2 1 Analytical Teachniques Titrtion 2013 Part 1 of 25

7/27/2019 2 1 Analytical Teachniques Titrtion 2013 Part 1 of 25

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Subtopic 2.1: Volumetric Analysis 1

To identify the constituent(s) and to

determine its amount in the sam le

Titrimetric

analysis

Gravimetric

analysis

Instrumental

analysis

Examples

1. Acid-base Titration

2. Redox Titration

3. Indirect Titration

4. Back Titration

Examples

To analyse chlorideion in water.

Ag+(aq)+ Cl

-(aq)

AgCl(s)

Examples

1. Atomic Absorption

Spectroscopy (AAS)2. UV-Visible

Spectroscopy

3. Mass Spectroscopy4. Nuclear Magnetic

Resonance (NMR)

Chromatography

To identify the

organic compounds

in a mixture

Qualitative Analysis

To identify the constituent(s) in

the sam le

Flame Test

To identify

the metal

ions

Topic 2 - Analytical Techniques

Quantitative Analysis


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Assumed Knowledge

Reactions

Oxidation is defined as the gain of oxygen, the loss of electrons, or the increase in oxidation number. (2, 3, 4) Reduction is defined as the loss of oxygen, the gain of electrons, or the decrease in oxidation number. (2, 3, 4) Oxidation and reduction are complementary processes. (2, 3, 4) When electrons are transferred in a reaction, half-equations can be written for the gain and the loss of electrons; full

equations can be written by combining two half-equations. (2, 3, 4)

Chemical Calculations

The quantities of different substances can be conveniently compared by the use of the mole as a unit. (2) The molar mass of a substance can be derived by the addition of the relative atomic masses of the elements present, with

the answer expressed in grams. (2)

The amount of a substance (in moles) is related to the mass, m (in grams), and the molar mass, M (in g mol -1). (2) The molar concentration (or molarity), c (in mol L -1), of a solution is related to the amount of solute, n (in moles), and the

volume of the solution, V(in litres). (2)

The concentration of a solution can be related to the mass of solute (in grams) and the volume of the solution (in litres).(2)

The relative amounts (in moles) of substances reacted or produced during a reaction are indicated by the coefficients inthe balanced equation for the reaction. (2)

Given the equation for a reaction, the quantity of one reactant or product involved in a chemical reaction can be used todetermine the quantity of another. (2)

Numerical answers are limited by the least accurate data used in the calculation. (2) Large and small quantities are more conveniently expressed by means of scientific notation. (2) Calculators frequently display scientific notation in different ways. (2) Substances vary in density, which is the mass of material per unit volume. (2)


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Basic Terms Symbols Units

1.Amount of substances or

no. of molesn mol

2. Relative Atomic Mass A r -

3.Relative molecular/formula

MassM r -

4. Molar Mass M gmol-1

5. Mass m g

6. Volume V mL

7. Concentration

(i)Molarity

Concentration in molL-1

cin molL

-1molL

-1

(ii) Concentration in gL-1

cin gL-1

gL-1

(iii) Percentage in w/w %w/w %

(iv) Percentage in w/v %w/v %

(v) Percentage in v/v %v/v %

(vi) Parts per million cin ppm ppm

(vii) Parts per billion cin ppb ppb

M in analytical chemistry is abbreviation for 3 chemical terms:

1. Molarity - widely used in SPM or O-level or equivalent however in SAM, we used c to represent it.2. Molar mass3. To represent concentration in molL-1 e.g. 0.123 M means 0.123 molL-1.


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Significant figures in chemistry calculation

number Scientific notation Significant figures

0.123 1.23 x 10-1

3

0.0123 1.23 x 10-2

3

0.0120 1.20 x 10-2

3

123 1.23 x 102

3

120.0 1.200 x 102

4

120 1.2 x 102

2

500 5 x 102

1

Consider these questions

(a) 123 x 31 = (b) 123 x 31.00 =

(c)

(d)

(e) 1111 x 44.0 = (f)

(g) [Answer of (f)] x 0.21= (h) [answer of (g)] x 0.010000 =


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For Solid For Solution

M

mn

cVn

Concentration, c

V

nc 1molL

Mcc 11 molLgL

V

mc 1gL

Given a chemical reaction,

aA + bB dD + eE Where, A, B, D, E = substancesa, b, d, e = coefficient

b

a

n

n

B

A

d

b

n

n

D

B

a

e

n

n

A

E

a

d

n

n

A

D

100%xm(sample)

m(analyte)w/w%

gin

gin Parts per million

ppm = mgL-1

= mgkg-1

= gg-1

=

100%xV(sample)

m(analyte)w/v%

mLin

gin

Parts per billion

100%xV(sample)

