In forming a chelate with a metal ion, a mixture of free EDTA (abbreviated Y^4-)
ID: 903024 • Letter: I
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 add 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-, alpha Y^4- is the fraction of EDTA in the Y^4- form, and [EDTA] is the total concentration of free (unbound) EDTA. K_f' is the "conditional formation constant." How many grams of Na_2 EDTA 2H_2 O (FM 372.23) should be added to 1.96 g of Mg(NO_3)_2 2H_2 O (FM 184.35) in a 500-mL volumetric flask to give a buffer with pMg^2+ = 8.00 at pH 10.007 (log K_f for Mg-EDTA is 8.79 and alpha Y^4- at pH 10.00 is 0.30.)Explanation / Answer
Solution:
[MgY2+]=1.96g/[(184.35g/mol)(0.5L)=0.01 M
alphaY4^-1=K1K2K3K4K5K6/D =3.7394393E-02
Kf = Kf X (alphaY4^-1)
given log kf of Mg(EDTA) =8.79
Kf of Mg(EDTA) =10^8.79
Kf'=(3.7394393E-02)(10^8.79)=(0.019288 M)/(X*(10^-8.96) ---> X=0.010532 M
(0.010532 M*0.5 L) + (0.019288 M*0.5L)=0.015916 mole
0.015916 mole* 372.23g/mol=5.55 g