chapter 7 let the titrations begin. titration n titration –a procedure in which one substance...

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.


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 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.


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.


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.


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+


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.


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


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


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.


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


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


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)


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


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


At the equivalence point– Stoichiometric amount of Ag+ as I-

so all has precipitated out as AgI(s)• Regular Ksp calculations


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.


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.


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