V(analyte)v/v%

mLin

mLin

ppb= gL-1

= gkg-1

= ngg-1

=

In a dilution process,

Dilution Factor = original

diluted

V

V

dilutedoriginal nn

Important Formulae


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c (gL-1

)

x M (gmol-1

)

% w/v

10x 1000

c (mgL-1

)c (ppm)

c (gL-1

)

x 1000

c (ppb)

n (mol)

x V (L)

m (g)

x M (gmol-1

)

% w/w % w/v

x

As long as it is 10

-6

(i.e. per million)

c(mgkg-1

) c(gg-1

)

c(ngmg-1

) c(gMg-1

)

c(kgGg-1

) c(gkL-1

); etc

As long as it is 10-9

(i.e. per billion)c(gkg-1

) c(ngg-1

)

c(pgmg-1

) c(mgMg-1

)

c(gGg-1

) c(gML-1

); etc

V (mL)

density (gmL-1

)

x

% v/v

x

c (ppm)

c (ppm)

Please note

1 L = 1 kgfor dilute

aqueous

solutions

c (molL-1

)


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For 0.475 g of MgCl2

(i) Calculate the molar mass for 0.475 g of MgCl2.

M(MgCl2) = M(Mg) + [2 x M(Cl)]

= 24.31 + 2 x 35.45

= 95.21 gmol-1

(ii) Calculate the number of moles for 0.475 g of MgCl2.

n(MgCl2) = m(MgCl2) / M(MgCl2)

= 0.475g / 95.21gmol-1

= 4.988971747 x 10-3

= 4.99 x 10-3

mol

(iii) Calculate the number of moles of Cl

in 0.475 g of MgCl2.

MgCl2Mg2+ + 2Cl

n(Cl

) / n(MgCl2) = 2 / 1

n(Cl

) = 2 x 4.988971747 x 10-3

= 9.98x 10

-3mol

(iv) Calculate the mass of Cl


m(Cl

) = n(Cl

) M(Cl

)

= (9.98x 10

-3mol)(35.45gmol

-1)

= 0.353718096g

= 0.354g

(v) Calculate the percentage of Cl

(by mass) in 0.475 g of MgCl2.

% of Cl

= [m(Cl

) / m(MgCl2)] x 100%

= (0.354g / 0.475g) x 100%

= 74.4669677%

= 74.5%

(vi) Calculate the number of moles of Mg2+


Similar to (iii)

n(Mg2+) = 4.99 x 10

-3mol

(vii) Calculate the mass of Mg2+


Method 1 : Similar to (iv)

Method 2:

m(Mg2+) = m(MgCl2) m(Cl

)

= 0.475

0.354 g

= 0.121 g

Element M, gmol-

Mg 24.31

Cl 35.45


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(viii) Calculate the percentage of Mg2+

(by mass) in 0.475 g of MgCl2?

Method 1 : Similar to (v)

Method 2:

% Mg2+

= 100% - % of Cl

= (100

74.5)%= 25.5 %

(ix) Calculate the concentration of 0.475 g of MgCl2 dissolved in a 50.0 mL volumetric flask by distilled water in g/L,

mol/L and % w/v.

c in gL-1

= m(MgCl2) in g / Vin L

= 0.475g / 0.0500L

= 9.5 gL-1

c in molL-1

= n(MgCl2) / Vin L

= 0.00499 mol / 0.0500L

= 0.0998 molL-1

% w/v

= [m in g / Vin mL] x 100%

= (0.475g / 50.0mL) x 100%

= 0.950%

(x) If now this 0.475 g of MgCl2 is dissolved in a 1000 mL volumetric flask by distilled water. Calculate its concentration

in mg/L.

c in mgL-1 = m(MgCl2) in mg / Vin L

= 475mg / 1.000L

= 475 mgL-1

(xi) What is the concentration of Mg2+

in the second solution, in mg/L?

c in mgL-1 = m(Mg2+) in mg / Vin L

= 121mg / 1.000L

= 121 mgL-1

(xii) What is the concentration of Cl

in the second solution in ppm then in ppb?

c in mgL-1

= m(Cl

) in mg / Vin L

= 354mg / 1.000L= 354 mgL

-1

354 mgL-1

= 354 ppm = 354000 ppb

(xiii) What is relative molecular mass of MgCl2?

