Question

I would like to know how to solve the Test Yourself portion of the question: At what pH does the fraction of EDTA in the form Y^4- = 0.50?

LUNI UI PA. PH The fraction of concentration of 0.10 MPLE What Does Qy* - Mean? call free EDTA in the form Y" is called ay. At pH 6.00 and a formal tion of 0.10 M, the composition of an EDTA solution is 1 89 X 1020 M [HY'] = 8.9 X 10-4 M [HY] = 2.8 x 10-7M -1 = 2.8 X 10^M [H2Y2-1 = 0.057 M [HY) = 0.043 M [Y ] = 1.8 X 10 M Find ay. Solution ay is the fraction in the form Yº-: dy- = [Y4-1 [H.Y2+] + [H Y*1 + [H.Y] + [HRY] + [H,Y"] + [HY] + [Yº-1 1.8 X 106 (8.9 x 10-20) + (8.9 X 10-14) + (2.8 X 10-?) + (2.8 x 10-5) + (0.057) + (0.043) + (1.8 X 100) = 1.8 X 105 TEST YOURSELF At what pH does Qyt- = 0.50? (Answer: pH =pK6 = 10.37) EDTA Complexes Th matal with a ligand is called the formation

Answer 1

. A From Henderson - Hasselbalch equation, we know, pH = pkatlog [salt] 8. [feed] For, EDTA, dissociation of to y 4- involves a : HYR 74.- + H+.. ÞH 3 plia + log [14-] [HY3-] Here, for, 284-2015 [44] [14]+[H¥3-] :[443-7 = 1 5.; $H = pka + log 1 [logize] So; þ H ¿ pka a 10.37 So when [44] = [Hy?] n then x44 2015 and pH = pkaa 10:37 ...