Chapter 7
Let the Titration Begin
1
Burets Through History
2
1
Titration Setup
3
Titrations in Analytical Chemistry
Volumetric analysis is the analytical chemistry technique
where one measures the volume of reagent needed to react
with the analyte
There are some general principles that apply to all
volumetric analysis procedures
A couple specific types of titrations will be studied in this
chapter.
–Precipitation titrations
–Spectrophotometric titrations
4
2
Titrations
The titrant (reagent) is added to the analyte until the
reaction between them is complete.
–Titrant is usually added using a buret.
Assumptions in a titration:
–The equilibrium constant is large (K>>1)
–Reaction time is short (kinetics is fast)
Types of titrations
–Acid-base (Ch. 12)
–Oxidation-reduction (Ch. 16)
–Complex formation (Ch. 13)
–Precipitations (Ch. 7)
5
End of the Analysis
The equivalence point is where the quantity of added
titrant has reacted stoichiometrically with the analyte.
5C 2 H 2 O 2 2MnO -4 6H   10CO 2 2Mn 2 8 H 2 O
–5 mol of oxalic acid reacts with 2 mol of permanganate
in acidic solution
The equivalence point is the ideal end of the titration, but
what we typically measure is the end point, where a
sudden change in a property of the solution is observed.
–Excess MnO4- gives a slight purple color in solution
6
3
Process of the Titration
Methods for determining the end point:
–Use of an indicator, a compound that exhibits a
property change at the endpoint (usually color)
–Detecting a voltage change between electrodes
(electrochemical)
–Measuring a change in absorption of light
(spectrophotometric)
Error
–The titration error is the difference between the end
point and equivalence point
–We try to estimate this with a blank titration
7
Calibration of the Titration
If the titrant is prepared by dissolving pure reagent in a
known volume of solution, then we can calculate its
concentration and the reagent is called a primary
standard (> 99.9% pure)
Many reagents are not available as primary standards. In
these cases we determine the concentration (standardize)
by first titrating against a primary standard. The titrant is
called a standard solution.
8
4
Types of Titration
Titrant can be added to analyte until the reaction is
complete (direct titration)
–If we added MnO4- to oxalic acid--direct
5C 2 H 2 O 2 2MnO -4 6H   10CO 2 2Mn 2 8 H 2 O
An excess of standard reagent can be added to analyte
and the excess can be titrated with a second reagent
(indirect or back titration)
–If we added excess MnO4- to oxalic acid, and then
titrated the excess MnO4- with Fe2+
9
Titrimetric Analysis
Always remember, the key step is to relate moles of
titrant to moles of analyte
Determination of calcium content in urine
–1) Ca2+ is precipitated as calcium oxalate
Ca 2 C 2 O 24- H 2 O  Ca(C 2 O 4 ) 
H 2 O(s)
–2) Solid is dissolved to give ions in solution
Ca(C 2 O 4 ) 
H 2 O(s)  Ca 2 (aq) C 2 O 24-(aq) H 2 O
–3) Standardization of permanganate solution
–4) Titration of oxalate with standard permanganate
5C 2 H 2 O 2 2MnO -4 6H   10CO 2 2Mn 2 8 H 2 O
10
5
Determination of Ca in Urine
Standardization: 0.3562 g of Na2CO3 is dissolved in
250.0 mL water. If 10.00 mL of the solution needs 48.36
mL if KMnO4 for titration what is the M of MnO4solution?
0.3562 g Na 2 C 2 O 4
M Ox 2- 
/ 0.2500 L 0.01063 M
134.00 g mol -1 Na 2 C 2 O 4
moles Ox 2- (0.0163 M Ox)(0.010 L) 0.106 mmol
2 mol MnO -4 
(0.106 mmol Ox 2- ) 0.04253 mmol MnO -4
moles MnO -4 
5 mol Ox 2- 


0.4253 mmol MnO -4
M MnO -4 
8.795 10 4 M
48.36 mL
11
Determination of Ca in Urine (2)
Analysis: Calcium in a 5.00 mL urine sample was
precipitated, redissolved and required 16.17 mL of
standard MnO4- for titration What is the concentration of
Ca2+?
moles MnO -4 (0.01617 L)(8.795 10-4 M) 0.01422 mmol MnO -4
5 mol Ox 2- 
(0.01422 mmol MnO -4 ) 0.0355 mmol
moles Ox 2- 
2 mol MnO - 
4 


