points in the titration of 25.00 mL of 0.02015 M Fe^2+ by 0.03650 EDTA at a 7.00 pH: 12.50 mL pFe^2+ = The equivalence point, V_0 pFe^2+ = 18.00 mL pFe^2+ = Incorrect. First, find the conditional formation constant of the Fe-EDTA complex at a 7.00 pH. Confirm in your text that for a 7.00 pH, and Calculate the conditional formation constant by multiplying these two values. Find the formal concentration of FeY^2- by multiplying the dilution factor at v_0 = 13.80 mL by the initial concentration of Fe^2+ Construct an equilibrium table for the reaction The initial concentration of FeY^2- equal to the format concentration of FeY^2-. Set the equilibrium expression equal to K_f'. Solve the

Explanation / Answer

logKf = 14.32

Kf = 2.09 x 10^14

alpha[Y4-] for EDTA at pH 7 = 5 x 10^-4

Kf' = Kf.alpha[Y4-] = 1.045 x 10^11

(a) 12.50 ml EDTA added

initial moles of Fe2+ = 0.02015 M x 25 ml = 0.504 mmol

moles of EDTA added = 0.03650 M x 12.50 ml = 0.456 mmol

remaining [Fe2+] after complexation = (0.504 - 0.456) mmol/37.5 ml = 0.0013 M

pFe2+ = -log[Fe2+] = -log(0.0013) = 2.893

(b) At equivalence point

Volume of EDTA added = 0.504 mmol/0.03650 M = 13.81 ml

[FeY2-] formed = 0.504 mmol/38.81 ml = 0.013 M

Fe2+ + EDTA <==> FeY2-

let x amount has dissolved

Kf' = [FeY2-]/[Fe2+][EDTA]

1.045 x 10^11 = 0.013/x^2

x = [Fe2+] = 3.53 x 10^-7 M

pFe2+ = -log(3.53 x 10^-7) = 6.45

(c) 18 ml EDTA added

Excess EDTA = 0.03650 M x (18 - 13.81)ml/43 ml = 0.0035 M

[FeY2-] formed = 0.504 mmol/43 ml = 0.012 M

Kf' = [FeY2-]/[Fe2+][EDTA]

1.045 x 10^11 = 0.012/[Fe2+](0.0035)

[Fe2+] = 3.281 x 10^-11 M

pFe2+ = -log(3.281 x 10^-11) = 10.48

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