Please answer urgently thanks In forming a chelate with a metal ion, a mixture o
ID: 944866 • Letter: P
Question
Please answer urgently thanks In forming a chelate with a metal ion, a mixture of free EDTA (abbreviated Y4 ) and metal chelate (abbreviated Y^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 K_1 = [MY^n-4]/[M^n+][EDTA] wher K_1 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 1.69 g of Ba(NO_3)_2 (FM 261.35) in a 500-mL volumetric flask to give a buffer with pBa^2+ = 7.00 at pH 10.00? (log K_f, for Ba-EDTA is 7.88 and alpha y^4- at pH 10.00 is 0.30.)Explanation / Answer
Kf' = Kf.alpha[Y4-] = [BaY2-]/[Ba2+][EDTA]
Kf' = 7.58 x 10^7 x 0.30 = 2.27 x 10^7
pBa2+ = -log[Ba2+] = 7
[Ba2+] = 1 x 10^-7 M
Initial [Ba2+] = 1.69/261.35 x 0.5 = 0.013 M
2.27 x 10^7 = 0.013 - x/1 x 10^-7x
2.276x = 0.013 - x
x = 0.004 M
mass of Na2EDTA.2H2O = 0.004 x 0.5 x 372.23 = 0.744 g