Chapter 7
Let the
Titrations Begin
Titration

Titration
– A procedure in which one substance (titrant) is
carefully added to another (analyte) until
complete reaction has occurred.
• The quantity of titrant required for complete
reaction tells how much analyte is present.

Volumetric Analysis
– A technique in which the volume of material
needed to react with the analyte is measured
Titration Vocabulary

Titrant
– The substance added to the analyte in a
titration (reagent solution)

Analyte
– The substance being analyzed

Equivalence point
– The point in a titration at which the quantity of
titrant is exactly sufficient for stoichiometric
reaction with the analyte.
Titration Vocabulary

End point
– The point in a titration at which there is a
sudden change in a physical property, such as
indicator color, pH, conductivity, or absorbance.
Used as a measure of the equivalence point.

Indicator
– A compound having a physical property (usually
color) that changes abruptly near the
equivalence point of a chemical reaction.
Titration Vocabulary

Titration error
– The difference between the observed end point
and the true equivalence point in a titration

Blank Titration
– One in which a solution containing all reagents
except analyte is titrated. The volume of
titrant needed in the blank titration should be
subtracted from the volume needed to titrate
unknown.
Titration Vocabulary

Primary Standard
– A reagent that is pure enough and stable
enough to be used directly after weighing.
Then entire mass is considered to be pure
reagent.

Standardization
– The process whereby the concentration of a
reagent is determined by reaction with a known
quantity of a second reagent.
Titration Vocabulary

Standard Solution
– A solution whose composition is known by virtue
of the way it was made from a reagent of known
purity or by virtue of its reaction with a known
quantity of a standard reagent.

Direct Titration
– One in which the analyte is treated with
titrant, and the volume of titrant required for
complete reaction is measured.
Titration Vocabulary

Back Titration
– One in which an excess of standard reagent is
added to react with analyte. Then the excess
reagent is titrated with a second reagent or
with a standard solution of analyte.
Titration Calculations


Titration Calculations rely heavily on
the ability to perform stoichiometric
calculations.
Examples
Titration Calculation
Examples

How many milligrams of oxalic acid
dihydrate, H2C2O4 . 2H2O (FM =
126.07), will react with 1.00 mL of
0.0273 M ceric sulfate (Ce(SO4)2) if
the reaction is:
H2C2O4 + 2Ce4+  2CO2 + 2Ce3+ + 2H+
Titration Calculation
Examples

A mixture weighing 27.73 mg
containing only FeCl2 (FM = 126.75)
and KCl (FM = 74.55) required 18.49
mL of 0.02237 M AgNO3 for
complete titration of the chloride.
Find the mass of FeCl2 and the weight
percent of Fe in the mixture.
Titration Calculation
Examples

The calcium content of urine can be determined by the
following procedure
– Step 1. Ca2+ is precipitated as calcium oxalate in basic solution:
Ca2+ + C2O42- + H2O  Ca(C2O4) . H2O(s)
– Step 2. After the precipitate is washed with ice-cold water to
remove free oxalate, the solid is dissolved in acid, which gives
Ca2+ and H2C2O4 in solution.
– Step 3. The dissolved oxalic acid is heated at 60oC and
titrated with standardized potassium permanganate until the
purple end point of the following reaction is observed:
5H2C2O4 + 2MnO4- + 6H+  10CO2 + 2Mn2+ + 8H2O
Titration Calculation
Examples


Standardization: Suppose that 0.3562 g of
Na2C2O4 is dissolved in a 250.0 mL volumetric
flask. If 10.00 mL of this solution require 48.36
mL of KMnO4 solution for titration, what is the
molarity of the permanganate solution?
Calcium in a 5.00-mL urine sample was
precipitated, was redissolved, and then required
16.17 mL of standard MnO4- solution. Find the
concentration of Ca2+ in the urine
Titration Calculation
Examples

A solid mixture weighing 1.372 g containing only
sodium carbonate and sodium bicarbonate required
29.11 mL of 0.734 M HCl for complete titration:
Na2CO3 + HCl  2NaCl(aq) + H2O + CO2
NaHCO3 + HCl  NaCl(aq) + H2O + CO2
Find the mass of each component of the mixture.
Titration Calculations

Kjeldahl Nitrogen Analysis
– Digest the organic compound in boiling
H2SO4 to convert Nitrogen to NH4+
• Hg, Cu, or Se will catalyze the digestion
process
• Raise the b.p. of H2SO4 (338oC) by adding
K2SO4 to increase the rate of the reaction
Titration Calculations

Kjeldahl Nitrogen Analysis
– Treat the ammonium compound with base
and distill as ammonia into a standard
acidic solution
• NH4+ + OH-  NH3(g) + H2O
• NH3 + H+  NH4+
– The moles of acid consumed equal the
moles of NH3 liberated
• H+ (HCl) + OH-(NaOH)  H2O
Titration Calculation
Examples

A typical protein contains 16.2 wt%
nitrogen. A 0.500-mL aliquot of protein
solution was digested, and the liberated
ammonia was distilled in 10.00 mL of
0.02140 M HCl. The unreacted HCl
required 3.26 mL of 0.0198 M NaOH for
complete titration. Find the concentration
of protein (mg protein / mL) in the original
sample.
Spectrophotometric
Titrations

