CHAPTER 11
Titrations:
Taking Advantage of Stoichiometric
Reactions
Terms Defined
• Titrimetry – includes a group of analytical
methods based on determining the quantity of a
reagent of known concentration that is required to
react completely with the analyte. The reagent
may be a standard solution of a chemical or
electric current of known magnitude.
• Volumetric Titrimetry – a type of titrimetry in
which the standard reagent is measured
volumetrically.
• Gravimetric Titrimetry – titrimetry in which
the standard reagent is measured by mass
instead of volume.
• Standard Solution – a reagent of known
concentration that is used in titrimetric
analysis. A primary standard is an ultrapure
compound that serves as a reference
material in titrations.
• Equivalence point – point in a titration
when the amount of added titrant is
chemically equivalent to the amount of
anlayte.
• End point – point in a titration when a
physical change occurs that is associated
with the condition of chemical equivalence.
• Back Titration – a process in which the
excess of a standard solution used to
consume an analyte is determined by
titration with a second standard solution.
Back titrations are usually required when
the rate of reaction between the analyte and
reagent is slow or when the standard
solution lacks stability.
Equivalence and End Points
• Experimentally, the equivalence point is difficult
to observe. We near the equivalence point when
we observe physical changes in the reaction with
the aid of an indicator or an instrument such as a
pH meter. (Titration methods are used throughout
the lab sessions. Of particular interest are Lab
27C-7 Standardization of Sodium Hydroxide
against Potassium Hydrogen Phalate and 27 J1Potentiometric Titration)
Titration Curves
• Titration curves can be
used to determine the
end point of a titration
• One such curve is the
Sigmoidal curve. It is
a useful tool to
illustrate changes in
concentration and pH
of the analyte. (Fig 112 a)
Primary Standard Requirements
•
•
•
•
•
•
•
High purity
Thermally stable
Anhydrous
Large molar mass
Fast stoichiometric reaction with analyte
99.999% reaction completion
Available at moderate cost
Concentration
The concentration of a standard solution
used in titration is vital to the accuracy of
the titrimetric method. In order to obtain
the proper concentration, either the direct
method or standardization is used.
• Direct method – carefully weighed quantity
of primary standard is dissolved in solvent
and diluted to known volume
• Standardization – the titrant to be
standardized is used to titrate either a) a
weighed quantity of primary standard b) a
weighed quantity of a secondary standard –
a compound whose purity has been
established by chemical analysis and that
serves as the reference material c) a
measured volume of another standard
solution.
• When discussing concentration, it is
important to note that concentration is most
often expressed in terms of molarity, M,
which is moles/Liter or mmol/mL. Moles,
molar mass and volumes can be determined
by understanding certain relationships
between molarity and the standards and
analytes used as follows:
Moles A = Volume A (L) X Concentration A
(mol/L)
Mass A = Moles A X Molar mass (g/mol)
Examples
p. 261 11-6
a)
b)
c)
d)
2.00 L X 2.76 x10-3 mol/L X 1000 mmol/mol = 5.52 mmol
750 mL X 0.0416 mol/mL = 31.3 mmol
4.20 g CuSO4/ 106 g soln X 1.00 g soln/mL soln X 1mmol/0.1596 g CuSO4
X 250 mL soln = 6.58 X 10-3 mmol
3.50 L X 0.276 mol/L X 1000 mmol/mol = 966 mmol
p. 261 11-8
a)
b)
26.0 mL X 0.150 mmol/mL X 0.342 g/mol X 1000 mg/g = 1.33 X 103 mg
2.92 L X 5.23 x10-3 mol/L X 34.02 g/mol X 1000 mg/g = 520 mg
Examples
Page 262 11-22 Back titration
VHClO4 = 27.43 mL HClO4 = 1.0972 mL HClO4
VNaOH 25.00 mL NaOH mL NaOH
Volume of HClO4 needed to titrate 0.4793 g of Na2CO3 is:
40.00 mL HClO4 – 8.70 mLNaOH X 1.0972 mL HClO4
mL NaOH
= 30.45 mL
Thus:
0.4793 g Na2CO3/ 30.45 mL HClO4 X 1 mmol Na2CO3/0.10599 g Na2CO3 X
2mmol HClO4/ mmol Na2CO3 = 0.2970 M HClO4
Page 263 11-27
MAg2O3 = 197.84 g/mol
Amount Ag+ required for Ag3AsO4= total mmol Ag+ - # mmol KSCN
40.00mL X 0.07891 M – 11.27 mL X 0.1000M = 2.0294 mmol Ag+
2.0294 mmol Ag+ X 1mmolAs/3mmol Ag X 1mmol Ag2O3/2mmol Ag X
0.19784 g Ag2O3/mmolAs2O3