Question

Ethylenediaminetetraacetic acid (EDTA) is a hexaprotic acid with the following structure and pK values. HOOC- COOH - NHA THN PK 0.000 PK 1.500 PK 1.991 PK, 2.669 PK 6.160 PK, 10.259 HOOC COOH For simplicity, the fully protonated form of the acid is commonly represented as HY”. The fully deprotonated form of EDTA (Y") reacts with a variety of metal cations to produce complex ions, M" (aq) +Y+ (aq) MY(aq) K= aK where K is the formation constant for the metal-EDTA complex and a is the fraction of Y' in the EDTA solution at a given pH. Given that K = 891 for the sodium-EDTA complex, calculate the equilibrium [Na], [Y"], and [NaY') in a solution prepared by adding 0.5844 g NaCl to 1 L of 0.100 M EDTA buffered to pH 10.

Answer 1

HY3 H® + 84. Kag = 10 10.259 - [47] [144] [443] 10 10.259 -10.259 [y4] . 10:259 lolo -0.259 0.55081 Oss Sin [HYS THt7 10 Hy would e at b²=10, the above equilibrium establish and dose atleast 5 pooton immediately, [Y] = 0.55081 = 0.355175 = a. 184-] = 0.355175x0.100 = 0.035sm [EDTA] Tissosla Estal Nad cap.) + Y Cape) → Millo Nay3 (ap) k=a.kg = 0.355175X 891 k= 316.46 0.0-n 0.035s 26 K=316-46= x 0.035sx (0.01-) = 11-24 0.5844g 58.44 g/mul IL = 0.01000 mol/L, 0.0 0·1124 12.24 -= 0.009183 (Nat] = 0.01 -0.009183 20.000817 m. Equilibrium (Nat] = 8.17x10 m 7 [4.] = 3.55x152m [ Nay3] 2.9. 18 x103 m. I Answers