Question

Report the process for utilizing EDTA as a metal ion buffer to prepare a Ga^3+ solution with a buffered free [ Ga^3+] of 5x10^-8 M at pH= 7.0

Answer 1

Given,

[Ga3+]=5*10^-8 M

Ga3+ + EDTA^4-[Ga(EDTA)]^-

Ga3+ and EDTA^4- (Y^4-)react in 1:1 molar ratio.

Formation constant of this complex,Kf

log Kf=21.7

Kf=10^21.7=5.012*10^21=[GaY-]/[Ga3+][Y4-]

But Y^4- fraction available at pH=7 is Y^4-=3.8*10^-4

K'f=[GaY-]/[Ga3+](Y^4-) [Y4-]

Conditional formation constant=K'f=Kf*Y^4-=(5.012*10^21)(3.8*10^-4)=19.045*10^17

K'f=[GaY-]/[Ga3+][Y4-]

At equilibrium, [Ga3+]free=5*10^-8 M

So, [Ga3+]o(initial concentration) -x =5*10^-8 M where x=equilibrium concentration of Ga3+

x=[Ga3+]o-(5*10^-8 M )

[GaY-]=x

[Y4-]=x

19.045*10^17=x/[[Ga3+]o-(5*10^-8 M )]*x

or,[Ga3+]o-(5*10^-8 M )=1/(19.045*10^17)=0.0525*10^-17

[Ga3+]o=0.0525*10^-17+5*10^-8 M

So [Ga3+]o=initial concentration of Ga3+ =must be > 5*10^-8 M to prepare a buffered solution with EDTA with free Ga3+ =5x10^-8 M

EDTA concentration=C(EDTA)*Y4-=[EDTA]eq

C(EDTA)=0.0525*10^-17M/Y4-=0.0525*10^-17M/3.8*10^-4=1.382*10^-15M

So EDTA concentration=1.382*10^-15M