volumetric analysis - pw

Author: J R Reid

Volumetric Analysis

IntroductionThe EquipmentThe ProcessCalculations

Dr.S.Kumar

Author: J R Reid

Volumetric analysis

A titration is a lab procedure where a measured volume of one solution (burette) is added to a known volume (flask) of another solution until the reaction is complete

Dr.S.Kumar

Introduction

Often in chemistry we need to work out the concentration of a solution. There are a number of methods we could use, but they all involve working out the amount of the substance in a certain volume.Volumetric analysis involves using volumes of liquids to analyse a concentration. To do this we need the following things:

A chemical of a known concentration that will react with our ‘unknown’ concentration chemicalAn indicator that will tell us when all the chemical has been reactedA number of pieces of equipment that we can use to measure volume accurately

Titrimetric methods include a large and powerful group of quantitative procedures based on measuring the amount of a reagent of known concentration that is consumed by the analyte. Titrimetry is a term which 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.

There are three main types of titrimetry: volumetric titrimetry, gravimetric titrimetry, and coulometrtic titrimetry.

Volumetric titrimetry is used to measure the volume of a solution of known concentration that is needed to react completely with the analyte.

Gravimetric titrimetry is like volumetric titrimetry, but the mass is measured instead of the volume.

Coulometric titrimetry is where the reagent is a constant direct electrical current of known magnitude that consumes the analyte; the time required to complete the electrochemical reaction is measured.

The benefits of these methods are that they are rapid, accurate, convenient, and readily available.

Quantitative Classical Chemical Analysis

Titrations

Acid-base Precipitation Complexometric Redox

Gravimetry

04/10/23 Lecture Notes: Dr.Santosh Kumar

Quantitative Classical Chemical Analysis

Titrations

Acid-base Precipitation Complexometric Redox

Titrations involving iodine (I2)

Iodimetry

Iodometry

Iodometric titration of copper

Gravimetry

DichromatometricPermanganimetric


Titrations Examples

Acid-base Quantification of acetic acid in vinegar

Complexometric Quantification of chloride (Cl-) in water

Precipitation Water Hardness (Calcium and magnesium)

Redox Quantification of hydrogen peroxide (H2O2)


Titration example

Analyte Titrant Indicator

Acid-base Quantification of acetic acid in avinegar

Acetic acid (CH3COOH)

NaOH (sodium hydroxide)

Phenolphthalein

Complexometric Water Hardness (Calcium and magnesium)

Calcium and magnesium (Ca 2+ , Mg 2+)

EDTA Eriochrome black TMurexide

Precipitation Quantification of chloride (Cl-) in water

Chlordie AgNO3 (silver nitrate)

Mohr, Volhard,Fajans

Redox Quantification of hydrogen peroxide (H2O2)

Hydrogen peroxide (H2O2)

KMnO4 (potassium permanganate)

No indicator


Equivalence Point: Occurs in a titration at the point in which the amount of added titrant is chemically equivalent to the amount of analyte in a sample.

Back- Titration: This is a process that is sometimes necessary in which an excess of the standard titrant is added, and the amount of the excess is determined by back titration with a second standard titrant. In this instance the equivalence point corresponds with the amount of initial titrant is chemically equivalent to the amount of analyte plus the amount of back- titrant.

More Defining Terms

Titration: This is performed by adding a standard solution from a buret or other liquid- dispensing device to a solution of the analyte until the point at which the reaction is believed to be complete.

End point: The point in titration when a physical change occurs that is associated with the condition of chemical equivalence.

Equivalence Points and End Points

Indicators are used to give an observable physical change (end point) at or near the equivalence point by adding them to the analyte. The difference between the end point and equivalence point should be very small and this difference is referred to as titration error. To determine the titration error: Et= Vep - VeqEt is the titration error

Vep is the actual volume used to get to the end point

Veq is the theoretical value of reagent required to reach the end point

One can only estimate the equivalence point by observing a physical change associated with the condition of equivalence.

