Dr. M Afroz Bakht
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Acid-Base Titration
Precipitation and Complex-formation Titration
Oxidation-reduction Titration
hours
Electrochemical methods
Ultraviolet/visible spectrophotometry
Introduction to chromatographic separation
6 hours
4 hours
4
4 hours
6 hours
4 hours
The Student will be able to :
1. Check the feasibility of titrimetric reactions. The student
shall be also able to choose the suitable indicator and
derive the titration curve.
2. Calculate the percentage purity of solid (powdered)
samples and the concentration of liquid samples.
3. Obtain the absorption spectrum of the light-absorbing
compound and define its max.
4. Know the principles of chromatographic separation of
pharmaceutical compounds in a mixture.
5. Check the feasibility of electrochemical methods of
analysis.
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Continuous Assessment:
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First Assessment Test
Second Assessment Test
Term Activity*
Laboratory Test
Final Laboratory Test
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Final Examination:
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Final Paper test Final Exam
15%
10%
10%
10%
15%
40%
Fundamentals of Analytical Chemistry, Douglas A. Skoog
and Donald M. West. Fourth edition. Sanders College
Publishing, Philadelphia (1984).
 Analytical Chemistry, Douglas A. Skoog; Donald M. West,
F. James Holter, Standey R. Crouch, 7th ed. Harcourt
College Publishers (2000).
 Principles of Quantitative Chemical Analysis, Robert de
Levie. McGraw Hill, New York (1997).
 Vogel’s Textbook of Quantitative Inorganic Analysis, 4th
ed. J. Baisett, R.C. Denney, G.H. Jefferg and J. Mendham,
Longman, Essex (1978)
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Analytical chemistry deals with methods used
for determining the composition of various
materials.
The process of material identification called
Qualitative Analysis .
The process of material quantitation called
Quantitative Analysis .
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Identification
What is the identity of the substance in the sample?
Quantitation
How much of the substance x is in the sample?
 Detection
Does the sample contain substance X or not?
 Separation
How the species of interest can be separated?
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Classification of Quantitative methods
a-According to the quantity to be analyzed
1- Micro methods
used for the determination of quantities less than 1
mg.
 2- Semi-micro methods
used for determination of quantities ranging from 1100 mg.
 3- Macro methods
used for determination of quantities more than 100
mg.
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b-According to technique
I- Volumetric or Titrimetric methods
Analysis by volume.
 II- Gravimetric methods
Analysis by weight.
 III- Instrumental methods (Physicochemical
methods)
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Electrochemical methods
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Spectroscopic methods
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Separation methods
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It is the quantitative chemical analysis carried
out by determining the volume of a solution of
accurately known concentration which is
required to react quantitatively with measured
volume of solution of the substance to be
analyzed.
The solution of accurately known
concentration is called the standard solution
(titrant).
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The process of adding standard solution
gradually to the sample until the reaction is
just completed is termed as titration.
The point at which the reaction is completed
is called end point or equivalence point.
The concentration of the substance to be
analyzed is calculated from the volume of the
standard solution.
1- Physical change produced by the standard
solution itself (Self indicator).
2-The Addition of a substance known as
indicator.
(Compound which has different colors at
different conditions).
The reaction between the sample and the
standard solution must be simple and can be
represented by a chemical equation.
 The reaction must be instantaneous (relatively
fast or rapid). Sometimes catalyst is needed.
 The substance to be determined should react
completely with the titrant in stoichiometric
manner (definite ratio).
 The end point of the reaction can be detected
easily. (indicator is available).
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I- Neutralization Reactions
(Acid-Base Reactions)
II-The Precipitation Reactions
(Precipitimetry)
III- Complex Formation Reactions
(Complexometry)
IV-Electron-transfer Reactions
(Redoximetry)
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These are solutions of exact known
concentration
• Types of standard solutions
1-Molar standard solution (M)
• It is the solution which contains the gram molecular
weight of the substance in 1L of solution.
1M solution contains 1 x gm m.wt of substance/L of
solution.
