introduction to analytical chemistry -...
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1
Introduction to Analytical Chemistry
ANALYTICAL CHEMISTRY: The Science of Chemical
Measurements.
ANALYTE: The compound or chemical species to be measured,
separated or studied
TYPES of ANALYTICAL METHODS:
1.) Classical Methods (Earliest Techniques)
a) Separations: precipitation, extraction, distillation
b) Qualitative: boiling points, melting points, refractive
index, color, odor, solubilities
c) Quantitative: titrations, gravimetric analysis
2.) Instrumental Methods (~post-1930’s)
a) Separations: chromatography, electrophoresis, etc.
b) Qualitative or Quantitative: spectroscopy, electro-
chemical methods, mass spectrometry, NMR, radio-
chemical methods, etc.
2
Classification of volumetric methods:
There are four general classes of volumetric methods:
1-Acid-Base:
Many compounds, both inorganic and organic, are either
acids or bases and can be titrated with a standard solution of
a strong base or a strong acid. The reactions involve the
combination of hydrogen and hydroxide ions to form water.
The end points of these titrations are easy to detect, either by
means of indicator or by following the change in pH with a pH
meter. The acidity and basicity of many organic acids and
bases can be enhanced by titrating in non-aqueous solvent, so
the weaker acids and bases can be titrated.
2- Precipitation:
In the case of precipitation, the titrant forms an insoluble
product with the analyte. An example is the titration of chloride
ion with silver nitrate solution.
3- Complexomietric:
In complexometric titrations, the titrant is a complexing
agent and forms a water-soluble complex with the analyte, a
metal ion. The titrant is often a chelating agent.
3
4- Reduction-oxidation:
These "redox" titration involve the titration of an oxidizing
agent with a reducing agent, or vice versa. An oxidizing agent
gains electrons and a reducing loses electrons in a reaction
between them.
Expressions of concentration:
Concentrations of solutions:
Standard solutions are expressed in terms of molar
concentrations or molarity (M).
Molar solution is defined as one that contains one mole
of substance in each liter of a solution.
Molarity of a solution is expressed as moles per liter or as
millimoles per milliliters.
Moles = (moles/liter) x liters = molarity x liters
millimoles = molarity x milliliters
mmole = M x ml
liters in solution of Volume
solute of M ole (M ) M olarity
The equivalent weight:
The equivalent weight is that weight of a substance in grams
that will furnish one mole of the reacting unit. Thus, for HCI, the
equivalent weight is equal to the formula weight:
4
mol) / (eq 1
(g/mol) HC1 w t.Formula eq.w t.HCl
The milliequivalent weight is one thousandth of the eq.wt.
A normal solution contains one gram eq.wt. of solute in one
liter of solution:
No. of gram-equivalents
liters of No.
sequivalent - gram No. N
smilliliter of No.
sequivalent - milligram No.
By rearrangement of these equations we obtain the
expression for calculating other quantities:
No of gram eq. = N x No. of liters
No. of milligram eq. =Nxml = NxV
Back titration:
In back-titrations, a known number of millimoles of reactant is
taken in excess of the analyte. The unreacted portion is titrated
for example, in the titration of antiacid tablets with a strong
acid such as HC1.
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Lecture 1 (2 hrs) …./……/……….: (Acid-base)
Theoretical bases of neutralization reactions, Electrolyte and the theory
of electrolytic Dissociation, Strong and weak electrolytes, Law of mass
action , The dissociation of water, Hydrogen ion exponent (pH), Acids
and bases, Acid–base equilibrium (pH calculations), Solution of strong
acids and strong bases, Solution of weak acids and bases, pH of salts
Theoretical bases of neutralization
reactions
Electrolyte and the theory of electrolytic Dissociation:
Aqueous solutions of substances differ in their behavior
when submitted to an electric current. Some of them allow
the current to pass, i.e. they conduct the electric current,
these are termed "electrolytes"; while other do not allow the
current to pass, i.e. they yield non–conducting solutions, and
are called "non–electrolytes". The first class includes mineral
acid, caustic alkalies and salts, while the second class is
exemplified by cane sugar, glycerin and ethyl acetate.
Pure water is a bad conductor of electricity, but when acid
like HCl, a base such as KOH or a salt like Na2SO4 is dissolved
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in water its conductivity is greatly improved. At the same
time, it is noticed that the solute decomposes by the
passage of the electric current into its components at the
cathode and the anode. These components of the
electrolyte are called "ions".
The whole phenomenon was called by Arrhenius in 1887
"ionisation" or "dissociation"
Strong and weak electrolytes:
According to the theory of ionisation, the extent of ionisation
increases with dilution, and at very great dilution it is practically
complete. On the other hand, for each concentration there is
a state of equilibrium between the undissociated molecules
and the ions; the process being a reversible one.
NaCl ⇌ Na+ + Cl–
K2SO4 ⇌ 2K+ + SO42–
Na2HPO4 ⇌ 2Na+ + H+ + PO43–
The balance can be shifted to the right or the left according
to conditions.
Arrhenius therefore introduced a quantity " ", called "the
degree of dissociation" defined as follows
7
ondissociati before molecules solute of number Total
ddissociate molecules solute of Number
When the ionisation is complete, i.e. when all molecules have
dissociated, the value of "a" will be unity or very nearly so. The
electrolyte will be then called "strong electrolyte". On the other
hand this value for weak electrolytes is very far from unity.
Law of mass action
The law is concerned with reactions involving equilibrium
between the reactants and productants. Since this the case in
the dissociation of most electrolytes, this law is the basis for
many calculations in neutralimetric analysis.
The law reads:
"The rate of a chemical reaction is proportional to the active
masses of the reacting substances."
In dilute solutions, where conditions approach ideal state
"active mass" may be expressed by the concentrations of the
reacting species, that is gram–molecules or gram–ions per liter.
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In a reaction: A + B C + D
The velocity of the forward reaction [Vf] and of the backward
reaction [Vb] can expressed.
