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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