2M solution contains 2 x gm m.wt of substance/L of
solution.
M/10 solution contains 0.1 x gm m.wt of substance/L of
solution.
-Examples
Molar standard solution (M)
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1M solution of H2SO4 contains 98.07 gm/L of solution.
2M solution of H2SO4 contains 196.14 gm/L of solution.
M/10 solution of H2SO4 contains 9.8 gm/L of solution.
1M solution of NaOH contains 40 gm/L of solution.
2M solution of NaOH contains 80 gm/L of solution.
M/10 solution of NaOH contains 4 gm/L of solution.
1M solution of Na2CO3 contains 106 gm/L of solution.
2M solution of Na2CO3 contains 212gm/L of solution.
M/10 solution of Na2CO3 contains 10.6 gm/L of solution.
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Solution which contains gm equivalent weight /L of solution.
Equivalent Weight
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Eq.Wt of acids = m.wt / no. of replaceable H+
Example
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Eq.Wt of bases = m.wt / no. of replaceable OHExample
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Eq.Wt of HCl = m.wt / 1
Eq.Wt of H2SO4 = m.wt / 2
Eq.Wt of NaOH = m.wt / 1
Eq.Wt of Ba(OH)2 = m.wt / 2
Eq. Wt For Salts = m. wt/( number of metal x its
charge )
Examples
-N.B.
NaCl eq. wt = m.wt / 1
CaCl2 eq. wt = m.wt / 2
Equal volumes of equal normalities contain equal number of
molecules, that means equal normalities react 1 to 1 ratio.
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Acid-Base Titrations In Aqueous Solution
Solutions
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Solution is a homogenous mixture of two or more substances.
The component (solid, gas or liquid ) present in small quantity is called
the solute, while the one present in large quantities is called the solvent
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Solutions may be
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1- Saturated solutions .
2- Unsaturated solutions .
3- Supersaturated solutions .
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I-Electrolytes:
Electrolytes are substances when dissolved in water
undergo dissociation and give electricity-conducting
solutions.
Electrolytes may be :
1-Strong Electrolytes :
Substances when dissolved in water dissociate or
ionize to a high degree.
Examples of strong electrolytes .
Acid: HCl , HNO3 , H2SO4 , HBr , HI .
Base: NaOH , KOH , Ca(OH)2 , Ba(OH)2.
Salt: NaCl , CH3COONa , NH4Cl.
2-Weak Electrolytes:
Substances when dissolved in water dissociate or
ionize to a slight degree. Examples of week
electrolytes .
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- Acid: CH3COOH , HCN , H2S , H3BO3 , HF .
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- Base: NH4OH , N2H4 .
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- Salt: HgCl2 , CdCl2 , HgBr2, CH3COONH4 .
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II-Non-electrolytes:
Non-electrolytes are substances when dissolved in
water do not undergo dissociation and give a nonconducting solutions.
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Examples: Sugar , Glycerin , Ethyl acetate.
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Pure water is a bad conductor for electricity.
When an electrolyte is dissolved in water, it dissociates
into negatively charged ions (anions) and positively
charged ions (cations).
 Solutions conduct the electric current due to the presence
of ions.
 The degree of dissociation is directly proportional to the
degree of dilution.
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It is the ratio of the ionized fractions to the total
amount of the dissolved solute.
For each concentration there is a state of
equilibrium between the un-dissociated molecules
and the dissociated molecules (ions).
Molecule = Cation (+ve) + Anion (-ve)
CH3COOH =
H+
+
CH3COONH4Cl
=
NH4+
+
Cl-
The degree of dissociation characterizes the
chemical activity of the respective substance.
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Molecular equations represent the reaction species (reactant
and products ) as molecules .
NaOH + HCl →
NaCl +
H 2O
This equation shows that one mole of NaOH neutralize one mole of
HCl to form exactly one mole of NaCl and one mole of H2O .
In ionic equations, strong electrolytes are represented as ions
while weak electrolytes represented as molecules .