95.21 [no unit]


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Usually to determine the concentration for one of the reactants involved accurately.

For example Redox titration between standard Fe2+

solution with unknown concentration of KMnO4

where the purpose of this titration is to find out the concentration of KMnO4 accurately by using an

accurately known concentration of Fe2+

solution.

Step 1: To prepare a standard Fe2+

solution. [i.e. to prepare a solution with accurately known

concentration of Fe2+

]

Consider the formulae below:

2

2

2

Fe

Fe

Fe V

nc

2

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22424

Fe

O.6H)(SO)Fe(NH

O.6H)(SO)Fe(NH

V

M

m

Accurate concentration of Fe2+

needs the following input data:

Accurate2Fe

V [obtains by using Volumetric Flask.]

Accurate2Fe

M [calculates from periodic table.]

Accurate2Fe

m [obtains from electronic weighing machine and pure grade of

Fe(NH4)2(SO4)2.6H2O.]

Purpose of titration

As the mole ratio between the Fe2+

and Fe(NH4)2(SO4)2.6H2O is 1:1,therefore the no. of moles of Fe2+

can be obtained from m of the salt

divided by molar mass of the salt.


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Step 2: Titration between the prepared standard Fe2+

(aq) with unknown concentration of KMnO4 (aq)

Half Equations

Oxidation half equation : Fe2+

Fe3+

+ e

Reduction half equation : MnO4

+ 8H+

+ 5eMn2+

+ 4H2O

Overall equation : MnO4 + 8H+ + 5Fe2+Mn2+ + 4H2O + 5Fe

3+

Therefore, from the overall equation

5

1

n

n

2

4

Fe

MnO

5

1

Vc

Vc

22

44

FeFe

MnOMnO

4

22

4

MnO

FeFe

MnO V

Vc.

5

1c

To obtain an accurate c HCl,the other variables must be accurate too.

Accurate

4MnOV can be obtained by using a burette.

Accurate2Fe

V can be obtained by using a volumetric pipette.

Accurate2Fe

c is obtained from step 1.

KMnO4

Fe2+


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Volumetric

ApparatusPurpose Rinsing Filling Delivering

Volumetric

Pipette

To transfer an accurate

fixed volume of

solution. (Fixed volume

such as 1, 2, 5, 10, 20,

25, 50, 100 mL & etc.)

First with distilled water

follow by the solution to

be transferred.

(If we used distilled

water only, the

concentration of the

solution will be diluted)

1. Overfill using a filler bulb or pipettefiller (above the calibration mark)

2. Remove the pipette from the solution& wipe the tip dry with tissue.

3. Use forefinger or the filler bulb toadjust solution level (bottom of

meniscus) aligns with calibration

mark at eye level

4. Make sure no air bubbles in thepipette

1. Hold pipette vertically & touch the tip of thepipette on the side (inner) of receiving vessel

(usually is conical flask or volumetric flask) to

allow smooth delivery.

2. Let solution drain under gravity (free flow) tothe receiving vessel

3. Hold for 10 seconds when solution has drained.4. Don't blow or shake outthe final drop in the

pipette.

Pipette Fillers (for practical)

Bulb-type

This bulb-type safety pipet filler is designed for efficient one-hand operation. It

made from natural rubber for better resistance to chemical attack. Its three

chemically resistant valves control the evacuation of air, liquid uptake and liquid

dispensing. Expel air from the bulb by squeezing the A valve and the bulb at

the same time. Draw (suck) liquid into the volumetric pipette by squeezing the

S valve. To dispenseliquid squeeze the E valve.

Pump-type

This pipetting device provides an easy, precise way of filling and releasing

liquid. By rotating the knurled thumb wheel the liquid is drawn up. The soft

elastic chuck has a threaded collar that secures the plastic or glass pipette to

the device. Fast releases by pressing air inlet valve button, or gradual release

by gently turning wheel.


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Volumetric


Burette Used for the accurate

delivery of a variable

volume of solution

First with distilled water

follow by the solution to

be placed in the burette

Transfer the solution into burette through

a funnel.

