Idea Transcript
1200 EDTA Titration
Dr. Fred Omega Garces Chemistry 251 Miramar College 1
EDTA Titration
Aug ‘17
Lewis Acid - Lewis Base Chemistry Lewis Acid : e- acceptor (metals are good e- acceptor) Lewis Base : e- donor (Ligands with lone pair electrons) Ligands, atoms or cluster of atoms with lone pair electrons available to donate Complexing Agent: H2O, NH3, Cl- CN-
2
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Metal-Ligand: Lewis Acid - Lewis Base Metal; possesses open orbital to accept e- pair(s) through their empty d-orbital(s)
Ligand; possess e- pair to donate to metal
Combine to form LA-LB (Compound)
In general, the complex is more stable than the separated compound.
M
M
M
M
M
Complex - Metal ion bonded to a number of ligands
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Complex - Metal coordinating to Ligands Metal- Ligand compounds [MLn] i.e., [Ag(NH3)2]+ or [Co(NH3)6] Cl3 [ ] denotes atoms bonded to each other through covalent bonds. These atoms are contained in the coordination sphere. Coordinated atoms are those elements that are directly bonded to each other and are contained in the coordination sphere. Counter ions atoms or ions that are denoted outside bracket and are not part of the coordinate sphere. A coordinated compound behaves like an electrolyte in water: the complex ion and counter separates from each other. But the complex ion behaves like a polyatomic ion that is the ligands and central metal ion remain attached. 4
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Coordinated Complexes and Coordination Number Coordination Number
2
Geometry
3-D Model
Example [CuCl2]-, [Ag(NH3)2]+, [AuCl2]-
Linear F F
4
Square
Br
F F
Planar
4
[Ni(CN)4] 2-, [PdCl4]2[Pt(NH3)4] 2+, [Cu(NH3)4] 2+ [Cu(CN)4] 3-, [Zn(NH3)4]2+ [CdCl4] 2-, [MnCl4] 2-
Tetrahedral F
F
F S
6
5
F
Octahedral
F
EDTA Titration
F
[Cu(H2O)6] 3+, [V(CN)6] 4-, [Cu(NH3)4Cl2] +, [Co(en)3] 3+ Aug ‘17
Ligands Consider [Ag(NH3)2]+ Ligand
(contains the donor atom, directly bonded to metal)
:NH3 - ligand occupy one site in coordinate sphere (monodentate)
examples (Monodentate) N3- , X-, CN- , OH-, NH3 , pyridine, H2O
Polydentate ligand - known as chelating agents - ligand which has several donor sites that can multi-bond (coordinate) metal simultaneously (chelates) i.e.
ethylenediamine (en), oxalate, 1.10 phenanthroline,
carbonate, bipyridine, phenylpyridine, [EDTA]4- or (ethylenediaminetetraacetate), 6
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Ligands
Example of Typical Uni-, bi- and poly dentate Ligands
Name of Neutral and Anionic Ligands: Neutral:
Anionic -ends in "o"
Aqua H2O
Fluoro Fl-
Amine NH3
Chloro Cl-
Nitrosyl NO
Bromo Br-
Carbonyl CO
Iodo IHydroxo OHCyano CN-
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Bidentate Ligands Structure of Analytically useful chelating agents
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EDTA Titration
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Chelates Chelating Ligands have two or more donor atoms that simultaneously coordinate to a single metal ion. Polydentate - (Many toothed - ligand) Chelating agent (Claw) Sequestering agent - sequester - to set apart or separate
en
ethylenediamine (shown) - two toothed ligand:
i.e., [Co(en)3]3+ [Pt(en)2]2+
EDTA ethylenediaminetetraacetate (picture) hexadentate EDTA is the antidote for heavy metal poisoning 9
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Chelating agents in Living system Seven of 24 elements necessary for life, based on ability to formed complexes V, Cr, Mn, Fe, Co, Cu, Zn Fe - hemoglobin, chlorophyll (Mg) Co - Vitamin B Zn – Zinc finger protein
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Biological Coordinating Complexes Chlorophyll
Oxymyoglobin
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Ferrichrome
EDTA Titration
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Chelating effect A complex containing chelate ligands is more stable than that from a monodentate ligand. K'f, larger for chelating complex. Ni(H2O)62+ + 6NH3 ® Ni(NH3)6 + 6H2O
kf = 4•108
Ni(H2O)62+ + 3en ® Ni(en)3 + 6H2O
