Transcript 投影片 1

Titrations
Volumetric analysis

Procedures in which we measure the volume of reagent
needed to react with an analyte
Titration

Increments of reagent solution (titrant) are added to
analyte until reaction is complete.
-
Usually using a buret

Calculate quantity of analyte from the amount of titrant
added.

Requires large equilibrium constant
(Thermodynamic)
Requires rapid reaction
(kinetic)


aA + tT → products
A: analyte
T: titrant
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Titrations
Buret Evolution
Gay-Lussac (1824)
Blow out liquid
Mohr (1855)
Compression clip
Used for 100 years
Descroizilles (1806)
Pour out liquid
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Henry (1846)
Copper stopcock
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Mohr (1855)
Glass stopcock
2
Type of Titrations based on Measuring Techniques
i) Volumetric titrimetry: Measuring the volume of a
solution of a known concentration (e.g., mol/L) that
is needed to react completely with the analyte.
ii) Gravimetric (weight) titrimetry: Measuring the
mass of a solution of a known concentration (e.g.,
mol/kg) that is needed to react completely with the
analyte.
iii) Coulometric titrimetry: Measuring total charge
(current x time) to complete the redox reaction, then
estimating analyte concentration by the moles of
electron transferred.
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Type of Titrations based on Chemical Reactions
i) Acid-Base Titrations, example:
H+ + OH– → H2O
ii) Precipitation Titrations, example:
Ag+(aq) + Cl–(aq) → AgCl(s)
ii) Redox Titrations:
5 H2O2 + 2 MnO4– + H+ → 5 O2 + 2 Mn2+ + 8H2O
iv) Complexometric Titrations, example:
EDTA + Ca2+ → (Ca–EDTA)2+
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Type of Titration Curves
Type
Example.
y-axis
x-axis
Acid-base
HCl/NaOH
pH
V. NaOH
Precipitation
Ag+/Cl–
pAg+
V. Ag+
Complexation
Ca2+/EDTA
pCa2+
V. EDTA
Redox
MnO4–/Fe2+
Potential
V. Fe2+
V. = volume
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Type
Example
Spectrophotometric
apotransferrin/ Absorbance
Fe3+
Thermometric
H3BO4/
NaOH
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y-axis
x-axis
V.
Fe3+
Temperature V.
NaOH
TMHsiung@2007 5/42
Expressing concentration
Formality
Molarity (V & W)
Molality
Normality
%W/W
%W/V
%V/V
part per thousand (ppt) - X's 1000
parts per million (ppm) - X's 106
parts per billion (ppb) - X's 109
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relationship between titrant
and analyte
# Eqg titrant = # Eqg analyte
(V*N)titrant =(V*N)analyte
# Eqg titrant = (V*N)titrant
#molestitrant=(V*M)titrant
#molesanalyte=(V*M)analyte
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 Standardization: The process by which the
concentration of a reagent is determined by reaction
with a known quantity of a second reagent
 Primary standard: The reagent which is ready to be
weighted and used prepare a solution with known
concentration (standard).
Requirements of primary reagent are:
- Known stoichiometric composition and reaction
- High purity
- Nonhygroscopic
- Chemically stable both in solid and solution
- High MW or FW
 Secondary standard: A standard solution which is
standardized against a primary standard.
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Standardization

Required when a non-primary titrant is used
-
Prepare titrant with approximately the desired concentration
Use it to titrate a primary standard
Determine the concentration of the titrant
Titration
Standardization
titrant known
concentration
titrant unknown
concentration
analyte unknown
concentration
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analyte known
concentration
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Standardization of 0.1 M NaOH
1-selection the PS (e.g. KHP)
mg
m l N 
2-wheing the PS
EqW
10*0.1=mg/204.1 213.8
N1V1=N2V2
3-making solution
4-addind suitable indicator
5-titration
9.1ml
6-calculation 9.1*n=213.8/204.1
n=0.115
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 Blank Titration: Titration procedure is carried out
without analyte (e.g., a distilled water sample). It is
used to correct titration error.
 Back titration: A titration in which a (known)
excess reagent is added to a solution to react with
the analyte. The excess reagent remaining after its
reaction with the analyte, is determined by a
titration.
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Standardization
Example: To standardizing a KMnO4 stock solution, the primary
standard of 9.1129 g Na2C2O4 is dissolved in 250.0 mL volumetric
flask. 10.00 mL of the Na2C2O4 solution require 48.36 mL of
KMnO4 to reach the titration end point. What is the molarity (M) of
MnO4– stock solution? (FW Na2C2O4 134.0)
Solution:
2–
5C2O4
(aq)
+ 2MnO4–(aq) + 16H+(aq) → 10CO2(g) + Mn2+(aq) + 8H2O(l)
9.1129 g Na 2C2O 4 1 mol Na 2C2O 4
1 mol C2O 24 
10 mL