In the above equation NaOH and HCl are strong electrolytes and
present as ions in the solution , so that , the equation can be written
as follows:
Na+ + OH- + H+ + Cl- → Na+ + Cl- + H2O
OH- + H+
→
H 2O
In the reaction of NaOH (strong electrolyte ) and CH3COOH (week
electrolyte ) ,the equation is written as follows:
Na+ + OH- + CH3COOH → Na+ + CH3COO- + H2O
OH- + CH3COOH →
CH3COO- + H2O
In reversible reactions products are formed
from the reactants and the reactants are being
produced from the products.
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A + B = C + D
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Reactants ↔ products
 Under that condition the composition of the
reaction mixture becomes constant and the
system is said to be in a state of equilibrium
which is the state at which the rate of forward
reaction equal to the rate of backward
reaction .
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The rate of chemical reaction is directly proportional
to the product of the molar concentration of the
reacting substances.
For the reaction
A + B
Vf α [A] [B]
or
Vb α [C] [D]
or
=
C + D
Vf = K1 [A] [B]
Vb = K2 [C] [D]
- At equilibrium Vf = Vb
K1 [A] [B] = K2 [C] [D]
- K1 / K2 = k equilibrium (equilibrium constant)
Keq = [C] [D] / [A] [B]
- In case of :
aA + bB = cC + dD
Keq = [C]c [D]d / [A]a [B]b
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Le Chatelier Principle:
According to Le Chatelier principle, if a stress is applied to
a system in an equilibrium state , the equilibrium will be
shifted in such direction to minimize that stress.
Applications of Le Chatelier Principle:
In Precipitation :
A + B = AB (precipitate)
Precipitating agent B is used to precipitate the
compound AB by combining with A to form more AB and
the equilibrium is shifted to the right .
In Solubility :
In endothermic solution , the solubility of the solute
increases by heating (equilibrium shifted to right ) .
solute + solvent + heat = solution
In exothermic solution , the solubility of the solute
decreases by heating (equilibrium shifted to right ) .
solute + solvent
= solution + heat
1-Arrhenius theory:
-An acid forms H+ in water (upon ionization)
HCl
→ H+
+
ClHNO3
→ H+
+
NO3H2SO4
→ 2H+
+
SO4-- A base forms OH- in water (upon ionization)
Na OH → Na+
+ OHCa(OH)2 →
Ca+ +
+ 2OH-
N.B.
1- Not all acid-base reactions involve water
2- Many bases (NH3, and carbonate) do not contain
any OH-
2-Bronsted - Lowry theory:
-Acid is a proton donor H+
Acid → H+
+ Conjugate base
HCl → H+ +
Cl- Base is a proton acceptor H+
Base + H+ →
Conjugate acid
NH3+ + H+ →
NH4+
The conjugate base of an acid is the acid minus
the proton it has donated
The conjugate acid of a base is the base plus the
accepted proton
3-Lewis theory:
-Base is a substance containing an atom that
has unshared pair of electrons e.g. N , O , P , S
(base is an electron donor e.g. NH3, amines
like triethylamine).
-Acid is a substance that can accept that pair of
electrons e.g. AlCl3 , BCl3 , BF3
example of Lewis acid Lewis base
reaction:
H3N:
+
BF3 →
H3N:
→ BF3
Lewis Base + Lewis acid
H2O =
H+ +
OH-Water molecules ionize in very slight degree.
-According to the law of mass action
K = [H+ ] [OH- ] / [H2O ]
K [H2O ] = [H+ ] [OH- ]
Kw = [H+ ] [OH- ]
Kw = ionic product of water
-It was found that under normal experimental
conditions and at 250c Kw = [H+ ] [OH- ] = 10-14
-Since the dissociation of water gives rise to equal
number of H+ and OHKw = [H+ ]2 =10-14
[H+ ] =√ 10-14 = 10-7
pH is the measure of acidity or alkalinity solution .
