Chapter five

Volumetric analysis

Principle

Volumetric analysis is one of the most useful and accurate

analytical techniques especially used for smaller a mounts (i.e in

millmoles) of substance to be analyzed. It is based on the

measurement of the volume of the added solution of known

concentration which is sufficient to react with all of the analyzed

substance in the sample then the concentration of the substance

in the sample can be calculated

Titrant: known solution which is gradually added which is

often placed in a container knows as a buret

Sample solution: solution which is to be analyzed is usually

placed in conical flask

Titration process by which the titrant is gradually added to

sample solution Titration is stopped when the volume of titrant

added is sufficient to react completely with the analyzed

substance of the sample solution

Equivalent point

The point at which the volume of titrant added is sufficient to

react completely with sample solution.

Equivalent point: usually detected by an change in color of an

indicator

Indicator: chemical substance that its color change at

equivalent point and it is added to the sample solution before

starting the titration and this point is called the end point of

titration.

However, the end point and the equivalent point are identical

under ideal conditions.

Types of titration and application principles with typical

calculations of volumetric analysis operations are reviewed in

this chapter.

Starting point of volumetric analysis is preparation of

Standard solution

Solution which is used as titrant and whose concentration is

accurately known. It is prepared by dissolving an accurately

weighed quantity of highly pure material called as primary

standard and diluting to an accurately known volume in a

volumetric flask

Standardization

The standard solution is used to determine the exact

concentration of any prepared solution

Any titrant must be standardized before using in volumetric

analysis measurements

Back titrations

Sometimes a sharp end point cannot be obtained and a slight

excess of titrant is normally added to the sample solution in such

cases back titration is used

Back titration :- is used to determine the excess amount of the

titrant by titration with another standard solution.

From the amount of such a standard solution required for back

titration the excess added amount of titrant can be computed.

Dilution of solutions

It is frequently required to prepare a dilute solution of certain

normality or molarity from a more concentrated one by

qualitative dilution.

Example ( 1 )

Calculate the volume of 0.25M of K2Cr2O7 solution required to

prepare 500 ml of a 0.1 M solution

MFinal × VFinal =Minitial ×Vinitial or

Mf Vf = MiVi

Vi=

=

= 200 ml

Example (2)

Calculate the volume of 0.4M of Ba(OH)2 solution must be

added to 50ml of 0.3M NaOH in order to prepare a solution of

pH = 13.699 pOH= 14- 13.699 = 0.301 [OH-] = 0.5M

x= ml of Ba(OH)2

mmol of OH- = 2(0.4)x + 50(0.3) = 15 + 0.8x

MnO2 + 2Fe2+

+ 4H+

→ Mn2+

+ 2Fe3+

+ 2H2O

The excess ferrous ion is determined by back titration with

0.02M of KMnO4 requiring 15ml according to the following

reaction:-

5Fe2+

+ MnO4- + 8H

+ → Mn

2+ + 5Fe

3+ + 4H2O

Calculate the manganese content in the sample as Mn3O4?

excess Fe2+

used = 5 (MnO4- required in the 2 nd reaction) =

5(0.02×15)= 1.5 mmol

total Fe2+

used = 50 ×0.1 = 5.0 mmol

net Fe2+

reaction in 1st

reaction= 5-1.5= 3.5 mmoles.

MnO2 in the sample =

× 3.5 = 1.75 mmoles

MnO2 →

Mn3O4

mmoles of Mn3O4 =

(1.75) = 0.583 mmole

wt% Mn3O4 =

× 100 =

×100 = 16.7%

Classification of Volumetric analysis

1. Neutralization method or (Acid – Base titrations)

2. Precipitations titrations

3. Oxidation – Reduction titrations

4. Complexation titration

Neutralization method or (Acid – Base titrations)

This type of volumetric analysis methods is based on the

neutralization of hydrogen or hydroxide ion of acid and bases.

The end point of such titrations are either detected by means of

an indicators or by rapid change in pH which can be measured

by the use of pH meter.

Titration curves

Graphs of pH versus the added volume of titrant. The inflection

points in the titration curves are usually to be the end points of

acid-base titrations. Atypical titration curve of strong acid with

strong base is shown in figure

the values have been computed for titration of 100 ml of 0.1 M

HCl by 0.1 M NaOH

It is shown that pH changes slowly at first until the equivalent

point, arapid change in pH, there is nearly a vertical rise in the

region between pH 4-10, the pH changes slowly again after the

equivalence point due to the addition of excess amount of

NaOH, However, the magnitude of the vertical region depends

on concentration of both the acid and base as shown in figure.