1
134.0 g Na 2C2O 4 1 mol Na 2C2O 4 250 mL
2 mol MnO4
1
1000 mL





0
.
02250
M
MnO
4
1L
5 mol C2O 24  48.36 mL
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Ans
TMHsiung@2007 12/42
Unknown Analysis with a Blank Correction
Example: A 0.2865 g sample of an iron ore is dissolved in acid, and
the iron is converted entirely to Fe2+. To titrate the resulting solution,
0.02653 L of 0.02250 M KMnO4 is required. Also a blank titration
require 0.00008 L of KMnO4 solution. What is the % Fe (w/w) in
the ore? (AW Fe 55.847)
MnO4–(aq) + 5Fe2+ + 8H+(aq) → Mn2+(aq) + 5Fe3+ + 4H2O(l)
Net titrant vol  0.02653  0.00008 L  0.02645 L
0.02645 L titrant 0.02250 mol MnO4 5 mol Fe 2 




1
1 L titrant
1 mol MnO4
55.847 g Fe
1

 100%  58.01% Fe ( w / w)
2
0.2865 g sample
1 mol Fe
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Back Titration
1)Add excess of one standard reagent (known concentration)
2)Titrate excess standard reagent to determine how much is left
-
Add Fe2+ to determine the amount of MnO4- that did not react with oxalic acid
Differences is related to amount of analyte
Useful if better/easier to detect endpoint
MnO4–(aq) + 5Fe2+ + 8H+(aq) → Mn2+(aq) + 5Fe3+ + 4H2O(l)
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Back Titration
Example: The arsenic in 1.010 g sample was pretreated to
H3AsO4(aq) by suitable treatment. The 40.00 mL of 0.06222 M
AgNO3 was added to the sample solution forming Ag3AsO4(s):
H3 AsO4( aq )  3Ag ( aq)  3H  ( aq)  Ag3 AsO4( s )
The excess Ag+ was titrated with 10.76 mL of 0.1000 M KSCN. The
reaction was:
Ag (aq )  SCN (aq )  AgSCN( s )
Calculate the percent (w/w) As2O3(s) (fw 197.84 g/mol) in the
sample.
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TMHsiung@2007 15/42
0.06222m m ol AgNO3
 (40.00 m L AgNO3 

1 m L AgNO3


0
.
1000
m
m
ol
SCN
1
m
m
ol
Ag
10.76 m L SCN 

)


m L SCN
1 m m olSCN
1 m m olH 3 AsO4
1 m m olAs



3 m m olAg
1 m m olH 3 AsO4
1 m m olAs2O3
1 m ol As2O3
197.84 g As2O3



2 m m olAs 1000m m olAs2O3
1 m ol As2O3
1
 100%  4.612% As2O3 (w/w)
1.010 g sam ple
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TMHsiung@2007 16/42
Back Titration
In a back titration analysis of HCO3-, 25 mL of a bicarbonate solution is
reacted with 25.00 mL of 0.100 M NaOH. The excess NaOH was
titrated with 0.100 M HCl. This required 14.82 mL. What is the
concentration of bicarbonate in solution?
→ Na + CO
NaOH + HCl → NaCl + H O
NaOH + HCO3-
+
3
-
+ H2O
2
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Equivalence point