Primary Standards

A primary standard is a highly purified compound that serves as a reference material in all volumetric and mass titrimetric properties. The accuracy depends on the properties of a compound and the important properties are:

1. High purity

2. Atmospheric stability

3. Absence of hydrate water

4. Readily available at a modest cost

5. Reasonable solution in the titration medium

6. Reasonably large molar mass

Compounds that meet or even approach these criteria are few, and only a few primary standards are available.

Standard solutions

A standard is a solution of precisely known concentration It must be available in a highly pure stateIt must be stable in airIt must dissolve easily in waterIt should have a fairly high relative molecular wtIt should under go a complete and rapid reactionEndure a selective reaction with analyte

Non standard solutions

Sodium hydroxide absorbs carbon dioxide from atmosphereHCl can produce chlorine gas in reactions and liberate hydrogen when exposed to airNitric acid can act as an oxidising agent interfering with reactionsSulphuric acid absorbs water form the air

Example of titration and set up

http://wine1.sb.fsu.edu/chm1045/notes/Aqueous/Stoich/Aqua02.htm

http://wine1.sb.fsu.edu/chm1045/notes/Aqueous/Stoich/Aqua02.htm

Fact File 1: Introduction to iodometric and iodimetric titrations

Titrations:

Direct TitrationsIndirect TitrationsBack TitrationsIodometry

Fact File 1: Introduction to iodometric and iodimetric titrations

Titrations Example Type of reaction

Acid-base Quantification of acetic acid in vinegar ■ Direct Titration □ Indirect Titration □ Back Titration

Complexometric Water Hardness (Calcium and magnesium) ■ Direct Titration □ Indirect Titration □ Back Titration

Precipitation Quantification of Cl in Water

Mohr Method

■ Direct Titration □ Indirect Titration □ Back Titration

Fajans Method ■ Direct Titration □ Indirect Titration □ Back Titration

Volhard Method □ Direct Titration □ Indirect Titration ■ Back Titration

Redox Quantification of hydrogen peroxide (H2O2) ■ Direct Titration □ Indirect Titration □ Back Titration


EQUIPMENTEQUIPMENT

The Equipment

Volumetric analysis involves a few pieces of equipment that you may not have seen before:Pipette – for measuring accurate and precise volumes of solutions

Burette – for pouring measured volumes of solutionsConical flask – for mixing two solutions

Wash bottles – these contain distilled water for cleaning equipment

Funnel – for transfer of liquids without spilling

Volumetric flasks – a flask used to make up accurate volumes for solutions of known concentration

Apparatus used

BuretteVolumetric flask BeakerPipetteFunnelIndicatorWhite tile

Burette titration procedures

The Process - Preparation

Two solutions are used:The solution of unknown concentration;The solution of known concentration – this is also known as the standard solution

Write a balanced equation for the reaction between your two chemicalsClean all glassware to be used with distilled water. The pipettes and burettes will be rinsed with the solutions you are adding to them

Process – The Setup

The burette is attached to a clamp stand above a conical flaskThe burette is filled with one of the solutions (in this case a yellow standard solution)A pipette is used to measure an aliquot of the other solution (in this case a purple solution of unknown concentration) into the conical flaskPrepare a number of flasks for repeat testsLast, an indicator is added to the conical flask

Process – The Titration

Read the initial level of liquid in the buretteTurn the tap to start pouring out liquid of the burette into the flask. Swirl the flask continuously. When the indicator begins to change colour slow the flow.When the colour changes permanently, stop the flow and read the final volume. The volume change needs to be calculated (and written down). This volume is called a titreRepeat the titration with a new flask now that you know the ‘rough’ volume required. Repeat until you get precise results

Precautions when using equipment

Burette must be vertical, use and then remove funnel, check meniscus, rinse with de-ionised water and then given solution.In using a Pipette rinse with de-ionised water first and then with given solution. Check meniscus. Do not blow out remainder of liquid into flask and keep tip of pipette in contact with flask

Precautions

Conical flask should not be rinsed with solution it is to contain and swirl In using a Volumetric flask the last few cm³ must be added so that the meniscus rest on calibration mark Invert stoppered flask to ensure solution is homogeneous/uniform

Why is a conical flask, rather than a beaker, used in the experiment?

To allow easy mixing of the contents, by swirling.