1 mol Ca 2 
0.0355 mmol
moles Ca 2 
1 mol Ox 2- 


0.03555 mmol Ca 2 
2
M Ca 
0.00711 M Ca 2 
12
5.00 mL
6
Titration of a Mixture
A solid mixture weighing 1.372 g contains only Na2CO3
and NaHCO3 and required 29.11 mL of 0.7344 M HCl
for titration:
Na 2 CO 3 2HCl  2NaCl(aq) H 2 O CO 2
NaHCO 3 HCl  NaCl(aq) H 2 O CO 2
What is the mass of each component in the mixture?
13
Titration of a Mixture (2)
x= g Na2CO3;
g NaHCO3= 1.372-x
xg
(1.372 x) g
Moles Na 2 CO 3 
Moles NaHCO 3 
105.99 g/mol
84.01 g/mol
mol HCl used (0.02911 L)(0.7344 M) 0.02138 mol
From stoichiometry:
2 (mol Na 2 CO 3 ) mol NaHCO3 0.02138 mol
1.372 x 
 x 
2

0.02138
105.99   84.01 

x 0.724g Na 2 CO 3
1.372 - x 0.648 g NaHCO314
7
Kjeldahl Analysis
Kjeldahl nitrogen analysis (1883) is still commonly used
for determination of nitrogen in organic substances.
–Solid is digested (decomposed and dissolved)
boiling
organic C, H, N  CO 2 H 2 O NH 
4
H 2SO 4
–Nitrogen, as NH4+, is neutralized (1) and distilled into
HCl, reacting with excess HCl (2) and then excess HCl
is titrated with NaOH (3)
(1) NH 
4 OH  NH 3(g) H 2 O
(2) NH 3 H   NH 
4
(3) H  OH -  H 2 O
15
Kjeldahl Methods
Long-necked flask
(Kjeldahl flask)
used for digestion
to minimize
spattering losses.
Distillation unit;
right beaker
collects NH3 in
standard HCl.
16
8
Kjeldahl Example
A protein contains 16.2 wt% N. A 0.500 mL aliquot of
protein solution was digested and liberated NH3 was
distilled into 10.00 mL of 0.02140 M HCl. Unreacted
HCl needed 3.26 mL of 0.0198 M NaOH for titration.
What was the concentration of protein (mg/mL)?
How to do this?
–1) Total HCl present in beaker
–2) NaOH titrated against HCl (excess HCl)
–3) NH3 reacted (Total HCl- excess HCl)
–4) Nitrogen in protein; mass of protein
17
Kjeldahl Example (2)
Mol total HCl
(10.00 mL)(0.02140 mmol/mL) 0.2140 mmol
Mol NaOH (mol of excess HCl)
(3.26 mL)(0.0198 mmol/mL) 0.0645 mmol
Mol HCl used during distillation (mol NH3)
0.2140 mmol - 0.0645 mmol 0.1495 mmol NH 3
Mol N in protein = mol of NH3 (0.1495 mmol)
0.1495 mmol N
18
9
Kjeldahl Example (3)
Mass of N in protein
(0.1495 mmol)(14.00674 mg N/mmol) 2.093 mg N
Mass of protein
2.093 mg N

12.9 mg protein
0.162 mg N/mg protein
Concentration of protein
12.9 mg protein
mg protein

25.8
0.500 mL
mL
19
Spectrophotometric Titrations
We can use indicators to detect a color change at an
endpoint, but what if we want to monitor the reaction as
it takes place or if the color change is beyond the
sensitivity of the human eye?
We can use absorption of light to monitor the progress of
the reaction.
This combines spectroscopy (Ch. 18-20) with
volumetric analysis
20
10
Transferrin protein
Transferrin is a protein that
serves to transport iron in
biochemistry. Of course, this
means that transferrin can be
measured by titration with iron
(nitrilotriacetate).
Transferrin has a molecular
mass of 81,000.
Each
molecule binds two iron atoms
and the protein-Fe complex
absorbs radiation with a
maximum at 465 nm.
Apotransferrin 2Fe 3(Fe3) 2 transferrin
colorless
red
21
Titration of Transferrin with Ferric
Nitrilotriacetate
Absorbance can be
monitored during the
titration, but the volume
changes at each point.
So we must correct for
the dilution during the
titration:
Total Volume
Corrected Abs. 
(Observed Abs.)
Initial Volume
22
11
Correction of Absorbance
The absorbance measured after adding 125 L of ferric
nitrilotriacetate to 2.000 mL of apotransferrin was 0.260.
What would the corrected absorbance be?
2.125
Corrected Abs. 
(0.260) 0.276
2.000
23
Titration Curves (Precipitation)
How do concentration of analyte and titrant vary during the
titration?--Titration curve
–Understand the chemistry during the titration
–What parameters control the quality of the titration
What could affect the titration?
–pH
–Ksp
24
12
Titration Curves
The titration curve is a plot of how the concentration of a
reactant varies with titrant
We usually plot this as a power (p) function because of the
wide concentration ranges encountered:
pX -log10 
X
A titration curve has three distinct regions: before, at, and
after the equivalence point
25
Titration Curves
If you titrate 25.00 mL of 0.1000 M I- with 0.05000 M Ag+
and monitor Ag+:
I -(aq) Ag (aq)  AgI(s)
This is the reverse of the equilibrium:
AgI(s)  I -(aq) Ag (aq) K sp Ag  I - 8.3 10 17
–Because K for the titration reaction is large (1/Ksp =
1.2x1016), products are favored, so we’
d expect Ag+ to be
nearly used up during the titration until the equivalence
point when [Ag+] will increase strongly
The expected equivalence volume of titrant Ve:
 
