Question

Consider the titration of 50.00 mL of 0.0980 M EDTA with 0.200 M Ga(NO3)3 all properly buffered to pH 4.00. Calculate the pGA+3 after the following additions of Ga(NO3)3.

Log Kf Ga+3 = 21.7 alpha at pH 4 = 3.0x10^-9

I'm pretty sure the Kf is 1.5 x 10^13

a) 10.00 mL Ga(NO3)3

b) 24.50 mL Ga(NO3)3

c) 30.00 mL Ga(NO3)3

Answer 1

log Kf = 20.3

Kf = 2.00 x 10^20

alpha[Y4-] at pH 4 = 3.8 x 10^-9

Kf' = Kf.alpha[Y4-] = 7.582 x 10^11

a) 10 ml of 0.2 M Ga(NO3) added

moles of EDTA = 0.098 M x 0.05 L = 4.9 x 10^-3 mols

moles of Ga3+ = 0.2 x 0.01 = 2 x 10^-3 mols

excess EDTA = 2.9 x 10^-3 mols

[GaY2-] = 2 x 10^-3/0.06 = 0.0333 M

[EDTA] = 2.9 x 10^-3/0.06 = 0.0483 M

Kf' = [GaY2-]/[Ga3+][EDTA]

7.582 x 10^11 = (0.0333)/[Ga3+](0.0483)

[Ga3+] = 9.093 x 10^13 M

pGa3 = -log[Ga3+] = 12.04

b) 24.50 ml of 0.2 M Ga(NO3)2 added

moles of EDTA = 0.098 M x 0.05 L = 4.9 x 10^-3 mols

moles of Ga3+ = 0.2 x 0.0245 = 4.9 x 10^-3 mols

This is equivalence point

[GaY2-] = 4.9 x 10^-3/0.0745 = 0.0658 M

GaY2- <==> Ga3+ + EDTA

let x amount has dissolved then,

Kf' = [GaY2-]/[Ga3+][EDTA]

7.582 x 10^11 = (0.0658)/x^2

x = [Ga3+] = 2.946 x 10^-7 M

pGa3 = 6.53

c) 30 ml of 0.2 M Ga(NO3)2 added

moles of EDTA = 0.098 M x 0.05 L = 4.9 x 10^-3 mols

moles of Ga3+ = 0.2 x 0.030 = 6.0 x 10^-3 mols

excess Ga(NO3)2 = 1.1 x 10^-3

[Ga3+] = 1.1 x 10^-3/0.08 = 0.01375 M

pAg = 1.86