Vf = [A].[B]. Kf
Vb = [C]. [D]. Kb
Where Kf and Kb are the proportionality constants called
"Velocity constant” and brackets indicate concentration, at
equilibrium that is when Vf = Vb
Kf [A]. [B] = Kb [C]. [D]
And
[A][B]
[C][D]
K
K
b
f
Since Kf and Kb are both constants, the fraction Kf / Kb must
also be constant. Hence:
[A][B]
[C][D]K
Where K denotes the "equilibrium constant" of the reaction
(constant at a given temperature).
9
The dissociation of water
The dissociation of water is reversible and to a very limited
extent as illustrated by its very weak conductivity to an electric
current, and can be represented by the equation:
H2O ⇌ H+ + OH–
According to the law of mass action:
Since the fraction of water ionised is very minute or negligible,
the value of [H2O] is equal to (1) can be regarded as equation
1, and the equation can be written therefore:
[H+] . [OH] = Kw ……………………….. (2)
Kw is known as "The ionic product of water"
Under ordinary experimental conditions and at about 25oC;
the value of Kw is taken to be 1 x 1014 and it follows that;
[H+]. [OH] = 1014
Furthermore, since the dissociation of water gives rise to equal
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number of hydrogen and hydroxyl ions. Equation (2) could be
written:
[H+]2 = Kw = 1 x 1014 ………………………(3)
and in other words;
[H+] = 1410 = 10-7
it follows that, if the [H+] = [OH–] = 10–7 the solution is described
as "neutral", if [H+] is more than 10–7, that is 10–6, 10–5…, etc. the
solution is said to be "acidic" and if [H+] is less than 10–7, that is
10–8 , 10–9 ..etc., the solution is called "alkaline".
Hydrogen ion exponent (pH):
The hydrogen ion exponent was introduced as an easy method
for representing small changes in the ion concentrations, and
was called the "pH". pH is defined as equal to the logarithm of
the hydrogen ion concentration with a negative sign.
i.e , pH = –log [H+]
if , [H+] = 1 x 10–10
pH = –log 10–10 = 10
This method of stating hydrogen ion concentration has the
advantage that all degree of acidity and alkalinity between
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that of a solution molar (or normal) with respect to hydrogen
and hydroxyl ions can be expressed by a series of positive
numbers between 0 and 14. A neutral solution is one in which
pH = 7, an acid solution is one in which pH is less than 7 and an
alkaline solution in which pH is more than 7.
N.B. this method of expressing the concentration of hydrogen
ions as its negative exponent (pH) has been extended to
express other numberically small values as [OH–], Kw …… etc.
Thus pOH = –log [OH–]
pKw = –log Kw
The ionic product of water (1 x 10–14) could be thus expressed:
pKw = pH + pOH
pH = pKw – pOH
or pH = 14 – pOH
or pOH = 14 – pH
Acids and bases:
According to the classical theory of Arrhenius all acids when
dissolved in water dissociate giving rise to hydrogen ions as
positive ions. Bases, on the other hands, undergo dissociation
with the formation of hydroxyl ions [OH–] as the only negative
ions. The old definition of both acids and bases was laid,
therefore, on that observation. The acidity of a solution or its
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basicity can also be determined by measuring the amount of
either hydrogen ions or hydroxyl ions it contains, respectively.
The degree of dissociation of the dissolved acid or base can be
used to calculate the concentration of the ions present in the
solution.
According to the degree of dissociation acids can be divided
into two groups:
A) Strong acids, having a high degree of dissociation and
B) Weak acids, which are feebly dissociated.
Similarly strong bases have a high degree of ionisation. While
weak bases dissociate feebly.
Apart from monbasic acids, which dissociate in one stage,
polybasic acids dissociate in consecutive stages. Sulphuric
acid, for example, dissociate in two stages, in the first stage one
hydrogen is almost completely ionised, thus:
H2SO4 ⇌ H+ + HSO4–
In the second stage, the other hydrogen is only partially
ionised.
13
Phosphoric acid dissociates in three stages:
H3PO4 ⇌ H+ + H2PO4
H2PO4 ⇌ H+ + HPO4
2
HPO42 ⇌ H+ + PO4
3
These stages are called the primary, secondary and tertiary
dissociations, respectively, the first stage is the most complete
while the others are smaller and smaller.
The equilibrium, which exists in a dilute solution of an acid like
acetic acid (HAc) at constant temperature, is
HAc ⇌ H+ + Ac
Applying the law of mass action
Where K is called "dissociation" "ionisation" or "acidity constant"
The stronger the acid, the larger the acidity constant. For a
completely ionised acid, the acidity constant is assumed to be
1, and the mass action law does not help much in this case.
[HAc]
][Ac x ][H
K
14
In considering acids with more than one replaceable
hydrogen, such as sulphuric or phosphoric acids the
dissociation takes place on stages and not on one stage.
The corresponding mass action expressions are:
In the same case of phosphoric acid, it may be considered
that three acids are present. The first, H3PO4, corresponds to a
moderately strong acid, the second is a weak acid, while the
third is a extremely weak acid.
3.6x10K][HPO
]][PO[H PO H HPO
2.0x10K]PO[H
]][HPO[H HPO H POH
1.1x10K)PO(H
]PO][H[H POH H POH
K][HSO
]][SO[H SO H HSO
1K]SO[H
]][HSO[H HSO H SOH
13-3
4
444
7-2
42
4442
2-1
43
424243
-4
-24-2
4-4
142
-4-
442
2
332
22
21032
Also
x
15
Acid–base equilibrium
pH calculations
1. Solution of strong acids and strong bases
Since strong acids and strong bases are considered
completely ionised in solvents., the calculation of pH or pOH of
such reagents a simple matter, the [H+] or [OH–] is directly
related to the concentration of the substance. The following
examples are illustrative.
Example 1:
Calculate the pH value of a solution of a completely ionised
1.0 N solution of acid; or base. ?
[H+] = 1M
pH = –log 1 = 0 (zero)
similarly, in a completely ionised 1.0 N solution of base
[OH–] = 1 M
pOH = –log 1 = 0 (zero)
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Example 2
Calculate the [H+] and pH of 0.009 N hydrochloric acid?