-pH is the negative logarithm of the hydrogen ion
concentration
pH =-log [H+]
pH range
0 1 2 3 4 5 6
7
8 9 10 11 12 13
14
Acidic
Neutral
Basic
-pH is a number obtained by giving a positive value to the
negative power of 10 in the expression .
[H+] = 10-n
pH = n
[H+] = 10-5
pH = 5
Kw = 10-14
pKw = 14
-In general : for acids
pH = -log [H+]
for bases
pOH = -log [OH-]
pKw = pH + pOH
pH = pKw - pOH
pH of strong acids and strong bases
-Strong acid and strong base are completely
ionized so, concentration of acid or base
represents the concentration of [H+] or [OH-] .
For acids pH =-log [H+]
For bases pOH = -log [OH-]
pH = 14 - pOH
For examples:
pH of 0.1 M HCl (strong acid)
pH =-log [H+] = -log 10-1 = 1
pH of 0.1 M NaOH (strong base)
pOH = -log [OH-] = -log 10-1 = 1
pH = 14 - pOH =14- 1= 13
- pH of weak acids
-A Small quantity of weak acid is dissociated with the
formation of [H+] . e.g. CH3COOH
CH3COOH
= CH3COO- + H+
Ka = [CH3COO- ] [H+] / [CH3COOH]
Where:
[H+] = [CH3COO- ]
[CH3COOH]= Ca (concentration of acid)
Ka = [H+] 2 / Ca
[H+] 2 = Ka Ca
[H+] = √Ka Ca
pH = ½ pKa + ½ pCa
pH = ½ (pKa + pCa)
Examples
Calculate the pH of 0.1 M solution of acetic acid (Ka
=1.75x10-5)
pH = ½ pKa + ½ pCa
= ½ (-log 0.1) + ½ (-log 1.75 x 10-5)
= (0.5 x 1) + (0.5 x 4.757)
= 2.88
Calculate the pH of 0.25 M solution of formic acid (Ka
=1.76x10-4)
pH = ½ pKa + ½ pCa
= ½ (-log 0.25) + ½ (-log 1.76 x 104)
= (0.5 x 0.602) + (0.5 x 3.754)
= 0.301 + 1.877
= 2.18
pH of salts
-Salts of strong acids and strong bases e.g. NaCl
is neutral pH = 7
-Salts of strong acids and weak bases e.g. NH4Cl
pH = ½ (pKw - pKb + pCs )
-Salts of weak acids and strong bases e.g.
CH3COONa
pH = ½ (pKw + pKa - pCs )
-Salts of weak acids and weak bases e.g.
CH3COONH4
pH = ½ (pKw + pKa - pKb )
Buffer solutions are solutions which resist the change in the pH of
solution upon addition of small amount of strong acid or strong
base
- Types of buffer solutions
1-Acidic buffer solutions
Consists of weak acid and its salt of strong electrolyte.
e.g. acetic acid and sodium acetate (CH3COOH/CH3COONa)
-Upon addition of a strong acid:
sodium acetate react with it giving weakly ionized acetic
acid
H+ + CH3COONa → CH3COOH + Na+
-Upon addition of a strong base:
acetic acid react with it and unionized water is formed
OH- + CH3COOH → CH3COO- + H2O
2- Basic Buffer solutions
Consists of weak base and its salt of strong
electrolyte.
