In forming a chelate with a metal ion, a mixture of free EDTA (abbreviated Y^4-)

Question

In forming a chelate with a metal ion, a mixture of free EDTA (abbreviated Y^4-) and metal chelate (abbreviated MY^n - 4) can buffer the free metal ion concentration at values near the dissociation constant of the metal chelate, just as a weak acid and a salt can buffer the hydrogen ion concentration at values near the acid dissociation constant. This equilibrium M^n+ + Y^4- MY^n = 4 is governed by the equation K_f^ = alpha_Y^4- K_f = [MY^n - 4] / [M^n+] [EDTA] where K_f is the association constant of the metal and Y^4-, alphay^4- is the fraction of EDTA in the Y^-4 form, and [EDTA] is the total concentration of free (unbound) EDTA. K_f^i is the "conditional formation constant." How many grams of Na_2EDTA 2H_2O (FM 372.23) should be added to 1.78 g of Mg(NO_3)_2 2H_2O (FM 184.35) in a 500-mL volumetric flask to give a buffer with pMg^2+ = 8.00 at pH 10.00? (log K_f for Mg-EDTA is 8.79 and alpha y^4- at pH 10.00 is 0.30.)

Explanation / Answer

pMg2+ = 8.00 = -log[Mg2+]

[Mg2+] = 1 x 10^-8 M

logKf = 8.79

Kf = 6.166 x 10^8

Kf' = 0.3 x 6.166 x 10^8 = 1.85 x 10^8

concentration of Mg(NO3)2 = 1.78g/184.35 g/mol x 0.5 L = 0.0193 M

let x amount of EDTA be added

Kf' = 1.85 x 10^8 = 0.0193/(1 x 10^-8)(x - 0.0193)

1.85x - 0.036 = 0.0193

x = [EDTA] = 0.030 M

mass Na2EDTA.2H2O = 0.03 x 0.5 x 372.23 = 5.58 g

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