[H+] = 0.009 N
pH = – log (9.0 X 10–3) = 2.05
Example 3
Calculate the pH values of a solution of sodium hydroxide
whose [OH–] is 1.05 x 10–3?
pOH = – (log 1.05 x 10–3) = 2.98
pH = 14 – 2.98 = 11.02
Example 4
Calculate the hydrogen ion concentration of a solution of
pH 5.3?
pH = – log [H+]
5.3 = –log [H+]
[H+] = 5.01 x 10–6 M
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Example 5
Calculate the hydroxyl ion concentration of a solution of pH
10.75 ?
pOH = 14 – 10.75 = 3.25
[OH–] = the antilog of –3.25
[OH–] = 5.62 x 10–4 M
2. Solution of weak acids and bases
A) Calculation of pH of solution of weak acids
Weak acids do not ionized freely and only a small fraction of
the molecules is partially ionized. To calculate the pH of a weak
acid HA, the law of mass action is thus applied to its
dissociation equilibrium:
HA ⇌ H+ + A –
and
[HA]
][A ][H
aK
Where Ka is the ionisation constant or dissociation constant of
the acid.
If the acid is pure, its ionisation gives equal concentration of H+
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and A– ions, and since their activities may be assumed equal in
dilute solutions, therefore:
[H+] = [A–]
If the total acid concentration is Ca moles/liter (and can be
determined by titration with standard base), then the moles of
the unionized acid [HA] must be numerically equal to Ca – [H+],
so that the above equation becomes
][HC
][HK
a
2
a
Now, the acid is weak and slightly ionised, thus [H+] is very
small compared to Ca and can be neglected in the
denominator in the above equation. Therefore:
aaa
2
a CK][H and C
][HK
And
pH = ½ (pKa + pCa) (1)
Example: 1
Calculate the pH and [H+] of 0.10 N acetic acid (pKa=
4.76)?
pH = ½ (pKa + pCa)
= ½ x 4.76 + ½ (– log 10–1) = 2.38 + 0.5 = 2.88
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Example 2
Calculate the pH and [H+] of a 0.0045 M solution of
phenobarbital (pKa= 7.41)?
PCa = – log (4.5 x 10–3) = 2.35
pH = ½ (7.41 + 2.35) = 4.88
[H+] = antilog – 4.88 = 10-4.88
= 1.32 x 10–5 M
B) Calculation of pH of solution of weak bases:
The same method discussed in the previous section is readily
adaptable to the calculation of the pH of solutions of weak
mono–equivalent bases; e.g. ammonia to give:
pH = pKw – ½ (pKb + pCb) (2)
Example 1:
Calculate the pH, and the [H+] of 0.13 N ammonia solution
(pKb = 4.76)?
pH = 14 – ½ (4.76) – ½ (– log 1.3 + 1)
= 14 – ½ (4.76 + 0.89) = 11.18
log [H+] = 0.82 – 12
[H+] = antilog 0.82 x 10–12 = 6.6 x 10–12 M
20
Example 2:
Calculate the pH and the [H+] of a 0.0037M solution of
cocaine base (pKb = 5.59)?
pH = 14.00 – ½ (5.59 + 2.43) = 9.99
[H+] = antilog of – 9.99 = 1.03 x 10–10
3. Calculation the pH of salts
When salt is dissolved in water, the solution is not always neutral
in reaction. Interaction occurs with the ions of water and the
resulting solution may be neutral acid or alkaline according to
the nature of the salt.
A) Salts of strong acids or bases
An aqueous solution of such salts, for example potassium
chloride, is neutral in reaction. Neither the cation nor the anion
has the tendency to hold any hydrogen or the hydroxyl ions of
water; the related acids and bases being strong electrolyte
KCl K+ + Cl-
K+ + OH- ⇌ KOH
Cl- + H+ ⇌ HCl
The equilibrium between the hydrogen and hydroxyl ions in
water
H2O ⇌ H+ + OH–
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B) Salts of weak acids or bases (hydrolysis)
Salts of weak acids (or bases) react with water to give basic
(or acidic) solutions respectively. This phenomenon is known as
hydrolysis; it is the reverse of neutralization. A hydrolytic
reaction proceeds because of the tendency of the ions of salts
of weak acids (or bases) to react with the hydrogen ions (or
hydroxyl ions) of water, forming slightly ionized acids (or bases).
The reaction of these salts with water does not proceed to
completion; it reaches an equilibrium point and thus
represented by an equilibrium expression and an equilibrium
constant, known as hydrolysis constant, Kh. the extent to which
the hydrolytic reactions proceed; is related to the ionisation
constants of the acids or bases formed; the lower the ionisation
constant, the more is the extent of hydrolysis.
C) Salts of weak acids and strong bases:
Sodium acetate is an example of such salts. Its hydrolysis in
water may be represented by the following equation:
H2O + CH3COO ⇌ CH3COOH + OH
Since the sodium ion of sodium acetate does not react with
water it is not included in the equation. But acetate ion reacts
with water to form the slightly ionised acetic acid and to satisfy
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the acid constant Ka. This requires that some of the water
molecules ionise to maintain the ion product of water, Kw,
constant; and in doing so, produces more hydroxyl ions.
Consequently, when equilibrium is reached there is an
increased hydroxyl ion concentration and a decreased
hydrogen ion concentration. For this reason, a solution of a salt
of a weak acid and a strong base is alkaline in reaction, and:
pH = ½ (pKw – pCa + pKa) (3)
The degree of hydrolysis, h, of a salt, analogous to the degree
of ionisation of a weak acid or base, is the fraction of the salt
hydrolysis when equilibrium is established, thus;
SC
y h
Example 1:
Calculate the pH, the [H+], the [OH–], and the degree of
hydrolysis of a 0.1 M solution acetate (pKa = 4.76)?
pH = ½ (14.00 – 1.00 + 4.76) = 8.88
[H+] = 1.3 x 10–9 M
[OH+] 10–14/ 1.3x10–9 = 7.5x10–6 M
h = 0.1
7.5x10 6
7.5 x 10–5
23
% h hydrolysed 0.007%)0100x(7.5x10.1
7.5x10 56
In spite of the small degree of hydrolysis. Yet the [OH–] is 75
– fold greater in this salt solution than in pure water
Example 2:
Calculate the pH of a 0.165 M solution of sodium
sulphathiazole (pKa = 7.12)?
