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

Question

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 Y4-. K f=[MY^n-4]/[M^n+][EDTA] where Kf 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 Na2EDTA- 2H20 (FM 372.23) should be added to 1.96 g of Ba(NO3)2 (FM 261.35) in a 500-mL volumetric flask to give a buffer with pBa^2+ = 7.00 at pH 10.00? (log Kf for Ba-EDTA is 7.88 and alphaY^4- at pH 10.00 is 0.30.) mass Na2EDTA. 2H2O = g

Explanation / Answer

How many grams of Na2EDTA.2H2O(Formula mass = 372.23) should be added to
1.9 g of Ba(NO3)2(Formula mass = 261.35) to give a buffer of pBa2+ = 7.0 at pH = 10?
pKf (Ba-EDTA) = -7.88 and alphaY4- = 0.3 at pH = 10.

Let initial concentration of EDTA = x M

Initial concentration:
Volume = 500 ml = 0.5 l
Moles of Ba2+ = 1.9/261.35 mole
[Ba2+]    = 1.9/261.35/0.5 = 0.0145 M
[EDTA]    = x M
[Ba.EDTA] = 0

Complexation Equilibria:
Ba2+ + EDTA = Ba-EDTA

Change:
final Ba2+ concentraion, pBa2+ = 7.0
Ba2+ = 1E-7
change in Ba2+ = final - inital = 1E-7 - 0.0145 = -0.0145(app)

[Ba2+]    = -0.0145
[EDTA]    = -0.0145
[Ba.EDTA] = 0.0145

Complexation equilibria concentration:
[Ba2+]eq    = 1E-7 M
[EDTA]eq    = x-0.0145 M
[Ba.EDTA]eq = 0.0145

Kf' = [Ba-EDTA]eq/[Ba2+]eq*[EDTA]eq
Kf' = Kf*alphaY4- = 10^(7.88)*0.3 = 2.3E7

[Ba-EDTA]/[Ba2+]eq*[EDTA]eq = 2.3E7

[EDTA]eq = [Ba-EDTA]eq/([Ba2+]eq*2.3E7) = 0.0145/([1E-7]*2.3E7) = 0.0063 M
[EDTA]eq = x - 0.0145 = 0.0063
Initial concentration of EDTA = x = 0.0063+0.0145 = 0.0208 M
moles of [EDTA] = molarity*volume = 0.0208*0.5 = 0.0104 mol
Molar mass of Na2EDTA.2H2O = 372.23 g/mol
Mass of Na2EDTA.2H2O = 372.23*0.0104 = 3.88 g

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