Question

In forming a chelate with a metal ion, a mixture of free EDTA (abbreviated Y4–) and metal chelate (abbreviated MYn–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 is governed by the equation where Kf is the association constant of the metal and Y4–, ?Y4– is the fraction of EDTA in the Y4– form, and [EDTA] is the total concentration of free (unbound) EDTA. K \'f is the \"conditional formation constant.\" How many grams of Na2EDTA·2H2O (FM 372.23) should be added to 1.72 g of Ba(NO3)2 (FM 261.35) in a 500-mL volumetric flask to give a buffer with pBa2 = 7.00 at pH 10.00? (log Kf for Ba-EDTA is 7.88 and ?Y4– at pH 10.00 is 0.30.)

Answer 1

K'_f = alpha_y + a . K_f = [MY^n - 4]/[M^n+][EDTA] For Ba^2+ alpha_y^4+ . K_f = [Ba Y^2-]/[Ba^2+][EDTA] Therefore log alpha_y^4+ + log k_f = log[Ba y^2-] - log [Ba^2+] - log [EDTA] Therefore log(0.30) + 7.88 = log [Ba Y^2-] + 7.00 - log [EDTA] - 0.523 + 7.88 - 7.00 = log [Ba Y^2-] - log[EDTA] as log[Ba Y^2+] - log[EDTA] = 0.348 Now - log[Ba^2+] = p^Ba^2+ = 7.0 - [Ba^2+] = 10^-7 M Now initial as total Ba^2+ come. = (1.72/261.35)/(500/1000) = 0.012 M Therefore [Ba Y^2+] = 0.012 - 10^-7 = 0.012 Therefore log[EDTA] = log(0.012) + 0.348 Therefore [EDTA] = 10^-1.75 = 0.027 M Therefore Total EDTA come = [EDTA] + [Ba Y^2+] = 0.027 + 0.012 = 0.039 M

now 0.039 M of 500 mL EDTA solution = 0.039 x 500/1000 = 0.019 moles of EDTA = 0.0195 x 372.23

= 7.26 g of EDTA