pCs = – log 1.65 x 10–1 = 0.78
pH = ½ (14.00 – 0.78 + 7.12) = 10.17
D) Salts of weak bases and strong acids:
Consider the case of ammonium chloride; its hydrolytic
reaction is:
NH4+ + H2O ⇌ NH4OH + H+
For which
pH = ½ (pKw + pCs – pKb) (4)
The degree of hydrolysis, h, of a salt is y/Cs and the
percentage of the salt hydrolysed is: 100 x (y/Cs)
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Example 1:
Calculate the pH, the [H+] and the degree of hydrolysis of
a 0.05 N solution of ammonium chloride.?
pH = ½ (14.00 + 1.30 – 4.76) = 5.27
[H+] = 5.4 x 10–6 M
Degree of hydrolysis = 46
1.1x100.05
5.4x10
Percentage of hydrolysis = 100 x (1.1 x 10–4) = 0.011 %
Example 2:
Calculate the pH of a 0.025 M solution of ephedrine
sulphate (pKb = 4.64)?
pCs = – log 2.5 x 10–2 = 1.58
pH = ½ (14.00 – 1.58 – 4.64) = 5.47
E) Salts of weak acids and weak bases:
Such salts, for example ammonium acetate, undergo hydrolysis
in aqueous solutions. Provided the dissociation of the acid and
the base are not widely different, the hydroxyl and hydrogen
ions will be produced in approximately equal amounts.
And,
pH = ½ pKw + ½ pKa – ½pKb (5)
25
Lecture 2 (2 hrs) …../…./……. (Acid-base)
Buffer solution, Type of buffers, Henderson equation, Properties of buffer
mixtures, Buffer capacity
Buffer solutions
Buffers are mixtures of compounds which by their presence in
solution, resist changes in pH caused by addition of small
amounts of acid or base; or upon dilution. The resistance to a
change in pH is known as buffer action.
Types of buffers
One type of buffer solution is readily prepared by dissolving a
weak acid [HA] and its salt [A–] in water. Both components are
necessary. If to such a buffer solution, acid is added, its
hydrogen ions will be removed by the anions of the salt
component of the buffer
H+ + A – ⇌ HA
Although more of weak acid is thus formed, yet very little
difference will occur in the [H+] of the solution because the very
low ionisation of the weak acid under these conditions.
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If base is added to such a buffer solution, the hydroxyl ions will
be removed by the weak acid to form strongly ionised salt.
Again, little change is found in the [H+]:
OH– + HA ⇌ A– + H2O
In such a buffer solution, the weak acid [HA] is said to be in
reserve: unable to increase the [H+] of the solution but available
to neutralize any base that may be added. Likewise, the salt, is
in reserve; unable to contribute to the [OH–] of the solution, but
available to neutralize any acid that may be added.
Another type of buffer is prepared from a weak base [BOH]
and its salt [B+]. The mechanism of its buffering action is added
its hydrogen ions are removed by the base as the fully ionised
salt, with almost no change in the [H+]:
H+ + BOH ⇌ B+ + H2O
If base is added, its hydroxyl ions will be removed by the salt
cations, forming more of the very slightly ionised base, with very
little or no change in the [H+]:
OH– + B+ ⇌ BOH
27
The Henderson equation:
The pH of buffer solution and the change in pH upon the
addition of an acid or base, may be calculated by the use of a
certain equation called the Henderson equation.
A) Henderson equation for acidic buffer:
In an acidic buffer mixture, the acid is less ionised than if it
were alone, because of the presence of the highly ionised salt
of the acid which provides a high concentration of the
common anion.
Whatever the relative proportions of the acid and its salt in the
solution, their concentrations must always be related to the
hydrogen ion concentration of the acid according to the
ionisation constant:
[HA]
]][A[H Ka
Taking logarithms of both sides, and separating the involving
the hydrogen ion concentration ion the right-hand side:
[HA]
][A log ][H log K log
-
a
pka = pH – log [HA]
][A
28
pH = pKa + log [HA]
][A
………………………. (1)
Since HA is a weak acid and only slightly ionised, and its
ionisation suppressed by the presence of the common ion of its
salt A-, the quantity [HA] is almost equal to the original
concentration of the acid and [A-] equal to the concentration
of salt.
Equation 1 could read
pH = pKa + log ][
][
Acid
Salt……………………..(2)
Example 1:
Calculate the pH of a solution containing 0.10 N acetic acid
and 0.10 N sodium acetate?
According to Henderson equation:
pH = 4.76 + log 4.760.10
0.10
If the pH of this solution is compared with the pH of a solution
which contains 0.10 N acetic acid only; then
pH = ½ (pKa + pCa) = ½ (4.76 + 1) = 2.88
29
It is seen that the pH of acetic acid solution has been
increased almost 2 pH units; i.e. the acidity has been reduced
to about one – hundredth of its original value by the presence
of an equal concentration of a salt with common ion.
Example 2
Calculate the pH of the solution produced by adding 10.0 ml of
N HCl to 1 liter of solution which is 0.1 N in acetic acid and 0.1 N
in sodium acetate (Ka= 1.82 x 10-5)
pH = pKa + log ][
][
Acid
Salt
Neglecting the volume change from 1000 to 1010 ml, the HCl
reacts with acetate ion forming practically undissociated
acetic acid.
H+ + CH3COO ⇌ CH3COOH
[CH3COO] = 0.1 – 0.01 = 0.09
[CH3COOH]= 0.1 + 0.01 = 0.11
pH = 4.74 + log0.11
0.09 = 4.74 – 0.09 = 4.65
Hence the addition of strong acid, the pH change only by 4.74
– 4.65 = 0.09 pH unit, whereas, if 10 ml of N-hydrochloric acid
were added to liter of pure water (pH =7), the pH would have
changed from 7 to – log (0.01) = 2, i.e by 5 pH units. This
illustrates the action of the acetic acid – sodium acetate buffer
30
B) Henderson equation for basic buffer:
pH = pKw – pKb – log [BOH]
][B
Now, the base BOH is weak and only slightly ionised. Also, its
ionisation is repressed by the relative large concentration of
cations B+ from the highly ionised salt. Therefore, [BOH] is
numerically equal to the initial concentration of the base, and
[B+] is numerically equal to the initial concentration of the salt.
