EDTA Titration - Lecture notes 5 PDF

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EDTA Titration Lectures (Analytical Chemistry)...

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Complexometric Titration

In a complexometric titration with EDTA, αY4- is the fraction of EDTA in the Y4- form that forms complex. We do not carry a complexometric titration at pH = 1 and 13 because at these pH values, αY4- is too small. At pH = 13, it is likely to form M(OH) precipitate.

Conditional Formation Constants

Mn+ +

Y4- ↔ MY(n-4)+

Kf = [MY(n - 4)+] [Mn+] α4CT

K’f = MY(n-4)+ = α4KY4[Mn+]CT Where K’f is the conditional formation constant at certain pH value. α4 = [Y4-] CT

where CT is the total molar concentration of uncomplexed EDTA

CT = [Y4-] + [HY3-] + [H2Y2-] + [H3Y-] + [H4Y] The formation constant in Table for FeY- is 1025.1 = 1.3 x 1025 Calculate the concentration of free Fe3+ in a solution of 0,10 M FeY- at pH 4.0 and at 1.0. Fe3+ + EDTA ↔ FeY-

K’ = α4Kf

Where EDTA = Y4- + HY3- + H2Y2- + H3Y-, etc. At pH = 4.0: K’f = (3.8 x 10-9)(1.3 x 1025) = 4.9 x 1016 At pH = 1.0: K’f = (1.9x 10-18)(1.3 x 1025) = 2.5 x 107 Fe3+ + EDTA ↔ FeYInitial 0 0 0.100 Final x x 0.100 – x [FeY-] = 0.10 – x = K’f = 4.9 x 1016 at pH 4.0 [Fe3+][EDTA] x2 = 2.5 x 107 at pH 1.0 [Fe3+] = x = 1.4 x 10-9 M at pH 4.0 x = 6.4 x 10-5 M at pH 1.0

EDTA Titration Curves Calculate the pM for the reaction of 50.0 mL of 0.0500 M Mg2+ (buffered at pH 10.0) with 0.05 M EDTA. Mg2+

+ EDTA

↔ MgY2-

50 (0.05) = 0.05 Ve

Ve = 50 mL

K’f = α4 Kf = (0.36)(6.2 x 108) = 2.2 x 108

Region 1: before the equivalence point V = 5.0 mL 50 mL(0.05 mmol Mg2+) - 5 mL (0.05 mmol EDTA) [Mg2+] = ( mL ) ( mL ) = 0.0409 M (50 mL + 5 mL) 2+ pMg = 1.39

Region 2: at the equivalence point V = 50.0 mL All metal is in the form MgY2[MgY2-] = 0.05 (50 ) = 0.0250 M (50 + 50) Mg2+ Initial 0 Final x

+ EDTA ↔ MgY20 0.0250 x 0.0250 – x

0.0250 – x = 2.2 x 108 x2

x = 1.07 x 10-5 M

Region after the equivalence point (V = 51.0 mL) ) = 4.95 x 10-4 M

[EDTA] = 0.05 mmol EDTA (1.0 (50 + 51)

[MgY2-] = 0.05 mmol Mg2+ (50.0 ) = 2.48 x 10-2 M (101.0) K’f = [MgY2-] = 2.2 x 108 2+ [Mg ][EDTA] (2.48 x 10-2) = 2.2 x 108 2+ -4 [Mg ][4.95 x 10 ] [Mg2+] = 2.3 x 10-7 M

p Mg2+ = 6.64

pMg2+ = 4.97

Ammonia Complexes of Zinc Zn2+ and NH3 form the complexes Zn(NH3)2+, Zn(NH3)22+, Zn(NH3)32+ , and Zn(NH4) 2+. If the concentration of free unprotonated NH3 is 0.10 M, find the fraction of zinc in the form Zn2+. From appendix the stepwise formation constant for zinc complexes are (K1, k2, K3, K4) β1 = 2.18

