Chapter 12 EDTA Titrations. Antibiotic chelate captures its prey.
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Transcript of Chapter 12 EDTA Titrations. Antibiotic chelate captures its prey.
Chapter 12
EDTA Titrations
Antibiotic chelate captures its prey
Mn+ + nL MLnn+
Mn+: central metal ion (Lewis acid)L: ligand (Lewis base)
Complexometric titrations are useful for the determination of metal.
15A Forming Metal-Chelate Complexes
Complexes: Formation Constants
ligand: have a pair of unshared electronsEx: N, O, S
NH3
NH3
[H3N:Cu:NH3 ] 2+Cu2+ + 4:NH3
pale bule deep bule
::
23
23f
233
7
3
23f2
2333
3
3
3f1
33
][NH ][Ag
])[Ag(NHK
)Ag(NH2NHAg
102.5][NH ][Ag
])[Ag(NHK
)Ag(NHNH)Ag(NH
102.5][NH ][Ag
])[Ag(NHK
)Ag(NHNHAg
8
23
23
fd
i
323
7f2f1f
104.0])[Ag(NH
][NH ][Ag
K1
)(K constant onDissociati
)(K constant yInstabilit
2NHAg)Ag(NH
constant. stabilityor constant formatoin
102.5KKK
The most widely used chelating agent in titration is ethylenediaminetetraacetic acid (EDTA)
HO2CH2C
NH+CH2CH2H+N-O2CH2C
CH2CO2H
CH2CO2-
H4Y hexadentate
15D EDTA: (1) A hexadentate
(2) The chelon effect or chelate effect:
Multidentate chelating agents form stronger complexes (Kf ) with metal ions than bidentate or monodentate ligands.
指 multidentate 之 Kf 比 monodentate 之 Kf 為大, 即較穩定。
15D EDTA
(3) Neutral EDTA is a tetrabasic acid
Y][H][Y ][H
105.5K YHHY
Y][H][HY ][H
106.9K HYHYH
Y][H]Y[H ][H
102.2K YHHYH
Y][H]Y[H ][H
101.0K YHHYH
4
411-
a443
4
37-
a332
2
4
223-
a22
23
4
32-a134
(5) Five species EDTA as a function of pH
Effect of pH on EDTA equilibria
4HCaYYCa
][Y][HY]Y[H]Y[HY][HC
][CaC
YHYH
YHHYYCaCaY
242
432234YH
2YH
4H
3H
22
H3H422
4
4
15D-2 EDTA Complex with Metal Ions
(1) Forms strong 1:1 complexes regardless of the charge on the cation
(2) Chelate with all cations
(3) Since the anion Y4- is the ligand species in complex formation, the complexation equilibria are affected markedly by the pH.
(4) The formation constant (Table 15-5)
[Y4-] ][Ca][CaY
K
CaYYCa
2
2
f
242
15D-3 Equilibrium: pH dependent M-Y
YH44
a4a3a2a1a3a2a12
a2a13
a14
a4a3a2a14
a4a3a2a1
4
a4a3a2
3
a4a3
2
a444
YH
4
4
Cα Y
KKKK][HKKK][HKK][HK][H
KKKKα
KKKK][H
KKK][H
KK][H
K][H
1α1
][Y
C
Ex
Calculate the fraction of EDTA that exists as Y-4 at pH 10, and from this calculate pCa in 100mL of solution of 0.100 M Ca+
2 at pH 10 after adding 100 mL of 0.100 M EDTA.
Calculate the fraction of EDTA that exists as Y-4 at pH 10, and from this calculate pCa in 100mL of solution of 0.100 M Ca+2 at pH 10 after adding 100 mL of 0.100 M EDTA.
∵ pH = 10 ∴ [H+] = 1.0 × 10-10
1/α4 = 1 + [H+]/Ka4 + [H+]2/Ka3Ka4 + ……
= 1 + [1.0 × 10-10 /5.5 × 10-11]+[(1.0 × 10-10)2/(6.9 × 10-7)(5.5 × 10-11)] +……
= 1 + 1.82 + 2.6 × 10-4 + …... = 2.82
∴ α4 = 0.35 Ca+2 + EDTA [or Y-4] = CaY-2
X 4[EDTA} [(0.1 × 100)/200] – X ≒ 0.05
Kf = [CaY-2]/[Ca+2] α4 [CH4Y] = 0.05/(X)(0.35)(X) = 5.0 × 1010
X = 1.7 × 10-6 M
pCa = -log (1.7 × 10-6) = 5.77
The conditional formation constant (ex:15.5)
(1) We can use K’f to calculate the equilibrium concentrations of the different species at a given pH.