kf = 2•1018
Driven by Entropy: Note that in the above reaction, the entropy increases via the increase of number of moles in the overall reaction. In reaction (1) there are 7 moles of reactant changing to seven moles of product but in reaction (2) there are 4 moles of reactant changing to 7 moles of product, Kef is much larger for reaction (2). 12
EDTA Titration
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EDTA: Acid-Base Properties
Acid-Base Properties pk1 = 0.0 (CO2H)
pk4 = 2.69
pk2 = 1.5 (CO2H)
pk5 = 6.13
pk3 = 2.00 (CO2H)
pk6 = 10.37
(CO2H) (NH+) (NH+)
Which specie has the greatest concentration at: pH = 6, pH = 7, pH =11
Where D = [H+]6 + [H+]5 K1 + [H+]4 K1 K2 + 13
[H+]3 K1 K2 K3 + [H+]2
K1 K2 K3 K4 +
EDTA Titration
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EDTA: fraction in Y4- form The sequential dissociation of H6EDTA+2 (H6Y2+)to EDTA-4 (Y4-) is based sequential removal of H+ from H6EDTA+2. The fraction of the EDTA-4 (Y4-) can be calculated based on the concentration of the (Y4-) and the total concentration of EDTA this is calculated as a below. The graph shows a
Y4-
Y4-
shown as a
function of pH
Fraction of EDTA in Y4- form:
α
Y 4−
=
[Y4- ] [H 6Y2+ ] + [H 5 Y+ ] + [H 4 Y] + [H 3Y- ] + [H 2Y2- ] + [H 1Y3- ] + [Y4- ]
=
[Y4- ] [EDTA]
this can be simplified, using K 1 - k 6 for EDTA, to the following equation.
α
Y 4−
14
=
K 1K 2K 3K 4K 5K 6
[H+ ]6 + [H+ ]5K 1 + [H+ ] 4K 1K 2 + [H+ ]3K 1K 2K 3 + [H+ ]2K 1K 2K 3K 4 + [H+ ]K 1K 2K 3K 4K 5 + K 1K 2K 3K 4K 5K 6 EDTA Titration
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Reaction with Metal: Metal-EDTA formation constant, Kf Values of aY4+ for EDTA
Formation constant Table-
Conditional Formation Constant @ pH
Kf' = aY4+ Kf i.e., 0.040M Ca+2 at pH = 10.0 Kf' = aY4+ * Kef = (0.30) 1010.65 Kf' = 1.34 • 1010 15
EDTA Titration
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Conditional Formation Constant The formation constant or stability constant can be expressed by the equationFormation Constant: M
+ Y
n+
4-
! MY
n-4
Kf =
[MYn-4 ] [M n + ] [Y4- ]
The formation constant from the aY4- vs. pH graph, most EDTA is not in the Y4- form below pH 10.37. The formation constant can be simplified the conditional constantConditional Formation Constant: K’ f =
[MYn-4 ] [M n + ] [Y4- ]
where, K’ f = α 4- K f Y
=
[MYn-4 ] [M n + ] [EDTA]
also known as the conditional (or effective) formation constant for systems in which the pH is fixed (buffered)
This conditional constant expresses the EDTA complex formation as if the uncomplexed EDTA is all in one form. Mn+ + EDTA D MYn-4
Kef = aY4- Kf
At any given pH, aY4- can be determined from K’ f.
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EDTA Titration
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EDTA Titration Curve In the titration reaction-
Mn+ + EDTA ! MYn-4
K’ f = a
Y4-
Kf
If K’f is large, then the reaction is complete at each point in the titration. The titration is a graph of pM (or –log [Mn+] ) versus volume EDTA added (M is metal) . Before equivalence point: There is excess Mn+ left in solution after the EDTA has been consumed. The concentration of the free metal ion is the excess, unreacted Mn+. Ignore the dissociated form of MYn-4 since it is negligible. At the equivalence point: EDTA equals the metal in solution. The pM is determine by the slight dissociation of MYn-4 to Mn+ and EDTA. After equivalence point: There is an excess EDTA and almost all metal ions are in the form of MYn-4. The concentration of free EDTA can be calculated by determining the excess EDTA after the equivalence point. The concentration of Mn+ is calculated by common ion effect equilibrium. 17
EDTA Titration
Aug ‘17
Homework 11.7 12.7 Consider the titration of 25.0 ml of 0.0200 M MnSO4 by 0.01000 M EDTA at pH 8.00. Calculate the pMn2+ at the following volumes of EDTA and sketch the titration curve. 0 mol, 20mL, 49 mol, 50.00mL 50.1 mL 60 mL Reaction: Mn2+
+ EDTA D MnY-2
0.0mL: [Mn ] = 0.0200M, pMn +2
2+
K'f = a
Y4-
⋅K f = (4.2 ⋅ 10-3 ) ⋅ ( 1013.89 ) = 3.3 ⋅ 1011
= -log[0.0200] → pMn2+
The eq point is 50.0mL.
