Transcript Titration
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