(2) K’f value holds for only a specified pH.(3) HgY-2 之 Kf > PbY-2 之 Kf > CaY-2 之 Kf ; Kf 不受 pH 值之影響,
Kf’ 則受 pH 值之影響,上述三者在 pH 值≦ 9.0 時, Kf
’ 開始變小,也就是 EDTA 的滴定需在 (pH > 9.0) 之鹼性溶液中進行
YH2
2'f4f
YH42
2
f
44]C[Ca
][CaYKαK
C]α[Ca][CaY
K
15D-4 EDTA Titration Curves The end point break depends upon
1) [Mn+]2) [L1]3) [pH] selectivity4) Kf
The smaller Kf, the more alkaline the solution must be to obtain a k’f of 106.
15D-4 The titration rxn:Mn++ EDTA MYn-4 K’f = 4Kf
Three regions:(1) Before(2) At(3) Afterequivalence point
The Kf for CaY-2 is 5.0 × 1010. At pH 10, α4 is c
alculated to be 0.35 to give a Kf` of 1.8 × 1010.
Calculate the pCa in 100 mL of a solution of 0.100 M Ca+2 at pH 10 after addition of (a) 0 mL (b) 50 mL (c) 100 mL (d) 150 mL of 0.100 M EDTA.
The Kf for CaY-2 is 5.0 × 1010. At pH 10, α4 is calculated to be 0.
35 to give a Kf` of 1.8 × 1010. Calculate the pCa in 100 mL of a
solution of 0.100 M Ca+2 at pH 10 after addition of (a) 0 mL (b) 50 mL (c) 100 mL (d) 150 mL of 0.100 M EDTA.
(a). pCa = -log [Ca+2] = -log 0.1 = 1.00 (b). pCa = -log [0.1 × 100 – 0.1 × 50]/150 = 1.48 (c). [0.05 - X]/[X][X] = 1.8 × 1010
X = 1.7 × 10-6 M = [Ca+2]
pCa = -log [1.7 × 10-6] = 5.77 (d). [CaY-2] = 10 mmol/250 mL = 0.04 M
[CH4Y] = [0.1 × 150 – 0.1 × 100]/250 = 0.02 M
0.04/[Ca+2](0.02) = 1.8 × 1010 = Kf`
[Ca+2] = 1.1 × 10-10 M; pCa = -log 1.1 × 10-10 = 9.95
Various pH: K’ becomes smaller as the pH decreases
Kf: cation with larger formation const provide good end point even in acidic media.
Minimum permissible pH for a satisfactory end point
以 EDTA 滴定一系列金屬離子,當各自之 Kf`≒106 且能給出一個 sharp end p
oint 之最小 pH 值。 整條曲線依金屬之 Kf 可約略分為三區,
於 pH < 3 可滴定者, 3 < pH < 7 可滴定者, pH > 7 可滴定者。
當 pH < 3 時只有第一區之 metals 能以 EDTA 滴定來分析, 3 < pH < 7時,則第一、二區之 metals 理論上都可以 EDTA 滴定來分析。但當 pH 為強鹼時,雖然理論上所有的金屬都可被 EDTA 來滴定,但卻無法區分不同之金屬,此外也有不少金屬在強鹼時會產生氫氧化物的沈澱,也無法以 EDTA 滴定來分析
15D-5 How do other complexing agents affect EDTA titration
At the highest pH range, all the metals will react, but not all can be titrated directly due to precipitation of hydroxides. Ex: Fe3+ or Th4+
[ammonia] influence
15D-6 Detection of the End Point:
IndicatorsO,O’-dihydroxy azo type (metallochromic indicators)EX: Eriochrome Black T: (EBT)
-O3S N N
OHOH
O2NH3In
H2In- HIn2- + H+ pKa=6.3 red blueHIn2- In3- pKa=11.6 orangepH > 7MIn + HY3- HIn2- + MY2-
red blue
In general, the metal-indicator complex should be 10 to 100 times less stable than the metal-titrant complex’
The formation constants of the EDTA complexes of calcium and magnesium are too close to differentiate between them in an EDTA titration, so they will titrate together.
Eriochrome Black T cannot be used to indicate the direct titration of calculium alone with EDTA, because the indicator forms too weak a complex with calcium to give a sharp end point.
Calcium can actually be titrated in the presence of magnesium by raising the pH to 12 with strong alkali; Mg(OH)2 precipitates and does not titrate.
EBT(pH<6) H2In- red
(pH 6~12) HIn2- blue(pH>12) In3- orange