= 1.70
19.00mL: [Mn+2] = [0.50mmol Mn2+ - 0.19 mmol EDTA] / 44 mL = 7.05 ⋅ 10-3 M → pMn2+
= 2.15
29.00mL: [Mn2+ ] = [0.50mmol Mn2+ - 0.29 mmol EDTA] / 54 mL = 3.89 ⋅ 10-4 M → pMn2+
= 2.41
49.90mL: [Mn2+ ] = [0.50mmol Mn2+ - 0.499 mmol EDTA] / 74.9 mL = 1.34 ⋅ 10-5 M → pMn2+
= 4.87
50.00mL: [0.500 mmol Mn2+ ] → [0.500 mmol MnEDTA2- ] → (0.500 mmol MnEDTA2- ] / 75mL = 6.67 ⋅ 10-3 M Mn2+ x
+ EDTA ! MnY-2 x 0.00667- x
→
(0.00667 − x) / x2 = a
Y4-
⋅K f = 3.3 ⋅ 1011 → x = 7.036 ⋅ 106
pMn2+
= 6.85
50.10mL → 5.01 mmol EDTA or 0.01 mmol excess EDTA after 5.00 mmol Mn+2 reaction [EDTA] = [0.10mL / 75.1mL] (0.0100 M) = 1.332 ⋅ 10-5 M [MnY-2] = [25.0 / 75.1] (0.0200 M) = 6.658 ⋅ 10-3 M [MnY-2] [MnY-2] K'f = → [Mn2+ ] = = 1.51 ⋅ 10-9 M [EDTA] ⋅ [Mn2+ ] [EDTA] ⋅K'f
→ pMn2+ = 8.82
52.00mL → 5.2 mmol EDTA or 0.020 mmol excess EDTA after 5.00 mmol Mn+2 reaction [EDTA] = [2mL / 77 mL] (0.0100 M) = 2.60 ⋅ 10-4 M [MnY-2] = [25.0 / 77] (0.0200 M) = 6.49 ⋅ 10-3 M K'f =
[MnY-2] [EDTA] ⋅ [Mn2+ ]
→ [Mn2+ ] =
[MnY-2] [EDTA] ⋅K'f
= 7.58 ⋅ 10-11 M
→ pMn2+ = 10.12
65.00mL → 6.5 mmol EDTA or 0.15 mmol excess EDTA after 5.00 mmol Mn+2 reaction [EDTA] = [15mL / 90 mL] (0.0100 M) = 1.67 ⋅ 10-3 M [MnY-2] = [25.0 / 90] (0.0200 M) = 5.56 ⋅ 10-3 M K'f =
18
[MnY-2] [EDTA] ⋅ [Mn2+ ]
→ [Mn2+ ] =
[MnY-2] [EDTA] ⋅K'f
= 1.01 ⋅ 10-11 M
→ pMn2+ = 11.00
EDTA Titration
Aug ‘17
Homework 11.7 11.7 Consider the titration of 25.0 ml of 0.0200 M MnSO4 by 0.01000 M EDTA at pH 8.00. Calculate the pMn2+ at the following volumes of EDTA and sketch the titration curve: 0.00 mL, 20.00mL, 49.00 mL, 50.00mL 50.10 mL 60.00 mL
Reaction: Mn2+ + EDTA D MnY-2
K‘f = aY4- K’f = (4.2•10-3) ( 1013.89) = 3.3•1011 The eq point is 50.0mL.