e.g. ammonium hydroxide and ammonium
chloride (NH4OH/NH4Cl)
-Upon addition of a strong acid:
H+ + NH4OH → NH4+ + H2O
-Upon addition of a strong base:
OH- + NH4Cl → NH4OH
+
Cl-
pH of acidic buffer:
pH = pKa + log Cs/Ca
pH of basic buffer:
pOH = pKb
pH = pKw
pH = pKw
OR pH = pKw
+
-
log Cs / Cb
pOH
pKb - log Cs/Cb
pKb + log Cb/Cs
Calculate the pH of a buffer solution consisting of 1 M
CH3COOH and 1 M CH3COONa where Ka=1.75x1 0-5
pH = pKa + log Cs/Ca
= -log 1.75 x 1 0-5 + log 1/1 = 4.76 + 0 = 4.76
Calculate the pH of a buffer solution consisting of 0.5
M NH4OH and 0.3 M NH4Cl where Kb = 1.8 x 10-5
pH = pKw – pKb + log Cb/Cs
= 14 – log 1.8 x 1 0-5 + log 0.5/0.3
= 14 – 4.745 + 0.222 = 9.477
Calculate the pH of a buffer solution consists of 1 M
CH3COOH and 1 M CH3COONa after addition of 0.1 mol
of HCl to one L of solution where Ka=1.75x10-5
after addition of 0.1 mol HCl , it will react with an
equivalent amount of CH3COONa forming the same
amount of CH3COOH
HCl + CH3COONa → CH3COOH + NaCl
0.1 mol 0.1 mol
0.1 mol 0.1 mol
Ca = 1+ 0.1 = 1.1
Cs = 1 - 0.1 = 0.9
pH = pKa + log Cs/Ca
= -log 1.75 x10-5 + log 0.9/1.1 = 4.757 + ( 0.087) =
4.67
It is a magnitude of the resistance of a buffer to
change in the pH
B = ΔB / Δ pH
- B is a buffer capacity
- ΔB is a strong acid or base added
- Δ pH is the change in pH
Buffer capacity is directly proportional to
concentration of buffer components
Solution has equal concentration of acid or (base)
and its salt appears to have the maximum buffer
capacity
- Buffer solution with high B is of high efficiency
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Neutralization indicators are weak acids or weak
bases which change their color according to the pH
of the solution
The acid form (HA) of the indicator has one color,
the conjugate base (A–) has a different color.
In an acidic solution, [H+] is high. Because H+ is a
common ion, it suppresses the ionization of the
indicator acid, and we see the color of HA.
In a basic solution, [OH–] is high, and it reacts with
HA, forming the color of A–.
The change of color is not sudden but
takes place within small interval of pH(2
pH units or less)
It is preferred to select an indicator
which exhibits color change at pH close to
that of salt formed at the end point
1 – Color indicators
Organic dyes that exhibit different colors at different
pH values
e.g. Methyl Orange (M.O.) pH range 3.3-4.4 red to
orange or yellow, Phenolphthalein(Ph.Ph.) pH range
8.3-10 colorless to pink , and Methyl Red (M.R.)
pH range 4.4-6.3 red to yellow
2 - Turbidity indicators
Precipitation or turbidity appears at the end point
e.g. Isonitrosoacetyl-p-aminobenzene
3 – Fluorescence indicators
Certain compounds emit visible radiations when
exposed to ultraviolet light stop or intensify when
certain pH is reached and used to detect end point
when color or turbid solutions are titrated e.g.
Umbelliferone
1 – Ostwald Theory:
- Neutralization indicators are either weak acids or weak
bases
- The color of ionized form differs from that of non-ionized
form
- In acidic medium basic indicators ionized and changed in
color e.g. M.O.
- In basic medium acidic indicators ionized and changed in
color e.g. Ph.Ph.
2 – Chromophore theory
- Indicators are Organic dyes which contain an unsaturated
group called chromophore group e.g. C=C , N=N , C=N ,
NO, NO2 which is responsible for the color change.
- Accumulation of unsaturated groups leads to color
development
- Presence of auxochromes (-OH, -NH2 ) influence the color.
It is the pH units over which the indicator changes its
color.
The color change within the effective range is gradual.
Effective range for a good indicator shouldn’t exceed 2 pH
units. Example: M.O. 3.3-4.4, M.R. 4.4-6.3 Ph.Ph. 8.310
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Mixed indicators
Sharper color produced by using mixture of two indicators
have the similar pH range but contrasting color.
Example: mixture of thymol blue with cresol red has:
Violet color at pH 8.4
Blue color at pH 8.3
Rose color at pH 8.2
Screened indicators
When the color change isn’t easily detectable particularly in
artificial light, addition of another indicator obtain Sharper and
more pronounced color change
Example screened mixture of M.O. and indogocarmine has:
At pH 4 yellowish green (alkaline) and violet (acidic).