Titration curve of strong acid-strong base

Example (3)

Calculate the pH and plot the resulting titration of 50ml of 0.1M

HCl after the addition of 0, 10, 20, 40, 45, 48, 49, 50, 51, 55, 60,

80, and 100ml of 0.1M NaOH chemical reaction which

takesplace is

H+ + OH

- ↔ H2O

Initial conc. of [H+] in solution = 0.1M pH = 1

Initial mmoles of [H+] = 50 × 0.1= 5 mmoles

After the addition of 10ml NaOH

mmoles of OH-added = 10×0.1= 1.0 mmoles

remaining mmoles of H+ = 5-1 =4 mmoles

total volume of solution = 50 + 10 = 60 ml

conc. of remaining H+ =

= 0.067M pH= 1.76

and 50 on before the eq.point.

ml, NaOH total volume mmoles remaining acid [H+] pH

0 50 5 0.1 1.0

10 60 4 0.0667 1.176

20 70 3 0.0429 1.368

40 90 1.0 0.0111 1.954

45 95 0.5 0.0053 2.279

48 98 0.2 0.0020 2.690

49 99 0.1 0.0010 2.996

Equivalence point: correspond to addition of 50ml of NaOH,

[OH-] and pH can be calculated from the value of Kw

Kw = [H+][OH

-] = 1.0×10

-14

at equivalence point [H+]=[OH

-] = √ = 1×10

-7M pH= 7

after the equivalence point, excess NaOH is added to the

solution, pH of solution can be calculated from [H+],[OH

-], pH,

pOH, Kw

pKw = pH + pOH = 14.0

the following table show the values of following

mlNaOH total vol.ml mmol.excessOH- [OH

-] pOH pH

51 101 0.1 9.9×10-4

3.004 10.996

55 105 0.5 4.77×10-3

2.322 11.672

60 110 1.0 9.09×10-3

2.041 11.95

80 130 3.0 0.0231 1.637 12.36

100 150 5.0 0.0333 1.477 12.58

The results of these calculations can be plotted as shown in this

figure (2)

Acid – Base Indicators

Are highly colored organic dyes which exhibit a change in color

when the pH of solution changed between certain limits, usually

two pH units. This behavior can be explained from the fact that

most acid- base indicators are weak organic acids of general

form HIn which can be dissociated as following

HIn → H+

+ In-

color1 color2

Ka =

Hence, according to Henderson equation

pH = pKa + log

in practice the human eye sees color 1 when

=

and color

2 when

= 10, thus

when color 1 is seen pH = pKa + log

= pKa – 1

when color 2 is seen pH = pKa + log 10 = pKa + 1

pH = (pH)2 – (pH)1 = pKa+1 – pKa –(-1)= 2

which means a pH change of two units when the color is

changed from color 1 to color 2. Midway in the titration [HIn]=

[In-], hence the pKa of the indicator is close to pH at the

equivalence point. However, a mixture of two or more

indicators may be used in some cases to give better color change

at the end point.

Titration of weak acid with strong base or weak base with

strong acid

Titration of acetic acid with sodium hydroxid

HOAC + NaOH → H2O + NaOAC

Stage Equation Assumption

Before titration [H+] =√ Ka <

With titration pH = pKa + log

At equivalent point [OH-] = √

[OAC

-] > 100

After equivalence [OH-] = [ excess titrant]

point excess titrant

Example (8)

Calculate the pH and plot the resulting titration curve for

titration of 50ml of 0.1M HOAC by addition of 0, 10,25,50 and

60 ml of 0.1M NaOH Ka = 1.75× 10-5

At 0ml NaOH x = [H+] = [OAC

-] [ HOAC] = 0.1-x

= 1.75× 10

-5 Ka

0.1-x 0.1

[H+] = √ = √ = 1.32 ×10

-3M

pH = 2.88

after the addition of 10 ml NaOH

mixture of weak acid + salt

intial mmole of HOAC = 50 ×0.1 = 5 mmoles

mmoles of [OH-] added = 10 ×0.1= 1 mmoles

remaining mmoles of HOAC = 5-1 = 4 mmoles

total volume = 50 +10 = 60 ml

[ HOAC] =

= 0.0667 M

[OAC-] =

= 0.0167 M

pH = pKa + log

pH = - log 1.75 × 10-5

+ log

pH = 4.76 + 0.6 = 4.16

Similarly with addition of 25ml.

