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 MgY^2- 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) reaction will be

Mg+2 + EDTA --> MgY^-2

We know that Mg+2 will react with EDTa in 1:1 molar ratio

Moles of Mg+2 present = Molarity X volume = 0.08 X 12 / 1000 = 0.00096 Moles

Moles of EDTA added = Molarity X volume = 0.052 X 5 / 1000 = 0.00026

So moles of Mg+2 reacted = 0.00026

Moles of Mg+2 left = 0.00096- 0.00026 = 0.0007

Concentration = Moles / Total volume = 0.0007 X 1000 / 12 + 5 = 0.0411

So pMg = -log [Mg+2] = 1.386

2) The alpha 4 value for EDTA is 0.35 at pH 10.0, and the formation constant for MgY^2- is 4.9*10⁸

Kf' = 0.35 X 4.9X 10^8

At equivalent point [Mg+2] will react completely with [EDTA] to give 0.00096 Moles of MgY^2-

So , For 0.00096 moles of EDTA the volume required = Moles / Molarity = 0.00096 / 0.052 = 0.0184 L = 18.4mL

Total volume = 18.4 + 12 = 30.4 mL

concentration of MgY^2- = Moles / Volume = 0.00096 X 1000 / 30.4 = 0.0315 M

         Mg⁺² + EDTA --> MgY^-2

Initila 0           0               0.0315

equilib x           x              0.0315- x

Kf' =1.715 X 10^8 = [MgY^-2] / [Mg+2] [ EDTA] = 0.0315-x / x^2

1.715 X 10^8x^2 - 0.0315 +x = 0

x =1.08 X 10^-6

pMg = 5.96

3) Given alpha 4 = 0.35

On addition of 25 mL of EDTA the moles of EDTA added = Molarity X volume = 25 X 0.052 / 1000 = 0.0013 moles

Moles of Mg+2 reacted = 0.00096

Moles of MgY-2 formed = 0.00096

so concentration = Moles of MgY-2 / Total volume = 0.00096 X 1000 / 25+12 = 0.0259

Moles of EDTA left = 0.0013 - 0.00096 = 0.00034

concentration of EDTA = 0.00064 X 1000 / 37 = 0.0173

            Mg⁺² + EDTA --> MgY^-2

Initila    0           0.0173               0.0259

equilib x         0.0173+ x             0.0259-x

Kf' = 1.715 X 10^8 = [MgY^-2] / [Mg+2] [ EDTA] = 0.0259-x / (0.0173+x )(x)

1.715 X 10^8 = 0.0259 -x / 0.0173x + x^2

1.715 X 10^8x^2 + 0.0173 X 1.715 X 10^8 x = 0.0259 -x

on calculating

pMg = Nearly 11