Thus, equation becomes:
pH = pKw – pKb – log [Base]
[Salt]
Example:
Calculate the pH of a solution containing 0.07 N ammonia, and
0.28 N ammonium chloride?
pH = 14.00 – 4.75 – log0.07
0.28 = 8.64
If the pH of this solution s compared with the pH of a solution
which contains 0.07 N ammonia only; being:
31
pH = 14.00 - ½(4.76 + 1.16) = 11.04
Therefore, the addition of the ammonium chloride has
decreased the ionisation of the base such that the pH was
decreased from 11.04 to 8.64.
Properties of buffer mixtures:
a) Effect of temperature: the change in pH value of buffers
with temperature is slight in case of acidic buffers (those
composed of weak acid and its salt). But the pH of most
basic buffers change more markedly with temperature;
owing to Kw which appears in the equation of these
buffers and which changes significantly with
temperature.
b) Effect of dilution: it is noted from Henderson equation
(for both acidic and basic buffer solutions) that the pH
depends on the ratio of the molar concentrations of the
two compounds in a buffer solution. Therefore dilution
should have no effect on pH because the volume term
cancels out; the concentration of each component
changes in a proportionate manner upon dilution.
32
Buffer Capacity
The buffer capacity is defined as the number gram equivalent
of strong acid or strong base required to change the pH of one
liter of buffer solution by one unit. Buffer capacity can be
calculated if the composition of the buffer is known.
N.B.
1. Higher the concentrations of the two components of the
buffer solution, the higher is the buffer capacity.
2. Also, the buffer capacity is a maximum when the two
components are present in equal concentrations,
The following illustrates these points.
Example:
Calculate the buffering capacity of a solution containing 0.1
gm equivalent of sodium acetate and acetic acid? PKa = 4.76
pH = 4.76 + log 1.00 = 4.76
To increase (or decrease) the pH by one unit, the ratio salt/acid
will have to alter by a factor of 10:
5.76 = 4.76 + log [acid]
[salt]
log [acid]
[salt] = 1; therefore
[acid]
[salt] = 10
To reach pH 5.76, then [salt] will have to increase to a
33
concentration of about 0.182-gram equivalent, and the [acid]
to decrease to about 0.018. The buffering capacity of the
buffer mixture is therefore; 0.100–0.082 gram equivalent towards
strong base, because the decrease in acid is brought about by
adding 0.082 gram equivalent of strong base.
Table 1 gives compositions of some buffer systems and the pH
ranges in which they are used
Table 1 Some Buffer System
Solution pH rang
Phethalic acid and potassium phethalate 2.2 – 4.2
Citric acid and sodium citrate 2.5 – 7
Acetic acid and sodium acetate 3.8 – 5.8
Sodium dihydrogen phosphate and and disodium
hydrogen phosphate
6.2 - 8.2
Ammonia and ammonium chloride 8.2 – 10.2
Borax and 9.2 - 11.2
34
Lecture 3 (2 hrs) ……./…../………. (Acid-base)
Acid-Base indicators , Oslwald Theory, Transition of indicators, Mixed
indicators, Screened indicators, Turbidity indicators, Universal indicators
Acid-Base indicators
Acid-base indicators are substances whose presence during
a titration renders the end-point visible. Thus, at a certain pH very
near, or at the equivalence point of the titration the indicator
produces in the system a changes which is easily perceptible to
the eye, and may consist of: a- Sharp transformation from one
colour to another or to colourless.
Most of the colour acid base indicators of practical value
are organic in nature. As the colour changes of these indicators
depend on the change of the pH, they must themselves be
acids or bases. The equilibrium between the indicator molecules
and their ions may be represented, as follows;
Acidic indicator Hln ⇌ H+ + In- (1)
Basic indicator InOH ⇌ OH- + In+ (2)
Where HIn is the unionised form of the acidic indicator, which gives
the acid colour; In- is the ionised form which produces the basic
colour; InOH is the unionised form of the basic indicator, which gives
35
the basic colour; In+ is the ionised form, which produces the acid
colour.
Oslwald Theory:
If the acid indicator is added to an acidic solution, the
concentration of the hydrogen ion term on the right-hand side
of equation (1) is increased and the ionization of the indicator is
repressed by the common ion effect. The indicator is then
predominantly in the unionised form of HIn, the acid colour. If on
the other hand, the indicator is added to a basic solution, the [H+] is
reduced by reaction of the acid indicator with the base, and
reaction (1) proceeds to the right yielding more ionised indicator
[In-] and the basic colour predominates.
The reverse is true for basic indicators; in basic solution, the
reaction (2) proceeds to the left and the basic colour is
prominent.
Objectives of Oslwald theory:
1. Phenolphthalein indicator has red colour in slightly alkaline
solution, when more alkali is added give colourless, while the
expected from the theory, the colour should be increased.
2. Slow colour change in some indicators, while ionic
reactions are usually instantanious.
3. Some indicators show their colour changes in non aqueous
36
media where ionisation is markedly decreased.
In other words, the change in colour of indicators is a
process of tautomerism, and the degree of ionisation of
indicators, as controlled by the pH, is the factor which
determined tautomer predominates. The nature of this process
may be demonstrated in the case of the nitrophenols. In basic
solutions, p-nitrophenol is present chiefly in the yellow ion; while in
acid solution it is present as the colourless nitro compound:
It should be noted that not all substances which show
tautomeric properties can be used as indicators. The tautomeric
change must be rapid, and must occupy a rather small range
of pH. Therefore consideration of the tautomeric equilibria
modifies the Ostwald equation. If the formula HIn represents the
normal indicator molecule and the formula HIn” represents the
molecule formed by rearrangement (the tautomer), then the
indicator salts is eventually produced from neutralisation of weak
equilibrium:
37
HIn HIn” H+ + In"– (3)
Considering the two equilibra separately:
a
"
eq. K][HIn"
]][In[H and K
[HIn]
][HIn"
(4)
Multiplying these two equations:
ind.K[HIn]
]"][In[H
This is called the indicator constant, and not die ionization
constant of the indicator. The equation can also be written in
the manner of Henderson equation for buffers:
pH = pKind. + log[HIn]
][In"
It is thus clear that any change in the pH, causes a change in the
ratio of the logarithm tern:
[HIn]
][In"
i.e colour] [acid
colour] [basic
So that at any pH value, both colours are present.