"LEFT HAND TURN, RIGHT HAND SWIRL"

1. Note the initial reading2. Add the solution slowly from the burette to the

conical flask.

3. Swirl the flask to ensure complete mixing4. When there is an indication that the end point

is near (indicator colour takes a longer time to

change) add drop by drop to the flask until

permanent colour change is achieved.

5. Note the final burette reading6. Repeat until 3 concordant titre values (within

0.10mL) are obtained.

Burette

reading (mL)

Titration Numbers

1 2 3 4 5

Final 26.5 26.6 26.7 26.8 26.7

Initial 0 0 0 0 0

Titre 26.5 26.6 26.7 26.8 26.7


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Volumetric


Volumetric flask

Designed to contain an

accurate volume at the

specified temperature

(25 or 20C)

To prepare anaccurately known

concentration of

solution i.e.

Standard Solution

To dilute a solutionaccurately.

(Fixed volume such as

5, 10, 25, 50, 100, 200,

250 & 500 mL. Also 1, 2

& 5 L)

Last rinse by the solvent

or distilled water.

Preparation of standard solution

1. Dissolve the accurately weigh solute/substance in minimumvolume of distilled water/solvent in a beaker.

2. Transfer the solution into the volumetric flask via a funnel [ORthe solid substance may transfer directly to the Volumetric flask]

3. Wash all traces of the solution [OR solid] into the flask by rinsingthe beaker and the funnel with distilled water/solvent.

4. Add distilled water to about 2/3 of the flask volume.5. Stopper & shake thoroughly to dissolve the solid (or mix the

solution for dilution.)

6. Add distilled water near to and lower than the calibration mark.7. Adjust the bottom of meniscus of solution till it aligns with

calibration mark at eye level by using a teat pipette.

8. Stopper the flask again & invert several times to mix thecontents to ensure a uniform concentration solution /

homogeneous solution

9. Label the flask with name of chemical, date prepared,concentration, person prepared.

For accurate dilution (replace the 1, 2 & 3 above with these)

1. Calculate the volume for volumetric pipette and volumetric flaskfrom the dilution factor. Example: 10 times dilution, 25.0 mL

volumetric pipette and 250.0 mL volumetric flask.

2. Pipette the solution till the bottom most of meniscus align withcalibration mark of volumetric pipette at eye level.

3. Make sure no air bubbles in the pipette, allow 10 secondsdraining time and also don't blow or shake out the final drop inthe pipette.

Transfer the solution to a

beaker before pipetting.


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3. A 10.00mL volumetricpipette is used to transfer the

solution into another 250.0mL

volumetric flask. So, for this

10.00mL of solution transferred,

c (Na2CO3) = 1.000 molL-1

m(Na2CO3) = 1.060 g

n (Na2CO3) = 0.01000 mol

5. A 25.00mL volumetricpipette is used to transfer the

solution into a conical flask for

titration with HCl. So, for this

25.00mL of solution

transferred,

c (Na2CO3) = 0.04000 molL-1

m(Na2CO3) = 0.1060 g

n (Na2CO3) = 0.001000 mol

The entire solid is

then transferred to

a 100.0mL

volumetric flask.

1. A beaker containing10.60g of Na2CO3.

M(Na2CO3) = 105.99gmol-1

So

n (Na2CO3) =0.1000mol

2. by adding d. water till the calibrationmark, for this 100.0mL solution

m(Na2CO3) = 10.60 g

n(Na2CO3) = 0.1000 mol

c(Na2CO3) =1.000 molL-1

4. by adding d. water till the calibrationmark, for this 250.0mL solution

m(Na2CO3) =1.060 g

n(Na2CO3) = 0.01000 mol

c(Na2CO3) = 0.04000 molL-1

6. For this 25.00mL solution

m(Na2CO3) = 0.1060 g

n(Na2CO3) = 0.001000 mol

c(Na2CO3) =0.04000 molL-1

HCl

Preparation and Dilution of Solution for Titration

Fill in the blanks by

following the sequence of

number in increasing order.