Why is the funnel removed from the burette after adding the acid solution?

So that drops of solution from the funnel will not fall into the burette.

Author: J R Reid

In using a burette, why is it important (a) to rinse it with a little of the solution it is going to contain. (b) to clamp it vertically. (c) to have the part below the tap full?

Solution (a) Rinsing

To remove any residual water, and so avoid dilution of the acid solution when it is poured into the burette.

Solution (b) clamp vertically

To enable the liquid level to be read correctly

Solution (c) Full tap

To ensure that the actual volume of liquid delivered into the conical flask is read accurately.

Author: J R Reid

The following procedures were carried out during the titration: The sides of the conical flask were washed down with deionised water.The conical flask was frequently swirled or shaken. Give one reason for carrying out each of these procedures.

To ensure that all of the acid added from the burette can react with the base.

To ensure complete mixing of the reactants

Why is a rough titration carried out?

To find the approximate end-point. This information enables the subsequent titrations to be carried out more quickly.

Why is more than one accurate titration carried out?

To minimise error by getting accurate readings within 0.1 cm3 of each other.

Calculations

Volume of acid Va (cm3) is the titration figure from buretteThe concentration of acid is Ca (mol)na is the mol of full balanced equation per litre

Volume of base is Vb (cm3). Usually placed in the conical flask.Cb is the concentration of the basena is the mol of full balanced per litre

Calculations

USE FORMULA Va. Ca = Vb. Cb na nb

Va = 37cm3

Ca is unknownna = 2Vb = 25 cm3

Cb = 0.1 molnb = 1

37 cm3× Ca = 25cm3 × 0.1mol

2 1

Ca = 25 × 2 × 0.1 = 0.13 mol/L

37

VVa. Ca a. Ca = = Vb. CbVb. Cb na na nb nb


EXAM QUESTIONS

Look out for dilution factors e.g vinegarChoice of indicatorType of vol flask given 1L OR 250 Cm³ as you have to adjust in your calculationsTake titre reading from burette and given vol of solution is taken from conical flaskUSE FORMULA Va. Ca = Vb. Cb

na nb

Author: J R Reid

Volumetric Analysis Calculations

Calculations – Mean Titre

We will have a number of titres for each solution we analysed. The first thing we do is to calculate the mean (average) titre:

Titres = 12.6ml 13.0ml 13.1ml 12.9ml

Mean = Sum of the titres / number of titres= (13.0 + 13.1 + 12.9) / 3= 13.0ml

Why did we discard the 12.6ml reading?

Calculations – The Unknown Concentration – Preparation

1. Write down the balanced equation e.g.H2SO4 + 2NaOH → Na2SO4 + 2H2O

2. Write down everything else we know. This will be:

a. Volume of liquid in the pipetteb. Mean titre (from burette)c. The concentration of the standard solutiond. Was the standard solution in the pipette or in

the burette?

Calculations – The ‘Unknown’ Amount

3. Now calculate the amount in the standard solution you used. Use the n = cv formula. Remember: the millilitres must always be converted into litres for these formulae

4. Now that you know how many moles of the standard you used, look at the balanced equation. Would you need more or less of the ‘unknown’ substance in a balanced reaction?

If more, then how much more – two times, three times?If less, then how much less – half as much, one third?

We can calculate the amount of the unknown:We multiply if we need more i.e. 2x, 3x, …etcWe divide if we need less i.e. ½ = divide by 2, …etc

Calculations – The ‘Unknown’ Concentration

5. Now we have the volume and amount of the ‘unknown’ substance. We can now rearrange our n = cv formula to say c = n/v

Remember: All the calculations must be in litres (not millilitres)The final value must have units (molL-1) written after it

Example:

1. H2SO4 + 2NaOH → Na2SO4 + 2H2O

2. - Standard solution = NaOH (in burette) = 0.1molL-1

- Unknown concentration = H2SO4 (from 20ml pipette)- Titres = 12.6ml, 13.1ml, 13.0, 12.9ml- Average titre = (13.1+13.0+12.9) / 3 = 13.0ml