In case of indicators in which the coloured ion is the cation and
not the anion it is possible to derive a relation between colour
and pH, similar to the above relation:
OHIn OHIn" In''- + OH–. ( 5 )
38
In case of indicators in which the coloured ion is the cation and
not the anion it is possible to derive a relation between colour
and pH, similar to the above relation:
OH In OH In" In"+ + OH–
ind.
"K
In] [OH
][In][OH
and pOH = pKind – log ][In
In] [OH
"
Therefore, pH = pKw – pKind – log In] [OH
][In"
in the other work, any change in the pH causes change in the
ratio:
In] [OH
][In" i.e
colour] [basic
colour] [acid
Transition range of acid – base indicators:
The most efficient transition range of acid base indicators,
corresponding to the effective buffer interval, is about 2 pH
units; i.e pKind. 1., the reason for the width of this colour range
may be explained as follows. The ability of the human eye is not
overly acute; and in general, the first change in the acid colour
of an indicator becomes visible when the ratio (basic colour)
39
/(acid colour) becomes 1/10. The pH value at which this colour
change is distinguished is given by the equation:
pH = pKind + log 10
1 = pKind –1
Conversely, the eye cannot detect a change in the basic
colour of the indicator until the ratio (basic colour)/(acid
colour) has become 10 /1, or:
pH = pKind + log1
10 = pKind +1
Therefore, when base is added to a solution of an indicator in
its acid form, the eye first visualizes a change in colour when pH
= pKind. – 1, and the colour ceases to change when pH = pKind. +
1. In other words, the indicator changes from the full acid
colour to the full basic colour through a range that extends 2
pH units, this is the effective transition range of the indicator,
and may be expressed as follows:
pH = pKind. 1
Between the two ratios: 1/10 and 10/1, one observes an
intermediate colour. An indicator therefore, does not change
colour suddenly at a definite pH, but changes colour gradually
over a certain pH range called the transition range of the
indicator.
40
Table summarizes the details -of some useful acid-base indicators.
Structurally, the indicators form three groups: phthaleins (e.g.
phenolphthalein); sulphonephthaleins (e.g. phenol red); and azo
compounds (e.g. methyl orange).
A range of visual indicators of acid-base titrations
Indicator
pKind
Low pH
colour
High pH
colour
Experimental colour
change range/pH
cresol red
-1.0
red
yellow
0.2-1.8
thymol blue
1.7
red
yellow
1.2-2.8
bromo-phenol blue
4.0
yellow
blue
2.8-4.6
methyl orange 3.7
red
yellow
3.1-4.4
methyl red
5.1
red
yellow
4.2-6.3
bromo-thymol blue
7.0
yellow
blue
6.0-7.6
phenol red
7.9.
yellow
red
6.8-8.4
phenolphthalein
9.6
colourless
red
8.3-10.0
alizarin yellow R
11.0
yellow
orange
10.1-12.0
nitramine
12.0
colourless
orange
10.8-13.0
The well-known indicator phenolphthalein is adiprotic acid and
is colorless. It dissociates first to a colorless form and then, on losing
the second hydrogen to an ion with a conjugated system; a red
colour results
41
colorless pink colorless
pH ≤ 8 pH 8 – 12 pH ≥ 12
Benzenoid structure Quinonoid structure Tribasic salt
Methyl orange, another widely used indicator, is a base and is
yellow in the molecular form. Addition of a hydrogen ion gives a
cation which is pink in color.
Yellow (azo – structure) Red
pH ≥ 4.4 pH ≤3.1
N.B. Increasing the concentration of indicators has a serious effect
on one-colour indicators such as phenolphthalein.
Let us write the equilibrium expression for the dissociation of the
colourless acid form of phenolphthalein as
aK[HIn]
]][In[H
or as [H+] = ka[HIn]/[In-]
42
Where Ka is the dissociation constant and [In-] is the minimum
detectable concentration of the pink base form, which we may
assume to be constant, hence
[H+] = Ka[HIn]
From the above equation, it is seen that the pH at which the
pink end point colour appears will depend upon the total
concentration of the indicator. If more indicator is present, the pH
at the end point will be lower. Conversely, if less indicator is present,
the end point pH will be shift toward hitter values.
In the case of the two colour indicators where the acidic and
the basic forms are both coloured, the transition range is
independent of the concentration. Thus, if the total concentration
of a two colour indicator is increased, the individual
concentrations of the acid and the base forms will increase
proportionally and the pH transition range should remain
unchanged even though the colour intensities are increased.
Mixed indicators:
These indicators are used when it is necessary to locate the
pH of an end point within close limits. This close adjustment of
pH may be obtained by using a suitable mixture of two
43
indicators, to produce a definite and characteristic colour
change within a very narrow range of pH. An example of such
mixed indicator is bromocresol green and methyl red; the
acidic and basic colours of the mixture are orange and green
respectively.
Screened indicators:
Screened indicator increase /the sharpness of the colour
change at the end point of. a titration. A screened indicator is
a mixture of an indicator and an inert dye whose colour does
not change with pH. The effect of the dye is to decrease the
range of wavelengths transmitted by the solution, so that the light
transmitted by the two coloured forms of the indicator is not
masked by ught of other wavelengths. An example is the so-
called "modified methyl orange" k is a mixture of methyl orange
with the inert dye xylene-cyanol F.I. This screened indicator is
purple-red in acid medium and green in alkaline medium, and
gray at its intermediate point.