One in which absorption of light is
used to monitor the progress of the
chemical reaction
Spectrophotometric
Titration Example

2Fe3+ + Apotransferrin  (Fe3+)2transferrin
–
Apotransferrin is clear
– (Fe3+)2transferrin is red
– Absorption increases as a result of red from Iron
attachment to the apotransferrin
• Saturated protein – no further color change – absorption
levels off
• Extrapolated intersection is the equivalence point
Spectrophotometric
Titration Example

Construction of Graph
– When constructing the graph of the
absorbance values versus the
concentration of Fe3+ you must take in
account the dilution factor
– Corrected absorbance = (total Volume / initial
Volume)(observed absorbance)
Corrected Absorbance
Calculations

The absorbance measured after
adding 125 L of ferric
nitrilotriacetate to 2.000 mL of
apotransferrin was 0.260. Calculate
the corrected absorbance that should
be plotted.
Titration Curve


Titration Curve is a result of plotting p[X]
(p[X]=-log [X]) versus the concentration of
the titrant
Precipitation Titration
– Concentration of analyte, concentration of
titrant and the Ksp influence the sharpness of
the endpoint

Acid-Base reaction and OxidationReduction reactions
– Have to calculate the theoretical endpoint to
determine the indicator to be used
Precipitation Titration Curve

Example – AgI(s) formation
– Ag+ + I-  AgI(s)
• Reverse of the Ksp reaction for the
dissociation of AgI(s) so the K for this
reaction equal 1/Ksp[AgI(s)] = 1.2 X 1016
– Large K value means that the equilibrium lies to
the right for this reaction
– Each aliquot addition of Ag+ titrant reacts
virtually entirely to form AgI(s)
Precipitation Titration Curve

Titration Curves typically exhibit 3
distinct regions for a single titrant
reacting with a single analyte.
– Before the Equivalence Point
– At the Equivalence Point
– After the Equivalence Point
Precipitation Titration Curve

Before the Equivalence Point
– Example of AgI(s) formation
• Most of the [Ag+] reacts entirely to give
AgI
• p[Ag+] =
– the amount of Ag+ left in solution
» The amount of I- present
» From Ksp determine the amount of Ag+
present
Precipitation Titration Curve

At the equivalence point
– Stoichiometric amount of Ag+ as I- so
all has precipitated out as AgI(s)
• Regular Ksp calculations
Precipitation Titration Curve

After Equivalence Point
– [Ag+] is determined by the Ag+ present
after the equivalence point with the
dilution factor taken into account
Precipitation Titration Curve
Equations

Calculate the Volume of Titrant needed to reach
equilibrium
MTVT = (Mole ratio)MAVA

Before the Equivalence Point
[A] = (Fraction of Titrant Remaining)(M of Analyte)(Original Volume of
solution / Total Volume of solution)
[T] = Ksp / [A]
pT = -log[T]
Precipitation Titration Curve
Equations

At Equivalence Point
[A][T] = Ksp
Calculate like in section 6-3

After the Equivalence Point
[T] = (Original Concentration of Titrant)(Volume of excess Titrant /
Total Volume of Solution)
pT = -log[T]
Precipitation Titration Curve
Example

25.00 mL of 0.1000 M I- was titrated with
0.05000M Ag+.
Ag+ + I-  AgI(s)
The solubility product for AgI is 8.3 x 10-17.
Calculate the concentration of Ag+ ion in
solution
(a) after addition of 10.00 mL of Ag+;
(b) after addition of 52.00 mL of Ag+;
(c) at the equivalence point.
Precipitation Titration Curve
Example

25.00 mL of 0.04132 M Hg2(NO3)2 was
titrated with 0.05789 M KIO3.
Hg22+ + 2IO3-  Hg2(IO3)2(s)
The solubility product for Hg2(IO3)2 is 1.3
x 10-18. Calculate the concentration of
Hg22+ ion in solution
(a) after addition of 34.00 mL of KIO3;
(b) after addition of 36.00 mL of KIO3;
(c) at the equivalence point.
Precipitation Titration Curve
Example

Consider the titration of 50.00 mL of
0.0246 M Hg(NO3)2 with 0.104 KSCN.
Calculate the value of pHg22+ at each of the
following points and sketch the titration
curve:
0.25VT
0.5VT
0.75VT
1.05VT
1.25VT
Titration Curve Shape

1:1 Stoichiometry of Reagents
– Equivalence point is the steepest point of a curve
• Maximum slope
• An inflection point

Other Stoichiometries
– The curve is not symmetric about the equivalence point
– The equivalence point is not at the center of the
steepest section of the curve

The less soluble the product, the sharper the
curve around the equivalence point
Titration Curve Shape
Titration of a Mixture


The less soluble product forms first
If there is sufficient difference in
solubility of products
– First precipitation is nearly complete
before the second one begins
• Separation by precipitation (section 6-5)
– Coprecipitation
• Alters the expected endpoints
Titration of a Mixture
Example
Chapter 7 - Homework

Problems – 2, 4, 8, 11, 12, 18, 22, 23,
28