β2 = 2.25

β3 = 2.31

β4 = 1.96

α Zn2+ = ______________1________________ 1 + β1 [L] + β2 [L]2 + β3 [L]3 + β4 [L]4 α Zn2+ = _____________________1______________________ 1 + 102.18 (0.1) + 104.43(0.1)2 + 106.74 (0.1)3 + 108.7(0.1)4 α Zn2+ = 1.8 x 10-5

EDTA Titration in the Presence of NH3

Consider titration of 1 x 10-3 M Zn2+ with 1 x 10-3 M EDTA at pH = 10.0 in the presence 0f 0.10 M NH3. Find p at 20, 50 and 60 mL EDTA. α4K’f = K”f = (1.8 x 10-5)(0.36)(3.2 x 1016) = 2.05 x 1011 Ve = 50.0 mL V = 20.0 mL (before the equivalence point) 50 mL(1 x 10-3 mmol Zn2+) - 20 mL (1 x 10-3 mmol EDTA) CZn2+ = ( mL ) ( mL ) = 4.286 x 10-4 M (50 mL + 20 mL) However, all zinc not bound to EDTA is bound to NH3. Free Zn2+ is [Zn2+] = αZn2+ CZn2+ = 1.8 x 10-5 (4.286 x 10-4) = 7.71 x 10-9 M p Zn2+ = -log Zn2+ = 8.11 V = 50.0 mL

[ZnY2-] = (1 x 10-3 M)(50 ) = 5 x 10-4 M (50 + 50) CZn2+ + EDTA ↔ ZnY2Initial 0 0 5 x 10-4 Final x x 5 x 10-4 - x K”f = 2.05 x 10-11 = 5.0 x 10-4 – x x2

x = 4.9 x 10-8 M = C Zn2+

[Zn2+] = αZn2+ CZn2+ = (1.8 x 10-5)(4.9 x 10-8) = 8.9 x 10-13 M V = 60.0 mL

(after equivalence point)

[ZnY2-] = 1.0 x 10-3 (50 ) = 4.5 x 10-4 M (50 + 60) EDTA = (10 ) (1.0 x 10-3M) = 9.10 x 10-5 M (50 + 60) [ZnY2-] CZn2+ [EDTA]

= K’’f = 2.05 x 1011

(4.5 x 10-4) = 2.05 x 1011 -5 CZn2+ (9.10 x 10 )

CZn2+ = 2.41 x 10-11

[Zn2+] = 2.41 x 10-11 (1.8 x 10-5) [Zn2+] = 4.49 x 10-16 M EDTA Titration Curve

p Zn2+ = 15.35

p Zn2+ = 12.05

Sample Problems 1. Calculate the equilibrium concentration of Ni2+ in a solution with NiY2- of 0.0150 M at pH 3.0 and 8.0. Ni2+ + EDTA ↔ NiY2K”f = [NiY2-] = 4.2 x 1018 2+ 4[Ni ][ Y ] a) K’f = α 4Kf = 2.5 x 10-11 (4.2 x 1018) = 1.05 x 108 1.05 x 108 = [NiY2-] = 0.0150 2+ 4[Ni ][ Y ] x2

x = 1.2 x 10-5 M

b) K’f = α 4Kf = 5.4 x 10-3 (4.2 x 1018) = 2.27 x 1016 2.27 x 1016 = 0.0150 x2

[Ni2+] = 8.1 x 10-10 M

2. Calculate the concentration of Ni2+ in a solution that was prepared by mixing 50.0 mL of 0.030 M Ni2+ with 50.0 mL of 0.050 M EDTA. The mixture was buffered to 3.0. [NiY2-] = 50 mL(0.030 mmol) = 0.0150 M 100 mL 50 mL (0.05mmol EDTA) - 50 mL (0.03 mmol Ni2+) [EDTA] = ( mL ) ( mL ) = 0.010 M ( 50 mL + 50 mL) [Ni2+]...