Question

If somebody could assist me with the below problem, that would be greatly appreciated! Thank you!

13-17. State (in words) what Qyt- means. Calculate aya- for EDTA at (a) pH 3.50 and (b) pH 10.50.

Answer 1

The structure of EDTA in fully protonated form and its pk values is HOOC –Học H2C–COOH +NH-HUC-CHU-NH HOOC–Học CH2COOH pk values are pK1=0.00 pK2 = 1.50 pK3 = 2.00 pK4 = 2.69 pK5 = 6.13 pK6 = 10.37 Among the deprotonated structures of EDTA, EDTA-4 (fully deprotonated structure) is the hexadentate lignad that forms the most stable complex with the metals. dyt is the fraction of such fully deprotonated form of EDTA among the all forms of EDTA It depends on pH value. Higher the pH, larger the fraction of EDTA as Y-4

Fraction of EDTA as Y-4 at pH = 3.50 HOọc –Học H.C—COOH +NH-HAC-CHU-NH –Học CH2 —COOH HOọc pk values are pK1=0.00 pK2 = 1.50 pK3 = 2.00 pK4 = 2.69 pK5 = 6.13 pK6 = 10.37 Fraction of EDTA as Y-4 | KK,K,K,K,K a [HT]® + [H] K, + [H*]*K,K, +[H*]K,K,K+[H+]KK,K,K, +[H*]K,K,K,K,K, +KK,K,K,K,K Given that | pH = 3.50=[H+]=10.50 pK, = 0.00=K, = 10-0.00 = 10° KK, = 10° x10-1.5 = 10-15 KK,K; = 10° x10-1-5 x 10-2.0 = 10-35 KK,K,K, = 10° x10-15 x10-2.0 x 10-2.69 = 10-6.19 KK,K,K,K, =10° x10-15 x10-2.0 x10-269x10-6.13 = 10-1232 K,K,K,K,K,K6 = 10° x10-1.5 x 10-2.0 x10-2.69 x10-6.13 x 10-10.37 = 10-22.69 10-22.69 Oy+ [10-350J* + [10-350j100.0 + [10-350*10-15 + [10-3.50 310-35 + [10-3:50 10-6.19 + [10-3:50 J10-12.32 +10–22.69 2.04174x10-23 = 04 = 7.50362x10-14 = 2.721x10-10 = 2.72x10-10 = 0,4 = 2.72x10-10

Fraction of EDTA as Y-4 at pH = 10.50 HOOC –Học H2C–COOH + NH–Học–CH,ẠNH –Học сн,— соон + V HOOC PK values are pK1=0.00 pK2 = 1.50 pK3 = 2.00 pK4 = 2.69 pk5 = 6.13 pK6 = 10.37 Fraction of EDTA as Y-4 KK,K,K,K,K [H+]° +[H] K, +[H*]*K,K, +[H]KK,K; +[H+] KK,K,K, +[H*]K,K,K,K,K, +K,K,K,K,K,K Given that pH =10.50 [H+]=10-10.50 PK, =0.00 = K = 10^0.00 = 10° KK2 = 10° x10-1.5 = 10-15 KK,K, = 10° x10-15 x 10-20 = 10-3.5 KK,K,K, =10° x10-15 x10-2.0 x10-2.69 = 10-6.19 K,K,K,K,K, =10° x10-15 x10-2.0 x10-2.69 x10-6.13 = 10-12.32 K,K,K,K K.K. = 10° x10-1.5 x10-2.0 x10-2.69 x10-6.13 x 10-10.37 = 10-22.69 10–22.69 Qyt [10-10.50 jø + [10-10.50 $ 100.0 + [10-10.50 j*10-15 +[10-10.50 j10-35 + [10-10.50 -10-6.19 +[10-10.507'10-12.32 +10–22.69 2.04174x10-23 aya 3.55536x10-23 10-23 = 0.57427 = 0.574 = Ox+ = 0.57