Question

Nitrilotriacetic acid is a tetraprotic acid and a tetradentate ligand, that unlike EDTA is biodegradable. It is used as a water softener in detergents to bind multiply charged cations such as Ca2+ and Fe3+. K1 = 0.100; K2= 1.55E-2; K3=3.02E-3; K4=3.47E-10. The active species for binding metals is the tetradeprotonated form, so calculate $%255Calpha_4$ α4for pH = 8.847. (the pH's have 3 significant figures in case the trailing zero is lost)

Answer 1

alpha[a^4-] = K1K2K3K4/[H+]^4 + K1[H+]3 + K1K2[H+]^2 + K1K2K3[H+] + K1K2K3K4

pH = -log[H+] = 8.847

[H+] = 1.422 x 10^-9 M

Feed values,

alpha[a^4-] = (0.1)(1.55x10^-2)(3.02x10^-3)(3.47x10^-10)/(1.422x10^-9)^4 + 0.1(1.422x10^-9)^3 + (0.1)(1.55x10^-2)(1.422x10^-9)^3 + (0.1)(1.55x10^-2)(3.02x10^-3)(1.422x10^-9) + (0.1)(1.55x10^-2)(3.02x10^-3)(3.47x10^-10)

= 1.624 x 10^-15/4.089 x 10^-36 + 2.875 x 10^-28 + 3.134 x 10^-21 + 6.656 x 10^-15 + 1.624 x 10^-15

= 0.196

So the answer would be, 0.196