CA 0.4569 G s C Ch University of Cincinnati CX ry q C WWW sapling learning.com /

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CA 0.4569 G s C Ch University of Cincinnati CX ry q C WWW sapling learning.com /ibiscms/mod/ibis/view.php?id-2366009 Apps For quick access, place your bookmarks here on the bookmarks bar. Import bookmarks now. Attempts score Print Calculator dic Tab 93 Calculate pFee at each of the following points in the titration O 2500 mL of 055 M Fe by 0.03676 Map EDTA at a 7.00 pH: Tutorial a) 10.00 mL Number 2.2 Tutorial 2.36 pFe 2.3 Tutorial 50 (b) The equivalence point, Ve Number 6.44 pFe 00 67 Incorrect. Find the formal concentration of FeY by multiplying the c) 18.00 mL dilution factor by the initial concentration of Fe 00 25.000 mL 0.02055 M 25.00 mL+ 18.00 mL Find the formal concentration of EDTA by multiplying the dilution factor of EDTA with the concentration of EDTA the solution that 18.00 mL 13.98 mL 0.03676 M 25.00 mL 8.00 mL Previous Give Up & View Solution Try Again Next SExit pyright 20 2016 Sapling Le tact 7:14 PM 3/16/2016 Available From 3/9/2016 07:00 PM 3/16/2016 11:55 PM Due Date Points Possible 100 Grade Category: Default Description: Policies: Homework heck y up on any question You can keep trying to answer each question until you get it right or give up 5% ble to ct att Textbook O Help With This Topic OWeb Help & Videos Technical Support and Bug Reports

Explanation / Answer

logKf for Fe2 = 14.32

Kf = 2.1 x 10^14

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

Kf' = 2.1 x 10^14 x 5 x 10^-4 = 1.05 x 10^11

(a) When 10 ml of 0.03676 M EDTA added

moles of Fe2+ = 0.02055 M x 25 ml = 0.514 mmol

moles of EDTA = 0.03676 M x 10 ml = 0.368 mmol

excess [Fe2+] = (0.514 - 0.368)/35 = 4.17 x 10^-3 M

pFe2+ = -log[Fe2+] = 2.38

(b) At equivalence point

Volume of EDTA added = 0.514 mmol/0.03676 M = 13.99 ml

[FeY2-] = 0.514/(25+13.99) = 0.0132 M

Kf' = 1.05 x 10^11 = 0.0132/x^2

x = [Fe2+] = 3.546 x 10^-7 M

pFe2+ = 6.45

(c) When 18 ml of 0.03676 M EDTA added

[FeY2-] = 0.514/(25+18) = 0.012 M

excess [EDTA] = (0.03676 x 18 - 0.514)/(25+18) = 0.0034 M

Kf' = 1.05 x 10^11 = 0.012/[Fe2+](0.0034)

[Fe2+] = 3.36 x 10^-11 M

pFe2+ = 10.47

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