[HOAC] =

= 0.0333 M [OAC

-] =

= 0.0333M

pH = 4.76 + log

= 4.76

At equil. Point = after the addition of 50 ml NaOH

mmoles NaOH = 50 0.1 = 5 mmoles

mmoles OAC- produced = 50 0.1 = 5 mmoles

total volume = 100 ml

[OAC-] =

= 0.05M

[OH-] = √

= √

= 5.35 10-6

M

pOH = 5.27 pH = 8.73

with the addition of 60 ml NaOH

mmole = 60× 0.1 = 6 mmole NaOH

total volume = 50 + 60 = 110 ml

[OH-] =

= 0.00909M pOH = 2.04 pH = 11.96

ml NaOH pH

0 2.88

10 4.16

25 4.76

50 8.73t

60 11.96

Titration curve of weak base (NH3) with strong curve with

reverse shape of titration curve reaction:-

NH3+ HCl → NH4Cl

Stage Equation Assumption

Before titration [OH-] =√ Kb <

With titration pH = pKa + log

At equivalent point [H+] = √

[NH4] > 100

After equivalence [H+] = [ excess titrant]

Example (9)

Calculate the pH and plot the resulting curve for the titration of

20 ml of 0.11M ammonia by the addition of 0, 5, 11, 15, 20, 22,

25, 30, 35, and 40ml of 0.1M HCl Kb = 1.79× 10-5

?

1- At 0 ml HCl NH3 + H2O → NH4OH

x = equil. Conc [OH-] and [NH4

+]

Kb = 1.79× 10-5

=

=

[OH-] =√ = √ = 1.4×10

-3M

pOH = -log [OH] = 2.85 pH = 11.15

Similar procedure of the previous example which give the

following result and titration curve

ml HCl pH

0 11.15

5 9.79

11 9.29

15 8.9

20 8.3

22 5.27 eq. point

25 2.19

30 1.82

35 1.64

40 1.54

Precipitation titrations

In this type of titration, titrant forms an insoluble product

(precipitate) from its reaction with the sample solution such as:

the titration of chloride ion with silver nitrate solution: end point

of this type of titration can be detected either a) with an

indicator or b) instrumental end point detection technique such

as spectrophotometry.

Two general methods are based on precipitation titrations

1. Mohr method

Silver nitrate solution is the most popular titrant for

precipitation titrations:-

Ag+

+ Cl- → AgCl (s) Ag

+ + Br

- → AgBr (s)

Ag+

+ I- → AgI (s)

End point location:-

By the addition of sodium chromate, dark-orange silver

chromate precipitate is formed upon the addition of excess

Ag+

to the solution within a pH between 6-5 and 10-3

Reaction

2Ag+

+ CrO42-

→ Ag2CrO4(s)

2. Volhard method

This method can be used to analyze bromide and iodide

solutions by the addition of an excess silver ion to

precipitate silver halide. Excess Ag+

remaining in the

solution after precipitation is back titrated with thiocyanate

to yield silver thiocyante

Ag+

+ SCN- → AgSCN (s)

End point location:-

Acidic solution of ferric ammonium sulfate as an indicator.

Excess thiocyanate ion reacts with ferric ion to form a red

complex of ferric thiocyanate :-

Fe3+

+ SCN- → FeSCN

2+ (red)

Example

A 50ml sample of bromide solution was analyzed by using

volhard method. It is found that 10ml of 0.1M AgNO3 is

required, and back titrated with 0.0832M potassium thiocyanate.

The end point of the back titration was reached after adding 5.34

ml of KCN solution. Calculate the bromide concentration (as

molarity) in the sample solution?

1) Ag+

+ Br- → AgBr (s) 2) Ag

+ + SCN

- → AgSCN (s)

mmoles of SCN- reacted in back titration reaction

=5.34×0.0832= 0.444 mmoles = excess mmoles of Ag+

reacted

total Ag+

reacted = 10(0.1) = 1.0 mmoles

Ag+

reacted with bromide = 1-0.444 =0.556mmol of Br-

[Br-] =

=

= 0.01 M

Oxidation – Reduction titrations (Red.Ox)