3. Amount of NaOH = cv = 0.1 x (13/1000) = 0.0013mol

4. Amount of H2SO4 = half of NaOH = 0.0013/2 = 0.00065mol

5. Concentration H2SO4 = n/v = 0.00065/(20/1000) = 0.325molL-1

Titration examples A

Titres: 12.1mL, 12.3mL, 12.1mL, 12.0mL

Known solution details: HCl in the burette, Concentration = 0.522 molL-1

Unknown solution details: NaOH15mL aliquots

Calculations:

HCl + NaOH → NaCl + H2O

Titration examples B

CH3COOH + NaOH → NaCH3COO + H2O

Titres: 17.6mL, 18.5mL, 17.4mL, 17.5mL

Known solution details: NaOH in the burette, Concentration = 0.103 molL-1

Unknown solution details: CH3COOH15mL aliquots

Calculations:

Titration examples C

2HCl + Na2CO3 → 2NaCl + H2O + CO2

Titres: 12.8mL, 12.8mL, 12.8mL, 12.9mL

Known solution details: HCl in the burette, Concentration = 0.555 molL-1

Unknown solution details: Na2CO3 25mL aliquots

Calculations:

Titration examples D

H2SO4 + 2NaOH → Na2SO4 + 2H2O

Titres: 12.1mL, 12.3mL, 12.1mL, 12.0mL

Known solution details: NaOH in the burette, Concentration = 1.04 molL-1

Unknown solution details: H2SO4 10mL aliquots

Calculations:

Level 3 – Spot the difference…

Here is an extract from a level 3 titration assessment. It shows an example of a chemical reaction that could be used in a titration:

Hypochlorite ions react with iodide ions according to the equation;

OCl + 2I- + 2H+ Cl + I2 + H2O The iodine produced is then titrated with standardised sodium thiosulfate

solution. It reacts according to the equation below.

I2 + 2S2O32 2I + S4O6

2–

Since starch turns blue in the presence of iodine, it is used as an indicator

for this final reaction. The overall equation for both reactions is:

OCl + 2H+ + 2S2O32 Cl + S4O6

2 + H2O

Example One:Jo has a solution of Hydrochloric acid (HCl) that she does not know the concentration of. She decides to use a standard solution of Sodium hydroxide (NaOH), 0.112molL-1, for the titrationShe uses a 15mL pipette to measure aliquots of NaOH into her flasks. She then titrates these with the HCl from the burette until the indicator turns from purple to colourlessThese are her results:

Titres of HCl: 23.3mL, 22.8mL, 22.9mL, 22.7mL

Balanced equation:

→

NaOHc =

n =

v =

HClc =

n =

v =

Example Two:Jim has a solution of Sodium carbonate (Na2CO3), but the label is missing. To work out the concentration he decides to use a standard solution of Hydrochloric acid (HCl) with a concentration of 0.322molL-1.He uses a 20mL pipette to measure aliquots of HCl into his flasks. He then titrates these with the Na2CO3 from the burette until the indicator turns from colourless to purpleThese are his results:

Titres of Na2CO3 : 18.2mL, 18.2mL, 17.9mL, 18.3mL

Balanced equation:

→

HClc =

n =

v =

Na2CO3

c =

n =

v =

Example Three:Jill has a bottle of vinegar (ethanoic acid solution – CH3COOH). The label says that it should have 3.1gL-1 of ethanoic acid in it. To work out the real concentration she decides to use a standard solution of Sodium hydroxide (NaOH) with a concentration of 1.22 x 10-2 molL-1.She uses a 25mL pipette to measure aliquots of vinegar into her flasks. She then titrates these with the NaOH from the burette until the indicator turns from colourless to purpleThese are her results:

Titres of NaOH : 22.1mL, 22.5mL, 22.6mL, 22.4mL

Balanced equation:

→

NaOHc =

n =

v =

CH3COOHc = molL-1

gL-1

n =

v =

Describe the preparation of a 5.0 L of 0.10 M Na2CO3 (105.99 g/mol) from the primary standard solution.

