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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+ + C2O4
2- + 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 Oxidation-Reduction 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 Hg2
2+ 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 pHg2
2+ 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