Quantity of added titrant is the exact amount necessary for stoichiometric reaction
with the analyte
Ideal theoretical result
Analyte
Oxalic acid
(colorless)
Titrant
(purple)
(colorless)
(colorless)
Equivalence point occurs when 2 moles of MnO4- is added to 5 moles of Oxalic acid
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End point
Occurs from the addition of a slight excess of titrant
Marked by a sudden change in the physical property of the solution
-
-
Change in color, pH, voltage, current, absorbance of light.
-
Analyte
Oxalic acid
(colorless)
End point ≈ equivalence point
Titrant
(purple)
(colorless)
(colorless)
After equivalence point occurs, excess MnO4- turns solution purple  Endpoint
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Titration Error
-
Difference between endpoint and equivalence point
Corrected by a blank titration
1) repeat procedure without analyte
2) Determine amount of titrant needed to
observe change
3) subtract blank volume from titration
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Calculation of ascorbic acid in Vitamin C tablet:
Standardization: Suppose 29.41 mL of I3- solution is required to react with 0.1970 g of pure
ascorbic acid, what is the molarity of the I3- solution?
Analysis of Unknown: A vitamin C tablet containing ascorbic acid plus an inert binder was
ground to a powder, and 0.4242g was titrated by 31.63 mL of I3-. Find the weight percent of
ascorbic acid in the tablet.
(i) Starch is used as an indicator:
(ii) starch + I3-  starch-I3- complex
ascorbic acid was oxidized with I3-:
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1 mole ascorbic acid  1 mole I3-
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Titration of a Mixture
Example: A solid mixture weighing 1.372 g containing only
sodium carbonate (Na2CO3, FW 105.99) and sodium bicarbonate
(NaHCO3, FW 84.01) require 29.11 mL of 0.7344 M HCl for
complete titration:
Na2CO3  2 HCl  2 NaCl( aq )  H 2O  CO2
NaHCO3  HCl  NaCl( aq )  H 2O  CO2
Find the mass of each component of the mixture.
Total mixture
Na2CO3
NaHCO3
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mass
1.372
moles
x
x/105.99 g/mol
1.372 - x
1.372 – x/84.01 g/mol
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TMHsiung@2007 22/42
Total mmol H  added
1L
0.7344 mol H 
 29.11 mL HCl 

 0.02138 mol H 
1 000 mL
1 L HCl
mmol H  consumed by Na2CO3
1 mol Na2CO3
2 mol H 
2x 

 x g Na2CO3 

 
mol H
105.99 g Na2CO3 1 mol Na2CO3  105.99 
mmol H  consumed by NaHCO3
1 mol NaHCO3
1 mol H 
1.372  x
 (1.372  x) g NaHCO3 

 
84.01 g NaHCO3 1 mol NaHCO3  84.01
2 x   1.372  x 
0.02138  


 105.99   84.01 
x g Na2CO3  0.724 g Na2CO3
Ans
(1.372  x) g NaHCO3  0.648 g NaHCO3
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mol H 


Ans
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TMHsiung@2007 23/42
Direct and back (indirect) titration of
Aspirin
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Precipitation Titrations
A titration in which the reaction between the analyte
and titrant involves a precipitation.
Ag+(aq) + Cl–(aq)  AgCl(s)
AgCl(s)  Ag+(aq) + Cl–(aq)
s = [Ag+]=[Cl–]
Ksp = 1.8×10–10
[Ag+]=[Cl–]=1.35x10–5
pAg = 4.89
pCl = 4.89
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TMHsiung@2007 25/42
Titration curve of
50.0 mL of 0.0500 M Cl– with 0.100 M Ag+
pCl
pAg
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TMHsiung@2007 26/42
Example: For the titration of 50.0 mL of 0.0500 M Cl– with 0.100
M Ag+. The reaction is:
Ag+(aq) + Cl–(aq)  AgCl(s) K = 1/Ksp = 1/(1.8×10–10) = 5.6 x 109
Find pAg and pCl of Ag+ solution added
(a) 0 mL (b) 10.0 mL (c) 25.0 mL (d) 35.0 mL
Solution:
(a) 0 mL Ag+ added (At beginning)
[Ag+] = 0, pAg can not be calculated.
[Cl–] = 0.0500, pCl = 1.30
N1V 1  N 2V 2
]
In eq point, 25 ml
N1V1=N2V2[Cl 50*0.05=0.1*V2
V1  V 2
(b) 10 mL Ag+ added (Before Ve)