Turbidity indicators:
Turbidity indicators are salts of weak organic acids or bases of
high molecular weight, which coagulate and settle out of the
solution at a definite pH value. It should be noted that not only
44
the pH of the solution influences the coagulation of the
indicator but also the temperature, the presence of other salts
and protective colloids, the speed of the titration, and the
presence of non electrolytes as glycerin, alcohol, etc.
Nevertheless, these turbidity indicators are useful in titrating
weak acids or bases.
Universal of multi-range indicators:
By suitable mixing of several indicators, the colour change
made to extend over a considerable portion of the pH range,
such indicators are called "universal" or "multirange" indicators.
They are not suitable for titration but indicate roughly the pH of
the solution e.g. a mixed indicators containing thymol blue,
methyl red, bromothymol blue and ph.ph. give the following
colors at various pH values:
pH 2 4 6 8 10
colour red yellow blue orange green
45
Lecture 4 (2 hrs) …./…/….. (Acid-base)
Neutralisation titration curves, Titration curves of strong acid versus strong
bases, Titration curves of strong bases versus strong acids, Titration curves
of weak acids versus strong bases, Titration curves of weak bases versus
strong acids
Neutralization curves
An insight into the mechanism of neutralization processes is
obtained by studying the changes in hydrogen ion concentration
during the course of the appropriate titration. The curve
obtained by plotting pH as ordinates against the percentage
of acid neutralized (or the number of nil of alkali added) as
abscissa is known as the neutralization curve.
1- Strong acid versus strong base:
In case of strong acid versus strong base, both the titrant
and the analyte are completely ionized. An example is the
titration of hydrochloric acid with sodium hydroxide.
H+ + Cl- + Na+ + OH- →H2O + Na+ + Cl-
The H+ and OH- combine to form H2O, and the other ions (Na+
and Cl-) remain unchanged, so the net results of neutralization is
46
conversion of the HC1 to a neutral solution of NaCl, To titration
curve for 100 ml of 1 M HC1 with 1 M sodium hydroxide solution. It
is a simple matter to calculate the pH values at different points
in the titration and from them to plot a titration curve, consider,
for example, the following points in the titration:
a) At the beginning of titration:
We have an acid concentration of 100x1 = 100 milliequivalent
per 100ml [H+]=lN
pH= -log [H+] = - log 1 = zero
b) During the titration:
For 50 ml of base: [H+] = 50x1/1 50 =3.33xl0-1, or pH=0.48.
For 75 ml of base: [H+] = 25 x 1/175 = 1.43 x 10-1, or pH = 0.94.
For 90 ml of base: [H+] = 10 x 1/190 = 5.27 x l0-2, or pH = 1.30.
For 99 ml of base: [H+]=1x1/199 =5.03x l0-3. or pH = 2.30.
For 99.9 ml of base: [H+ ]= 0.1 x 1/199.9 = 5.01x10-4. or pH = 3.30
c) At the equivalence point:
(The point at which the reaction is theoretically complete).
When acid and alkali have been added in exactly equivalent
point, the solution contains only NaCI and water. The pH value of
the solution is 7.00
47
d) Beyond the equivalence point:
The solution contains excess alkali:
With 100.1 ml base
[OH-] = 0.1/200 =5.00x10-4 or pOH 3.3 and pH =10.7
With 101 ml base
[OH-] = 1/201 =5.00x10-3 or pOH 2.3 and pH =11.7
The results show that as the titration proceeds the pH rises
slowly, but between the addition of 99.9 and 100.1 ml of alkali,
the pH of the solution rises from 3.3 to 10.7, i.e in vicinity of the
equivalence point the rate of change of pH of the solution is
very rapid. The appropriate indicator is one that changes
colour between pH 33 and pH l0.5.
48
Phenolphthalein methyl red and methyl orange are most
often used.
2. Titration curves of Strong bases versus Strong acids:
The derivation of the titration curves in this case is analogous to
that for strong acid versus strong base involving only the
calculation of the concentration of excess base or excess acid
at any point in the titration.
3. Titration curves of weak acids versus strong bases:
In the derivation of titration curves for a solution of a weak
49
acid versus a strong base, the ionisation equilibrium of the acid
must be taken into account; and four types of calculations
must be used corresponding to the four distinct parts of the
curve.
Consider the titration of 100.0 ml. of 0.10 N acetic acid (pKa =
4.76) with 0.10 N sodium hydroxide.
a) pH before adding titrant:
the pH of acetic is calculated using equation
pH = ½ (pKa + Ca)
pH = ½ (4.76 + 1.00) = 2.88
b) pH during titration:
The additions of base procedures a buffer of acetic acid and
sodium acetate. The pH of the solution is calculated from the
Henderson equation.
pH = pKa + log [Acid]
[Salt]
Thus, on adding 25 ml. of base, the molar concentrations of the
two components of the buffer will be:
[acid] = 75 x 125
0.1 [salt] = 25 x
125
0.1
pH = 4.76 + log 25
75 = 4.26
on adding 50 ml. of base, the molar concentrations of acid
and salt will be identical; so that the log term in the Henderson
50
equation will be omitted, and the pH = 4.76.
Similar calculations will delineate the curve in the entire buffer
region, as in the next figure and table
c) pH at the equivalent Point:
At the equivalence point, the acetic acid is quantitatively
converted to sodium acetate whose molar concentration is:
[NaAc] = 200
0.1 x 100
The pH of the solution is calculated from equation
pH = ½ (pKw – pCs + pKa)
pH = ½ (14.00 – 1.30 + 4.76) = 8.73
d) pH beyond equivalence point:
Beyond the equivalence point, the excess sodium hydroxide
present represses the hydrolysis of the acetate ion to such an
extent that its concentration becomes negligibly small.
On adding 50.1 ml. of base:
[NaOH] = [OH–] = 200.1
0.1 x 0.1 = 5.0 x 10–5
pOH = 4.3
pH = 14.00 – 4.30 = 9.7
The data from these calculations are given in table and plotted
in the next figure comparing this curve with that in the figure for
strong acid, it is apparent that both curves are identical in the
51
region beyond the equivalence point. But the pH of the
solution is higher in the curve for weak acid for all points up to,
and including, the equivalence point. This major difference
between the two curves results in a decrease in the magnitude
of the pH change in the region of the equivalence point.