Oxidation- reduction reaction. A substance is oxidized when it

loses electrons and is reduced when it gains electrons. In general

redox reactions take place in two half reactions. In one half,

oxidation takes place, and in other half, the reduction takes place

Oxidation reaction Fe2+

- e → Fe3+

reduction reaction Ce4+

+ e → Ce3+

over redox reaction Fe2+

+ Ce4+ Fe

3+ + Ce

3+

Chemical indicators: which are used to locate the end point of

redox reactions which are organic compounds which have

different colors in the oxidized form (In ) and the reduced form

(Inr) of the indicator

In + mH+ + ne

- = Inr or In + ne

- = Inr

color 1 color 2 color 1 color 2

m= can be either positive or negative

several substance can be analyzed by redox titration. In most

redox titration the end point can be located by direct titration

with the titrant, However, in the some cases the back titration

procedure is often used since an excess amount of titrant is

required to force the slow reaction rapidly to completion

Iodometric titration

Iodine ion is a weak reducing agent and when an excess of

iodide is added to a solution of an oxidizing agent, I2 is

produced in an amount equivalent to the oxidizing agent present

Cr2O72-

+ 6I- + 14H

+ ↔ 2Cr

3+ + 3I2 + 7H2O

The iodine is titrated with reducing agent, usually sodium

thiosulfite (Na2S2O3)

I2 + 2 S2O32-

→ 2I- + S4O6

2-

Analysis of oxidizing agent in this way is called "iodometric

method" End point for iodometric titration is detected with

starch- disappearance of the blue starch I2 color indicates the

end of titration

Example

A metal sample containing copper with mass of 0.2514 was

analyzed by idometric method. The sample was dissolved in

acid and an excess a mounts of potassium iodide was added to

the sample solution. The iodine produced was titrated with

26.32 ml of 0.10M sodium thiosulfate. Calculate the percent of

copper in the metal sample?

2Cu2+

+ 5I- → 2CuI + I3

-

I3-+ 2S2O3

2- → 3I

- + S4O6

2-

mmoles of thiosulfate = 26.3×0.1 = 2.63mmoles

I3- produced in 1

st reaction = 2.63 (

= 1.316 mmoles

Copper ion reacted in 1st

reaction = 1.316 × 2= 2.632mmoles

mass of copper in the sample=2.632 mmol×

=167.2 mg

wt% Cu=

× 100=

4- Complexation titration

Useful for determination of a large number of metals which from

slightly dissociated complexes in solution containing complexing agent

usually known as ligand which can be either a neutral molecule such as

water or ammonia or an ion such as chloride cyanide or hydroxide. The

complex can have either positive or negative charge, or it can be neutral.

A formation constant (Kf) can be written for each step and for the

overall reaction as shown in the following example

NH3 + Ag+ Ag(NH3)

+ Kf1 =

= 2×10

3

Ag(NH3)+ Ag(NH3)2

+ Kf2 =

= 8×10

3

Over all reaction :-

Ag+ +2NH3 Ag(NH3)2

+ Kf= Kf1. Kf2=

= 1.6×10

7

Complexation titration with EDTA

EDTA (ethylene diamine tetra acetic acid) or its salt is the most titrants

used as complexing agents about 95% of complexation titrations are

carried out with EDTA which has the following chemical structures

EDTA has four acidic dissociation constants since it is considered as

tetraprotic acid with the symbol H4Y. It is dissociated by the following

dissociation step:-

H4Y ↔ H+

+ H3Y- Ka1=

= 1×10

-2

H3Y- ↔ H

+ + H2Y

2- Ka2=

= 2.1×10

-3

H2Y2-

↔ H+ + HY

3- Ka3=

= 6.92×10

-7

HY3-

↔ H++ Y

4- Ka4=

= 5.5×10

-11

Equilibrium calculations can be used to determine the fraction ( ) of

each of the five forms of EDTA which is presented in solution at any

pH for example the following equation can be used to calculate y4-

at

any pH

=

= 1+

+

+

+

Where CT = total concentration of all forms of EDTA

CT = [H4Y] + [H3Y-] + [H2Y

2-] + [HY

3-] + [Y

4-]

Similar equation can be used to calculate α HY3-

, α H2Y2-

, α H3Y- and

H4Y-

Plot of the fraction of each form of EDTA as function of pH are shown

in the following figure (3)

Example

Calculate the fraction of EDTA that exists as Y4-

at pH 10 ?