N 2V 2  N1V 1
[ Ag ] 
K sp

V2
[ Ag ]  V 1 

-5

√Ksp=[Ag+]=[Cl-]=1.34*10
(d) 35 mL Ag+ added (After Ve)
[Cl  ] 
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[Cl ]
K sp
[ Ag  ]
TMHsiung@2007 27/42
Diluting effect of the titration curves
25.00 mL 0.1000 M I–
titrated with 0.05000 M Ag+
25.00 mL 0.01000 M I–
titrated with 0.005000 M Ag+
25.00 mL 0.001000 M I–
titrated with 0.0005000 M Ag+
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TMHsiung@2007 28/42
Ksp effect of the titration curves
25.00 mL 0.1000 M halide (X–)
titrated with 0.05000 M Ag+
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TMHsiung@2007 29/42
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TMHsiung@2007 30/42
Titration of a mixture (uncertainty concerned)
(a) 40.00 mL of 0.0502
M KI + 0.0500 M
KCl, titrated with
0.0845 M Ag+
(b) 20.00 mL of 0.1004
M KI titrated with
0.0845 M Ag+
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TMHsiung@2007 31/42
Example: A 25.00 mL solution containing Br– and Cl– was
titrated with 0.03333 M AgNO3. Ksp(AgBr)=5x10–13,
Ksp(AgCl)=1.8x10–10.
(a) Which analyte is precipitated first?
(b) The first end point was observed at 15.55 mL. Find the
concentration of the first that precipitated (Br– or Cl–?).
(c) The second end point was observed at 42.23 mL. Find the
concentration of the second that precipitated (Br– or Cl–?).
Solution:
b
(a)
Ag+(aq) + Br–(aq) cAgBr
N1V(s)1  N 2VK2= 1/Ksp(AgBr) = 2x1012
Ag+(aq) + Cl–(aq)  AgCl(s)
K = 1/Ksp(AgCl) = 5.6x109
N1
(V.55
2 
V0
1).3333
 N 2V325 N 2
15
Ans: AgBr precipitated first
0.3333
(42
 15.M
55)  25 N 2
N2 
0..23
02073
N 2  0.03557M
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2) Argentometric Titration
 Define Argentometric Titration:
 A precipitation titration in which Ag+ is the titrant.


–
–
–
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Argentometric Titration classified by types of
End-point detection:
Volhard method: A colored complex (back titration)
Fajans method: An adsorbed/colored indicator
Mohr method: A colored precipitate
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Mohr method
The Mohr method was first published in 1855 as a method for
chloride analysis.

In the precipitation of chloride by silver ion, chromate
ion (CrO42) is used as an indicator in the formation of Ag2CrO4, a
reddish-brown precipitate formed when excess Ag+ is present.
CrO42
+

Ag
+ Cl  AgCl(s)
Ksp= 1.8 x 10-10 (S = 1.34 x 10-5 M)
white precipitate
2Ag+ + CrO42 
Ag2CrO4(s)
Ksp= 1.2 x 10-12 (s = 6.7 x 10-5M)
red precipitate
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TMHsiung@2007 34/42
The titrations are performed only in neutral or slightly basic medium
to prevent silver hydroxide formation (at pH > 10).
2Ag+ + 2OH  2AgOH(s)  Ag2O(s) + H2O
precipitate
to prevent chromic acid formation (at pH < 7).
CrO42 + H3O+
HCrO4 + H2O
2 CrO42 + 2 H3O+
H2CrO4 + H2O
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TMHsiung@2007 35/42
Volhard METHOD
Back titration for determination of Cl-.. First published in 1874.
Reactions:
Ag+ +
(excess)
Cl 
AgCl(s)
Ksp = 1.82 x 10-10
white precipitate
H+, Fe3+
SCN + Ag+ 
titrant
Ksp = 1.1 x 10-12
white precipitate
SCN + Fe3+ 
Indicator
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AgSCN(s)
FeSCN2+
Kf = 1.4 x 10+2
red complex
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TMHsiung@2007 36/42
The titration is usually done in acidic pH medium
to prevent precipitation of iron hydroxides, Fe(OH)3.
Fe3+ +3(OH)- ⇄ Fe(OH)3 Ksp=1*10-39
If
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[Fe3+]=0.001 M
pH=?????
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TMHsiung@2007 37/42
Since SAgSCN <SAgCl , equilibrium will shift to the right causing a
negative error for the chloride analysis.
Ag+ +
Cl 
SCN + Ag+ 
AgCl(s)
Ksp = 1.82 x 10-10
AgSCN(s)
Ksp = 1.1 x 10-12
SCN- + AgCl  AgSCN + Cl-
To eliminate this error, AgCl must be filtered or add nitrobenzene
before titrating with thiocyanate; nitrobenzene will form an oily
layer on the surface of the AgCl precipitate, thus preventing its
reaction with thiocyanate.
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TMHsiung@2007 38/42
Fajans Method: An adsorbed/colored indicator.
Titrating Cl– and adding
dichlorofluoroscein as indicator:
Before Ve
(Cl– excess)
Greenish
yellow solution
After Ve
(Ag+ excess)
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TMHsiung@2007 39/42
5) Applications of argentometric titrations:
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