Amount Na2CO3 = n Na2CO3 (mol) = Volume solution x c Na2CO3 (mol/ L) = 5 L x 0.1 mol Na2CO3 = 0.5 mol Na2CO3

L Mass Na2CO3 = mNa2CO3=0.5 mol Na2CO3 x 105.99 g Na2CO3 =53 g Na2CO3

mol Na2CO3

The solution is prepared by dissolving 53 g of Na2CO3 in water and diluting to 5 L.

Example: Calculating the Molarity of Standard Solutions

How would you prepare 50mL portions of standard solutions that are 0.005 M, 0.002 M, and 0.001 M in a standard 0.01 M Na+?

To solve this the relationship Vconcd x cconcd = Vdil x cdil

Vconcd = Vdil x cdil = 50mL x 0.005 mmol Na+ /mL = 25mLcconcd 0.01mmol Na+ /mL

To produce 50mL of 0.005 M Na+, 25mL of the concentration solution should be diluted to 50mL.

Example: Calculating the Molarity using different algebraic relationships

The following two examples show the two types of volumetric calculations. The first involves computing the molarity of solutions that have been

standardized against either a primary standard or another standard solution. The second example involves calculating the amount of analyte in a sample

from titration data.

How to deal with titration data…

Example: Molarity of solutions that have been standardized

A 50mL volume of HCl solution required 29.71mL of 0.01963 M Ba(OH)2 to reach an end point with bromocresol green indicator. Calculate the molarityof the HCl.

Stoichiometric ratio= 2 mmol HCl/ 1 mmol Ba(OH)2Amount Ba(OH)2 = 29.71 mL Ba(OH)2 x 0.01963 mmol Ba(OH)2

mL Ba(OH)2Amount HCl = (29.71 x 0.01963) mmol Ba(OH)2 x 2 mmol HCl

1 mmol Ba(OH)2C HCl = (29.71 x 0.01963 x 2) mmol HCL

50mL solution = 0.023328 mmol HCl = 0.0233M mL solution

Example: Amount of analyte in sample from titration

Titration of 0.2121 g of pure Na2C2O4 (134 g/mol) required 43.31 mL of KMnO4. What is the molarity of the KMnO4 solution?

Stoichiometric ratio = 2 mmol KmnO4/ 5 mmol Na2C2O4Amount Na2C2O4= 0.2121 g Na2C2O4 x 1 mmol Na2C2O4 0.134 g na2C2O4Amount KMnO4 = 0.2121 mmol Na2C2O4 x 2 mmol KMnO4 0.134 5 mmol Na2C2O4C KMnO4 = ( 0.2121 x 2) mmol KMnO4

0.134 5 = 0.01462M 43.31 mL KMnO4

Example: Computing analyte concentrations from titration data

A 0.8040g sample of an iron ore is dissolved in acid. The iron is then reduced to Fe2+ and titrated with 47.22mL of 0.02242 M KMnO4 solution. Calculate the results of this analysis in terms of percent Fe (55.847 g/mol).

Stoichiometric ratio = 5 mmol Fe2+/ 1 mmol KMnO4Amount KMnO4 = 47.22mL KMnO4 x 0.02242 mmol KMnO4

mL KMnO4Amount Fe2+ = (47.22 x 0.02242) mmol KMnO4 x 5 mmol Fe2+

1 mmol KMnO4Mass Fe2+ = (47.22 x 0.02242 x 5) mmol Fe2+ x 0.055847 g Fe 2+

mmol Fe2+% Fe2+ = (47.22 x 0.02242 x 5 x 0.055947) g Fe 2+ x 100% = 36.77%

0.8040 g sample

Titration Curves

Example of a sigmoidal titration curve once calculations of data have been computed.

www.psigate.ac.uk/newsite/ reference/plambeck/chem1/p01173.htm

ReferencesSkoog, D., West, D., Holler, F.J., & Crouch, S. (2000). Analytical Chemistry: An Introduction. 7th ed. Thomson Learning, Inc: United States of America.http://wine1.sb.fsu.edu/chm1045/notes/Aqueous/Stoich/Aqua02.htmwww.psigate.ac.uk/newsite/ reference/plambeck/chem1/p01173.htmhttp://www2.hmc.edu/~karukstis/chem21f2001/tutorials/tutorialStoichiFrame.html

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