• 0.0mL: [Mn+2] = 0.0200M, pMn2+ = -log[0.0200] = 1.699
• 20.00mL: [Mn+2] = [.50mmol Mn2+ - .20 mmol EDTA] / 45 mL = 6.67•10-3 M g pMn2+ = 2.176
• 49.00mL: [Mn+2] = [.50mmol Mn2+ - .49 mmol EDTA] / 74 mL = 1.35•10-4 M
g
pMn2+ = 3.869
• 50.00mL: [0.500 mmol Mn+2] g [0.500 mmol MnEDTA2-] g (0.500 mmol MnEDTA2-] / 75mL = 6.67•10-3 M Mn2+
+ EDTA D
x
x
MnY-2 0.00667- x
g
(0.00667 – x) / x2 = aY4- Kf = 3.3•1011
g
x = 1.4•10-7 pMn2+ = 6.845
• 50.10mL 0.10mL excess EDTA [EDTA] = [0.10 / 75.1] (0.0100 M) = 1.332•10-5 M [MnY-2 ] = [25.0 / 75.1] (0.0200 M) = 6.658•10-3 M [Mn+2] = [MnY-2 ] / [EDTA] K’f = 1.51 •10-9 M
g pMn2+ = 8.82
• 60.00mL 10.00mL excess EDTA [EDTA] = [10.0 / 85.0] (0.0100 M) = 1.176•10-3 M [MnY-2 ] = [25.0 / 85.0] (0.0200 M) = 5.88•10-3 M [Mn+2] = [MnY-2 ] / [EDTA] K’f 19
= 1.5 •10-11 M EDTA Titration
g
pMn2+ = 10.814 Aug ‘17
Homework 11.7 11.7 Consider the titration of 25.0 ml of 0.0200 M MnSO4 by 0.01000 M EDTA at pH 8.00. Calculate the pMn2+ at the following volumes of EDTA and sketch the titration curve: 0.00 mL, 20.00mL, 49.00 mL, 50.00mL 50.10 mL 60.00 mL
Reaction: Mn2+ + EDTA D MnY-2
K‘f = aY4- K’f = (4.2•10-3) ( 1013.89) = 3.3•1011 The eq point is 50.0mL.
• 0.0mL: [Mn+2] = 0.0200M, pMn2+ = -log[0.0200] = 1.699
• 20.00mL: [Mn+2] = [.50mmol Mn2+ - .20 mmol EDTA] / 45 mL = 6.67•10-3 M g pMn2+ = 2.176
• 49.00mL: [Mn+2] = [.50mmol Mn2+ - .49 mmol EDTA] / 74 mL = 1.35•10-4 M
g
pMn2+ = 3.869
• 50.00mL: [0.500 mmol Mn+2] g [0.500 mmol MnEDTA2-] g (0.500 mmol MnEDTA2-] / 75mL = 6.67•10-3 M Mn2+
+ EDTA D
x
x
MnY-2 0.00667- x
g
(0.00667 – x) / x2 = aY4- Kf = 3.3•1011
g
x = 1.4•10-7 pMn2+ = 6.845
• 50.10mL 0.10mL excess EDTA
[MnEDTA-2 ] = All Mn2+ converts to [MnEDTA-2 ] in a volume of 25 + 50.10 mL (.500mmol / 75.10mL) = 6.658•10-3 M [EDTA4-] = Excess unreacted is 50.10 - 0.10mL, so concentration EDTA4- = 0.0100M x .10mL= .0010mol, conc = .0010mol/75.1 [Mn+2] = [MnEDTA-2 ] / [EDTA4-] K’f = 1.51 •10-9 M
g pMn2+ = 8.82 or use pM = p(aY4- K’f) + log (EDTA4-/MEDTA2-)
• 60.00mL 10.00mL excess EDTA [MnEDTA-2 ] = All Mn2+ converts to [MnY-2 ] in a volume of 25 + 60.0 mL (.500mmol / 85.0mL) = 5.88•10-3 M [EDTA] = 10mL unreacted so concentration EDTA4- = 0.0100M x 10.0mL= .010mol, conc = .010mol/85.0 [Mn+2] = [MnEDTA-2 ] / [EDTA] K’f = 1.5 •10-11 M 20
g pMn2+ = 10.814 or use pM = p(aY4- K’f) + log (EDTA4-/MEDTA2-)
EDTA Titration
Aug ‘17
Homework 11.7 11.7 Consider the titration of 25.0 ml of 0.0200 M MnSO4 by 0.01000 M EDTA at pH 8.00. Calculate the pMn2+ at the following volumes of EDTA and sketch the titration curve: 0.00 mL, 20.00mL, 49.00 mL, 50.00mL 50.10 mL 60.00 mL [MnSO4] = 0.0200M
Vol EDTA
mmol EDTA
Excess Chem.