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HCl + NaOH NaCl + H2Otitre value is affected by

1. c(substance) in the burette2. n(substance) in the conical flask

1

1

)(

)(

NaOHn

HCln

1

1

)()(

)()(

NaOHVNaOHc

HClVHClc

c(HCl) Calculated=)(

)()(

HClV

NaOHVNaOHc OR

c(NaOH) Calculated=)(

)()(

NaOHV

HClVHClc

Apparatus

Correct

solution for

final rinsing

(just before

we use it)

If rinse wrongly

with

Consequences of wrong

rinsing of apparatus

Titre value,

V(HCl)

Calculated

c(HCl) c(NaOH)

Volumetric

Pipette NaOH Distilled water

c(NaOH) pipettedV(NaOH) pipetted=

n(NaOH) delivered/pipetted

Burette HCl Distilled water

c(HCl) in buretten(HCl) to complete the titration=

V(HCl) to complete the titration

Volumetric

Flask

(for

dilution of

NaOH)

Distilled

waterNaOH

c(NaOH) preparedV(NaOH) delivered by v. pipette =

n(NaOH) delivered by v. pipette

HCl

NaOH

Effects of Wrongly Rinsed Volumetric Apparatus on Titration

Titre values will be affected by

1. c(substance) in the burette , titre value

2. n(substance) in the conical flask, titre value


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Apparatus

Correct

solution for

final rinsing

(just before

we use it)

If rinse wrongly

with

Consequences of wrong

rinsing of apparatus

Titre value,

V(HCl)Calculated

Conical

Flask

Distilled

waterNaOH

c(NaOH) in conical flask=

V(NaOH) in conical flaskn(NaOH) in conical flask

Beaker to

store HClHCl Distilled water

c(HCl) in buretten(HCl) for titration=

V(HCl) for titration

Beaker to

store NaOHNaOH Distilled water

c(NaOH)V(NaOH) pipetted=

n(NaOH) pipetted

Detail Explanation of rinsing procedure for Volumetric pipette

The volumetric pipette shall first rinse with distilled water 3 times and then the solution to be placed in

it 3 times (i.e. HCl in this case).

When the pipette was last rinsed with distilled water concentration of NaOH will be lowered (dilution by

distilled water in it / left over) and therefore the no. of moles of NaOH transferred into the conical flask

will become less. With lower no. of moles of NaOH in the conical flask, lower no. of moles of HCl will be

needed to completed the titration, i.e. lower titre value. With lower titre value, the calculated

concentration of HCl will become lower than the actual value.


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Many chemicals are impure or without accurately supplied concentration e.g1. KOH and NaOH is hygroscopic/deliquescent i.e rapidly absorbs moisture from the

atmosphere. Moreover, these strong bases can react with acidic oxide present in the air

i.e. CO2

2. Strong acids are supplied in concentrated form by manufacturers and have a variableconcentration from batch to batch.

Therefore the amount of substance cant be calculated or obtained ACCURATELY.The amount or no. of moles of substances that can be calculated from their mass are called primary

standard.

Criteria for a primary standard

Primary standards must obey the following criteria

1. obtainable in pure form (no. of moles can be obtained)2. has a known formula (including a known degree of hydration, so the no. of moles can be

obtained)

3. Do not absorb moisture or other chemicals in the air i.e. stable in air. (no. of moles can beobtained)

4. Has a reasonable high molar mass (to minimize weighing error)5. Completely soluble in the solvent (water) at room temp.

Please note the different between Primary Standardand Standard Solution.

Standard solution-solution with accurately known concentration

Primary standard is used to prepare standard solution.

For example, in practical 4, Fe2+ salt serves as a primary standard as it is used to dissolve in water tomake an accurately known concentration of solution i.e. standard solution.

Primary Standard


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Acid-base titration

Titration curve is not in subject outline!



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pH

1 4 7 10 14

Strong Acid Weak Acid Weak alkaline Strong AlKaline

Indicator Phenolphthalein Methyl orange

Colour inAcid colourless Red

Alkaline pink Yellow

Colour will change at pH 8.2 10.0 3.2 4.4

Suitable for

1. Strong acid vs strongalkaline

2. weak acid vs strongalkaline

1. Strong acid vs strongalkaline

2. strong acid vs weakalkaline

Indicators are either coloured weak organic acid or base, therefore the amount we used to identify the

end-point should be as minimum as possible, because it will increase or lower the titre value.

For example the acid in conical flask and you added a weak acid indicator, the titre value will increase.


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In term of Reduction Oxidation

Oxygen - +

Hydrogen + -Electron + -

Oxidation no. - +

Redox (Reduction-Oxidation) is a complementary process, means that both have to occursimultaneously.