at pH = 10 → [H+] = 10

-10

=1+

+

+

+

= 1+

+

+

+

= 1+1.82 + 2.63 ×10-4

+1.22×10-11

+1.22×10-19

= 2.82

αY -4

= 0.355

Example

Use the result of the previous example to calculate the equilibrium

concentration of ca2+

at pH = 10 if 100ml of solution of 0.1M Ca2+

is

added to 100ml of 0.1M EDTA

mmoles of Ca2+

added= 100 (0.1) = 10 mmoles

mmoles of EDTA added= 100 (0.1) = 10 mmoles

total volume = 100 +100 = 200 ml.

initial conc. of CaY2-

=

= 0.05M

the reaction of Ca2+

with EDTA can be represent by the equation

Ca2+

+ Y4- →

CaY2-

x x 0.05-x 0.05-x

x = equilibrium conc. of Ca2+

= [Y4-

] = CT Y4-

= 0.35 CT = 0.35x

3. Kf =

=

=1×10

-11

x=√

=1.2×10-6

M

Detection of the End point in complexation titration

Spectrophotometry and several electroanytical techniques can be used to

locate the end points of complexation titration. However, chemical

indicators can be used as in other types of titrations. metallochromic

indicators are widely used in complexation titration which from

complexes with metal ions and exhibit different colors according to the

following equation

Inm

↔ HIn-1-m

↔ H2In2-m

↔ ………etc

Consequently, the pH of solution should be controlled when using

metallochromic indicators.

Problems:-

1- The mohr method was used to determine sodium chloride

content in a 1.0004 gm sample. The sample was dissolved

in water and titrated to the end point with 32.36ml. of

0.1012M AgNO3. Calculate the wt% of NaCl in the

sample? Ans : 19.13 wt%

2- A 25ml sample of potassium iodide was analyzed by using

Volhard method. It is found that 20.0ml of 0.5015M

AgNO3 is required, and the excess silver ion was back

titrated with 32.35ml of 0.1038M KSCN solution.

Calculate the concentration of KI solution? Ans :0.267M

3- A 0.5247gm metal sample containing copper was analyzed

by iodometric method. The sample was dissolved in acid

and diluted with distilled water to about 50ml. An excess

amount of KI was added, and the liberated iodine was

titrated with 34.87ml. of 0.1234M sodium thiosulfate.

Calculate the percent of copper in the metal sample?

Ans: 52.11%

4- Calculate the conditional formation constant (Kf) for the

formation of Mn2+

-EDTA complex at pH= 8 K1=1×10-2

,

K2=2.16×10-3

, K3=6.92×10-7

, K4=5.5×10-11

?

Ans: αY -4

= 5.4×10-3

, Kf = 2.05×1011

5- Calculate the conditional formation constant (Kf) for the

formation of Fe2+

-EDTA complex at pH= 9 K1=1×10-2

,

K2=2.16×10-3

, K3=6.92×10-7

, K4=5.5×10-11

?

Ans: αY -4

= 5.21×10-2

, Kf = 1.093×1013

6- Calculate [OH-] and the pH of the solution prepared by

dissolving 7.82gm NaOH and 9.26gm Ba(OH)2 in water

and diluting to 500ml? Ans : 0.67,13.

7- Calculate the volume of 0.1M H2SO4 must be added to

50ml of 0.10M NaOH solution to give a solution of 0.05M

H2SO4 assuming additive volumes? Ans : 100ml

8- A 1.68 gm sample of iron ore is analyzed for iron content

by dissolving in acid, coverting the iron to Fe2+

then

titrating with 0.015M K2Cr2O7 solution according to the

reaction:-

6Fe2+

+ Cr2O72-

+ 14H+

→ 6Fe3+

+ 2Cr3+

+ 7H2O

If 35.6ml of potassium dichromate solution is required for

titration, calculate the iron content in the sample expressed

as wt% Fe2O3 Ans: 15.

9- A potassium permangate solution is prepared by dissolving

4.68gm KMnO4in water and diluting to 500ml. How many

mill liters of this solution is required to react with iron in a

0.50 gm sample of iron ore containing 35.6 wt% Fe2O3?

The reaction is

5Fe2+

+ MnO4 + 8H+

→ 5Fe3+

+ Mn2+

+ 4H2O

Ans: 7.5ml

10- Calculate the pH for the titration of 50ml of 0.1M

NaOH by the addition of 0, 10, 25, and 30ml of 0.2M

HCl? Ans:13, 12, 70, 7.0, 1.90

11- Calculate the pH for the titration of 50ml of 0.1M NH3

by the addition of 0, 10, 25, 50 and 60ml of 0.1M HCl?

Given that Kb =1.75×10-5

Ans: 11, 12, 9.85, 9.24, 5.27, 2.02