V Total mL
[Mn2+] M
pM
Vol MnSO4 = 25.00mL
0
0
Mn+2
25
0.020
1.699
mmol MnSO4 0.500mmol
20
.200
.300 mmol Mn+2
45
6.66e-3
2.176
[EDTA] = 0.0100M
49
.49
0.0100 mmol Mn+2
74
1.351e-4
3.869
[Ay+ Kf = Kf’ =[(0.50/75)- x] /x2
50
.50
-
75
1.43e-7
6.845
pM = p(aKf) + log [(Y4-/MY+2)]
50.1
.501
[EDTA] = 0.001 / 75.1
75.1
1.534e-9
8.814
60
.600
[EDTA] = 0.01 / 85
85
2.042e-11
10.690
Buffer equation pM = p(aKf) + log [(Y4-/MY+2)] Buffer equation
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EDTA Titration
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Auxillary Complexing Agent When metals are exposed to hydroxides, the metal hydroxide product is insoluble in solution. To prevent this so that metals can be titrated under basic conditions, auxiliary completing agent are used. The agent, is a ligand (i.e., ammonia, triethanolamine) that binds strongly to the metal to inhibit hydroxide products, but can easily be displaced by the EDTA ligand to form the EDTA complex. Consider the reaction of a metal with two auxiliary completing ligand L: M + L
! ML:
β1 =
[ML]
M + 2L ! ML 2:
[M] [L]
Fraction of metal ion in the uncomplex state −
αM =
β2 =
[ML 2] [M] [L]2
[M]
where all forms of the metal is C M = M + ML + ML 2
CM
Plugging equations and solving simultaneous equations −
αM =
[M] CM
=
[M] [M]{1+β1 [L] + β2 [L]2 }
=
1 1+β1 [L] + β2 [L]2
The K"f for a metal in the presence of an auxiliary agent, i.e., NH3 requires a new conditional constant to account for the fact that only some of the EDTA is in the form Y4- and only some of the metal is in the form M+n. 22
K"f = aM+n aY4- Kef EDTA Titration
Aug ‘17
Metal Ion Indicators The titration of EDTA usually involve a metal ion indicator, a mercury electrode or an ion-selective electrode. Metal indicators are similar to acid-base indicators in that the color of the solution changes when endpoint is reacted, in this case when the indicator binds to the metal. Examples are: MgIn
(red)
+ EDTA(colorless)
D MgEDTA(colorless)
+ In(blue)
Common metal indicators are:
Eriochrome Black T is blocked by: Cu2+, Ni2+, Co2+, Cr3+, Fe3+, Al3+
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EDTA Titration
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EDTA titration Techniques Direct titration involves analyte titrated with standard EDTA. The analyte is usually buffered to a pH so that the conditional formation constant for the metal-EDTA complex is large and color of indicator is distinct. Auxiliary complex agent can be added to prevent metal hydroxide precipitate. Back Titration: The back titration usually involve excess EDTA added to the analyte. The excess EDTA is then titrated with a standard solution of a second metal ion. This technique prevents precipitation formation of analyte. Displacement titration: Some metals do not have satisfactory indicator, i.e., Hg2+. Here displacement titration can work. M2+is treated with excess Mg(EDTA)2- to displace Mg2+, which is titrated with standard EDTA. The condition formation constant for M(EDTA)2- must be greater than the Mg(EDTA)2- otherwise the Mg2+cannot be displaced from Mg(EDTA)2Indirect Titration: Anions that precipitate with certain metal ions can be analyzed with EDTA by this method. For example, sulfates can be analyze by precipitation with excess Ba2+ at pH = 1.. The BaSO4 is washed and boiled at pH 10 and excess EDTA. Ba2+ is converted to Ba(EDTA)2- with the excess EDTA back-titrated with Mg+2. Masking: Some reagent may be added to prevent some component of the analyte to react with EDTA. When multiple metals are in solution and preferred binding is favored a masking chemicals such as cyanide can inhibit one metal from binding EDTA over the other. Demasking releases the metal from the masking agent. 24
EDTA Titration
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EDTA Titration Guide
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EDTA Titration
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