Question

1) For a complexation titration of 12.00 mL of 0.0800 M Mg2+ solution with 0.0520 M EDTA at pH 10.0, please determine the pMg after 5.00 mL of EDTA solution is added. The alpha 4 value for EDTA is 0.35 at pH 10.0, and the formation constant for MgY2- is 4.9*108.

2) For the same titration as described in the last question, what's the pMg at the equivalence point?

3) For the same titration as described above, what is the pMg after 25.00 mL of EDTA solution is added?

Answer 1

1) moles of Mg2+ = 0.080 M x 12 ml = 0.96 mmol

moles of EDTA added = 0.0520 M x 5 ml = 0.26 mmol

[Mg2+] remaining = 0.96 - 0.26/17 = 0.0412 M

pMg = log(0.0412) = 1.38

2) At equivalence point

moles of Mg2+ present = moles of EDTA added

Volume of EDTA added = 0.08 M x 12 ml/0.052 M = 18.46 ml

[MgY2-] = 0.08 x 12/30.46 ml = 0.0315 M

Mg2+ + EDTA <==> MgY2-

let x amount of MgY2- has dissolved

Kf' = Kf.alpha[Y4-] = 4.9 x 10^8 x 0.35 = 0.0315/x^2

x = [Mg2+] = 1.35 x 10^-5 M

pMg = 4.87

3) After 25 ml EDTA added

moles of Mg2+ = 0.080 M x 12 ml = 0.96 mmol

moles of EDTA added = 0.0520 M x 25 ml = 1.3 mmol

[EDTA] remaining = 0.34 mmol/37 ml = 0.0092 M

[MgY2-] = 0.96/37 = 0.026 M

Kf' = Kf.alpha[Y4-] = 4.9 x 10^8 x 0.35 = 0.026/[Mg2+](0.0092)

x = [Mg2+] = 1.65 x 10^-8 M

pMg = 7.78