(0.025L)(0.100M)
Ve 
50.00 mL
(0.05000 M)
26
13
Before the Equivalence Point
What is [Ag+] in flask once 10.00 mL of the Ag+ titrant
have been added?
–The initial mol of I- was:
I  (0.025L)(0.100M) 0.0025 mol
–The current mol of I- is (1:1 rxn with Ag+):

I  mol I -0 - mol Ag added
0.0025 mol - (0.0100 L)(0.050 M) 0.00200 mol

I 0.00200 mol/0.03500L 0.05714 M

–Concentration of Ag+ in equilibrium:
  
K sp 8.3 10-17
Ag   
1.4 10-15 M
I
0.05714

 


pAg  log Ag  - log 1.4 10-15 14.84
27
Shortcut Calculation
You can reduce the complexity of calculations a bit by
using a shortcut to calculate [I-]
–When 10 mL of the Ve (50 mL) have been added:
4 
25.00 

I  (0.1000 M)
0.0571 M
5
35.00


Frxn
Remaining

Conci

Dilution
[I-] is simply a function of the portion of the titration left to
go and how much the solution has been diluted.
28
14
Example (Before Equivalence)
Calculate pAg+ when VAg+ is 49.00 mL
1 
25.00 

I  (0.1000 M)
6.76 10
50
74.00

 

8.3 10
Ag K

I
6.76 10

sp


-17
4
 
4
M

M
1.2 10-13 M

pAg  log Ag  - log 1.2 10-13 12.91
Even with the titration 98% complete, the concentration of
Ag+ is negligible
29
At the Equivalence Point
At the equivalence point, we have added the
exact amount of Ag+ needed to react with I–pAg+ is found by solving for x from Ksp
Ag 
I K

-
sp
8.3 10 17 x 2
x 9.110 9
 


pAg  log Ag  - log 9.110-9 8.04
30
15
After the Equivalence Point
After equivalence, pAg+ is simply determined by
calculating the excess Ag+
–If we’
ve added 52.00 mL of Ag+ solution (77.00 total
mL):

mol Ag excess
(0.00200 L)(0.050 M) 0.000100 mol
Ag 0.000100 mol/0.07700L 1.30 10

 

-3
M

pAg  log Ag  - log 1.30 10-3 2.89
31
Shortcut Calculation
Again, you can shorten the process a bit:
–2.00 mL of excess Ag+ have been added:
2.00 
Ag (0.05000 M)

1.30 10
77.00


Conci

3
M
Dilution
[Ag+] is simply a function of how much excess has been
added and the dilution factor
32
16
Shape of the Titration Curve
If all the data is plotted, you get a
titration curve that looks similar to
this. The equivalence point is the
steepest point of the curve (1st
derivative) and an inflection point (2nd
derivative=0).
This is always true for systems
with 1:1 stoichiometry for the
reactants, whether it is a precipitation,
acid-base, compexation, or redox
titration. With other stoichiometries,
the curve will not be symmetric at the
equivalence point.
When we set conditions for a
titration, we try to make the steepest
part of the curve representative of the
equivalence point.
33
Effect of Ksp on the Curve
The less soluble the product of a
precipitation titration, the steeper the
titration curve will be at the equivalence
point. We have here the titration curves
for Ag+ with three halides, I-, Br-, and
Cl-.
Each titration curve is for a solution
of 25 mL of 0.10 M X- titrated with
0.050 M Ag+.
34
17
Example
25.00 mL of 0.04132 M Hg2(NO3)2 was titrated with
0.05789 M KIO3:
Hg 22 (aq) 2IO3- (aq)  Hg 2 (IO3 ) 2 (s)
Ksp for Hg2(NO3)2 is 1.3x10-18
Calculate the concentration of Hg22+ in solution (a) after
addition of 34.00 mL KIO3, (b) after addition of 36.00 mL
KIO3, and (c) at the equivalence point
35
Step 1
Volume of iodate needed to reach the
equivalence point (Ve):
2 mol IO3- 
(mol Hg 22 )
mol IO 
2