Universal or multi-range indicators
The pH range can be extended By suitable mixing certain
indicators.
Example: mixture of Bromothymol blue with Ph.Ph.
has:
Red color at pH 2
Orange color at pH 4
Yellow color at pH 6
Green color at pH 8
Blue color at pH 10
Titration curve is the plot of pH versus the volume of
titrant
Titration curves are constructed to
- study the feasibility of titration
- choosing indicator
1- Titration curve of strong acid Vs strong base
- e.g. HCl against NaOH .
1- At beginning , pH of acid.
pH = - log [H+]
2- During titration, pH of strong acid.
pH = pCa
3- At the end point, pH of salt of strong acid and
strong base (neutral).
pH = pOH = ½ pKw
4- After end point , pH of strong base.
pH = pKw - pCb
Both M.O. and Ph.Ph. are suitable
- e.g. CH3COOH against NaOH.
1- At beginning , pH of weak acid.
pH = ½ pKa + ½ pCa
2- During titration, PH of acidic buffer.
pH = pKa + log Cs/Ca
3- At end point , PH of salt of weak acid and strong
base.
pH = ½ pKw + ½ pKa - ½ pCs
4- After end point , PH of strong base .
pH = 14 -pCb
So that M.O. Isn’t suitable
The suitable indicator Ph.Ph. pH range 8.3-10
The suitable indicator is Ph.Ph. Not
- e.g. NH4OH against HCl.
1- At beginning , pH of weak base.
pH = pKw - ½ pKb – ½ pCb
2- During titration, pH of basic buffer.
pH = pKw - pKb + log Cb /Cs
3- At end point , pH of salt of strong acid and weak base.
pH = ½ pKw - ½ pKb + ½ PCs
4- After end point , PH of strong acid.
pH = pCa
So that M.O. or M. R. are used and Ph.Ph. Isn’t useful
N.B. Titration curve of weak acid against weak base and weak
base
against weak acid.
Titration curves in both cases are smooth and change of pH at end
point is very small . So such titrations must be avoided.
This titration curve shown in the figure involves 1.0 M solutions of
an acid and a base. Identify the type of titration it represents.
It means in a medium free of water and mainly used for
determinations of weak acids and weak bases
Solvent Properties and Role of Solvent in Non-Aqueos Titration
1– Relative acidity and basicity :
-According to Bronsted, acidity and basicity of substance are
relative to the solvent e.g. potassium acid phthalate when
dissolved in water acts as an acid while in glacial acetic acid acts as
base.
-Similarly solvent behaves as an acid when the dissolved substance
is more basic e.g. acetic acid + pyridine and behaves as a base
when substance is more acidic e.g. acetic acid and perchloric acid.
2– Leveling effect:
-It is the ability of solvent to increase the strength of weak acids or
weak bases to reach that of strong acid or base .
-Acidic solvents have leveling effect on bases and basic solvents
have leveling effect on weak acids.
Example : acetic acid on amines and liquid ammonia on acetic
acid.
3– Differentiating effect:
-It is the ability of solvent to differentiate
between the strength of acids or bases e.g.
glacial acetic and mixture of HNO3, HCl,
HClO4.
4– Autoprotolysis effect:
-It is self dissociation of solvent
HA + HA ↔ A- +H2A+
-Two molecules of solvent interact ,one as
proton donor and one as proton acceptor.
1– Aprotic Solvents:
-These are neutral, inert, can't donate or accept protons
e.g. hexane, benzene , nitrobenzene, chloroform .
2– Amphiprotic Solvents:
-They act as acids or bases (may donate or accept protons):
a-Neutral solvents:
They have tendency to accept or donate proton e.g.
methanol, ethanol.
b-Protogenic solvents:
They are more acidic than water and have tendency to
give proton than accept proton e.g. acetic acid.
c-Protophillic solvents:
They are more basic than water and have higher
tendency to accept proton than to give proton e.g.
ammonia.