Calculate pFe2 at each of the following points in the titration of 25.00 mL of 0

Question

Calculate pFe2 at each of the following points in the titration of 25.00 mL of 0.02245 M Fe2 by 0.03583 M EDTA at a 6.00 pH:

(a) 10.00 mL pFe2+ = 2.22

(b) The equivalence point, Ve pFe2+ = 5.72

(c) 18.50 mL pFe2+ = ??

help someone with the last part for C. iI already got the first two answers but couldn't figure out part C for that question. when i put an answer this is what it told me.

Incorrect. Find the formal concentration of FeY2– by multiplying the dilution factor by the initial concentration of Fe2 . (25.00 ml/25.00 ml +18.50) x (0.02245)

then.... Find the formal concentration of EDTA by multiplying the dilution factor of EDTA with the concentration of EDTA in the solution that is added.

(18.50 mL- 15.66mL)/ (25.00mL + 18.50mL) x (0.03583 M)

Construct an equilibrium table for the formation of the complex. Set the initial concentration of FeY2– equal to the formal concentration of FeY2–, and the initial concentration of EDTA to the formal concentration of EDTA. Equate the equilibrium expression to Kf' and solve for x, the concentration of Fe2 .

Explanation / Answer

The EP was at V = 25 mL x 0.02245 M / 0.03583 M = 15.66 mL

c)

The reactant in excess is EDTA

The volume in excess is 18.50 mL – 15.66 mL = 2.84 mL of 0.03583 M EDTA. It is diluted in the final volume of solution 25.00 mL + 18.50 mL = 43.50 mL.

[EDTA]excess = 0.03583 M x 2.84 mL / 43.50 mL = ….. M

You have also

[FeEDTA] = 0.02245 M x 0.025 L = ….. M   (after EP)

You have to replace these values in the Kf formula to find [Fe2+] and pFe2+.

(you have used the formula at b., but now [EDTA] = [EDTA]excess

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