1 mol Hg 2 

3
2(25.00 mL Hg 22)(0.04132 M)
Ve 
35.69 mL
(0.05789 M)
36
18
Example A
If V=34.00 mL, Hg22+ precipitation is not complete
35.69 34.00 
 25.00

Hg 

(0.04132 M)
8.29 10
35.69
25.00 34.00
2
2


Frxn
Remaining

4

M
Conci
Dilution
Factor
37
Example B
If V=36.00 mL, the precipitation is complete and we have
gone past the equivalence point
36.00 35.69 
IO 

(0.05789 M) 2.9 10
25.00 36.00
3


Dilution
Factor
Conci
10
Hg IOK  (21..93
10
2
2
sp
 2
3
18
4 2
)
4
M
1.5 10 11 M
38
19
Example C
At the equivalence point, there is exactly enough IO3- to
react with Hg22+
Hg 2 (IO3 ) 2 (s)  Hg 22 (aq) 2IO3- (aq)
x
Hg ( x)(2 x)
2
2
2x
2
K sp
4 x 3 1.3 10 18
 
x Hg 22  6.9 10 7 M
39
Titration of a Mixture
With a mixture of two ions that are titrated, the less soluble
precipitate will form first
When the equivalence point of the less soluble precipitate
is obtained, the concentration of titrant suddenly increases
and then the more soluble salt begins to precipitate
Therefore, we should observe 2 breaks in the titration
curve, one corresponding to each equivalence point
40
20
Titration Curve of Mixture
Titration curve for AgNO3 vs.
and Cl-. AgI precipitates first
while Cl- remains in solution.
As [Ag+] increases, than AgCl
precipitates.
I-
The system used 0.0502 M
KI and 0.050 M KCl with
0.0845M AgNO3 titrant.
41
Titration of a Mixture
Comparing the mixture curve to the pure curves for I- and
Cl-, we find the endpoint for AgI is slightly higher (more
volume) for the mixture.
–Random error (this could be positive or negative)
–Coprecipitation (systematically positive)
In coprecipitation some Ag+ precipitates with Cl- early,
carried out on the surface of AgI
42
21
Technique to Monitor Titration
The technique used to
monitor the mixture titration of
Ag+ with I- and Cl- was
electrochemistry. An electrode
specific to Ag+ was used to
monitor [Ag+]
The principles of
electrochemistry and electrode
operation will be discussed in
Chs. 14-15. As discussed, this
is just one method of
monitoring a titration.
Spectroscopy is another.
43
Detection of End Point
Methods for determining the end point for precipitation
titrations
–Indicators
–Electrodes (Ch. 15)
–Light scattering
We’
ll discuss indicators used for the Ag+ Cl- titration
equilibrium
–Titrations with Ag+ are called argentometric titrations
44
22
Indicators
Volhard titration
–Uses the formation of a soluble colored complex at
the end point
Fajans titration
–A colored indicator is adsorbed onto the precipitate at
the end point
45
Volhard Titration
Used as a procedure for titrating Ag+; determination of Clrequires a back-titration
–First, Cl- is precipitated by excess AgNO3
Ag (aq) Cl -(aq)  AgCl(s)
–Excess Ag+ is titrated with KSCN in the presence of
Fe3+

-
Ag (aq) SCN (aq)  AgSCN(s)
–When Ag+ has been consumed, a red complex forms
Fe 3(aq) SCN -(aq)  FeSCN 2(aq)
red
The Volhard titration can be used for any anion that
forms an insoluble salt with silver
46
23
Fajans Titration
The technique uses an adsorption indicator. Prior
to the equivalence point, there is excess Cl- in solution.
Some is adsorbed on the surface of the crystal, giving a
partial negative charge. After the equivalence point,
there is excess Ag+ in solution. Some adsorbs to the
surface imparting a partial positive charge to the
precipitate. Choosing an indicator with a partial
negative charge will cause it to adsorb to the surface.
-O
O
O
Cl
Dichlorofluorescein is
green in solution but pink
when absorbed on AgCl
Cl
CO2-
47
Common Precipitation Titrations
48
24
Light Scattering Detection
In a precipitation reaction, the mixture gets cloudier
Particles in solution scatter light (turbidity).
–Turbidity ceases to change after the equivalence point
because no more precipitate is formed.
–Scattering is particle-size dependent, so conditions
much be very well controlled
Methods
–Turbidimetry-measures transmittance of light
–Nephelometry-measures light scattering
49
25