Oxidising agent/oxidant will be reduced and Reducing agent/reductant will be oxidized.

Rule of oxidation number

1. Free elements/pure metal have oxidation no. of zero, e.g. Na, Ca, K, C

2. The oxidation no. for ions or polyatomic ions is equal to the charge on the ion e.g.

O2

Al3+

OH

Cr2O72

MnO4

Charge 2- 3+ 1- 2- 1-

Oxidation no. -2 +3 -1 -2 -1

3. The sum of oxidation no. of a compound is equal to zero, e.g. NaCl, MgO, KMnO4, K2Cr2O7

4. Some element have oxidation no. which are regarded as fixed in the compounds such asa. Group I elements+1b. Group II elements+2c. Oxygen usually has an oxidation no. of -2 except in peroxides such as H2O2, it has -1.d. Hydrogen usually has an oxidation no. of+1except in metal hydrides such as MgH2, it

has -1.

Redox


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Writing of half equations

Method 1 (harder)

Steps Cr2O72

Cr3+

1.Assign oxidation no. to the element being

oxidized or reduced.

2.Balance the element involved in the

oxidation or reduction

3.

Calculate thechange in total oxidation no.

and add electrons to balance this change to

the appropriate side.

4.Balance the charge by adding H

+to the

appropriate side.

5. Balance the equation by adding H2O to theappropriate side (if necessary)

Method 2 (easier)

Steps Cr2O72

Cr3+

1.Balance the element involved in the

oxidation or reduction

2.Add H2O to balance the O, (if no oxygen or

already balance, omit this step)

3. add H+

to balance the H

4. Add e

to balance the charge.

In this reduction reaction, the oxidation number of Cr in Cr2O72

has been reduced from +6 to +3 in Cr3+

ion.

Another important note is that acid-base titration is not the same with redox titration in term of

reaction involved.

Acid-base titrations involving neutrlisation reaction between the acid and base / alkaline.

Whereas the redox titrations involving the reduction and oxidation reactions occur simultaneously. The

acid used in the reaction i.e. Cr2O72-

above is not for neutralization reaction!


23/24


Exercises

Write balance half equations for the following conversions and state whether it is an oxidation or

reduction.

1. MnO4Mn2+MnO4

+ 5e

+ 8H

+Mn2+ + 4H2O

2. H2O2 H2OH2O2 + 2e

+ 2H

+ 2H2O

3. H2O2 O2H2O2 O2 + 2e + 2H+

4.

ClO

Cl

ClO

+ 2e

+ 2H

+ Cl+ H2O

5. SO32 SO42SO3

2+ H2O SO42 + 2e+ 2H+

6. SO2 SO42SO2 + 2H2O SO42 + 2e+ 4H+

7. H2S SH2S S + 2H+ + 2e


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If we have an analyte that has acid-base or redox properties and is soluble but also fits into one or more

of the following categories,

Toxic Volatile Gaseous and in a mixture of gases Fairly unreactive - that is too weak to have clearly defined end-point.

Then direct titration cannot be used to determine the amount of analyte. Indirect or back titration is themore suitable method.

Indirect Titration

Determining the concentration of an analyte by reacting it with an excess reagent. One of the product(s)

produced is then titrated with a second reagent. The concentration of the analyte in the original solution is then

related to the amount of reagent consumed.

Part A: Excess amount of L is added to the system to react with the K (analyte). At the end of the reaction, K is

fully reacted i.e. no more K in the system

K + L

M + NPart B: Titration is carried out to determine the amount of M produced by titrating with known concentration of

O.

M + O P + Q

Sample question: Redox titration Q29.

Back Titration

Determining the concentration of an analyte by reacting it with an accurately known number of moles ofexcess

reagent. The unreacted excess reagent left over is then titrated with a second reagent. The concentration of the

analyte in the original solution is then related to the amount of reagent consumed.

Part A: Excess amount and accurately known amount of NaOH is added to the system to react with the CH3COOH

(analyte) . At the end of the reaction, CH3COOH is fully reacted i.e. no more CH3COOH in the system.

CH3COOH + NaOH CH3COONa + H2O

Part B: Titration is carried out to determine the amount of unreacted NaOH from part A by titrating with known

concentration of HCl.

NaOH + HCl NaCl + H2O

S l i id b i i Q45 47

Indirect And Back Titrations

2 1 Analytical Teachniques Titrtion 2013 Part 1 of 25

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