Changes of pH during titration of 100 ml. of 0.1 N Acetic acid with 0.1 N
sodium hydroxide
NaOH added
ml.
pH NaOH added
ml
pH
0.0 2.88 99.9 7.7
10.0 3.9 100.0 8.7
25.0 4.3 100.1 9.7
50.0 4.7 100.2 10.0
90.0 5.7 101.0 10.7
99.0 6.7
52
Effect of concentration:
Using dilute reagents (e.g. 0.01 N), the change in pH associated
with the equivalence point becomes less, as seen in figure. The
initial pH is higher, but in the buffer region the two curves are
identical, because the pH of buffer solutions in almost
independent of dilution
Indicator choice; Feasibility of titration:
The above curves clearly indicate the limited choice of
indicators for the titration of a weak acid. Thus, methyl orange
cannot be used because its transition range does not fall within
the vertical part of the curves. Bromothymol blue is not
satisfactory, since its colour – change requires the addition of 1
or 2 ml. of excess titrant. A suitable indicator should be one
whose transition range exists in the basic region of the pH scale,
e.g. phenolphthalein.
4. Titration curves of weak bases versus strong acids:
The derivation of titration curves of weak bases with strong
acids is analogous to that described for the case of weak
acids and strong bases, except that with weak base as
indicator having an acid transition range is required. E.g.
titration of ammonia against HCl)
53
54
Lecture 5 & 6 (4 hrs) …./…./…….. (Acid-base)
Application of neutralization reactions, Direct titration methods,
Displacement titration, Biphasic titration, Residual titration
Application of Neutralization Titrations
Neutralization titrations are used to determine the
innumerable inorganic, organic and biological species that
posses inherent acidic or basic properties. Equally important,
however, are the many applications that involve conversion of
analyte to an acid or base by suitable chemical treatment
followed by titration with standard strong base or acid.
Determination of Inorganic Substances
1. Determination of ammonium salts (e.g. NH4Cl)
1. Formol titration:
When formaldehyde is added to solution of ammonium salt,
hexamethylene tetramine [(CH2)6N4] is produced accompanied by an
amount of acid equivalent to the ammonium salt present in the solution.
The librated acid can be titrated against standard alkali using
phenolphthalein (ph.ph.) as indicator.
4NH4C1 + 6HCHO → (CH2)6N4 + 4HCl + 6 H2O
55
2. Direct method:
In this method a solution of the ammonium salt is treated with
a solution of strong base (e.g. sodium hydroxide) and the
mixture distilled using the distillation apparatus. Ammonia is
quantitatively expelled, and is absorbed in an excess standard
acid. The excess of acid is back titrated in the presence of methyl-
orange as indicator.
NH4+ + OH− → NH3↓ + H2O
3. The indirect method
The ammonium salt (other than carbonate or bicarbonate) is
boiled with known excess of standard sodium hydroxide solution.
The bailing is continued until no more ammonia is evolved. The
excess of sodium hydroxide is titrated with standard acid, using
methyl orange as indicator.
2. Determination of Carbonate and Bicarbonate in Mixture
The analysis of such mixture requires two titrations, one with an
alkaline-range indicator, such as ph.ph., and the other with an acid-
range indicator, such as methyl-orange.
56
Na2CO3 + HC1 → NaHCO3 + NaCl . . . . . . . . . pH 8.3
MaHCO3 + HCl → CO2 + H2O + NaCl . . . . . . pH 3.8
A portion of cold solution is slowly titrated with standard
hydrochloric acid using ph.ph. as indicator. This volume of acid
(V1) corresponds to half the carbonate:
CO32- + H+ → HCO3
-
Another sample of equal volume is then titrated with the same
standard acid using methyl-orange, as indicator. The volume
of acid (V2) corresponds to carbonate + bicarbonate; hence 2
V4 = carbonate and V2 - 2Vi = bicarbonate.
Titration curve for 50 ml 0.1 sodium carbonate versus 0.1 M HCl
57
3- Determination of a Mixture of Carbonate and Hydroxide
(analysis of commercial caustic soda)
Two methods are used for this analysis. In the first method the
total alkali (carbonate + hydroxide) is determined by titration
with standard acid, using methyl orange as indicator. In a second
portion of solution the carbonate is precipitated with a slight
excess of barium chloride solution, and, without filtering, the
solution is titrated with standard acid using phenolphthalein as
indicator. The latter titration gives the hydroxide content, and by
subtracting this from the first titration, the volume of acid required
for the carbonate is obtained.
Na2CO3+ BaCl2 = BaCO3 (insoluble) + 2 NaCl
The second method utilizes two indicators. It has been stated
that the pH of half-neutralised sodium carbonate, i.e. at the
sodium hydrogen-carbonate stage, is about 8.3, but the pH
changes comparatively slowly in fee neighborhood of the
equivalence point; consequently the indicator color-change with
phenolphthalein (pH range 8.3-10.0) is not too sharp. This difficulty
may be overcome by using a comparison solution containing
sodium hydrogen-carbonate of approximately the same
concentration as the unknown and the same volume of indicator.
58
A simpler method is to employ a mixed indicator composed of
thymol blue and cresol red; this mixture is violet at pH 8.4, blue at
pH 8.3 and rose at pH 8.2. With this mixed indicator the mixture
has a violet colour in alkaline solution and changes to blue in the
vicinity of the equivalence point; in making the titration the acid is
added slowly until the solution assumes a rose colour. At this stage
all the hydroxide has been neutralised and the carbonate
converted into hydrogen carbonate. Let the volume of standard
acid consumed be v mL.
OH− +H+ = H2O
CO32- + H+ = HCO3
−
Another titration is performed with methyl orange, as indicator. Let the
volume of acid be V mL.
OH− +H+ = H2O
CO32- + H+ = HCO3
−
H2CO3 = H2O + CO2
Then V - 2(V-v) corresponds to the hydroxide, 2(V-v) to the
carbonate, and V to the total alkali. To obtain satisfactory results
by this method the solution titrated must be cold (as near 0°C as
is practicable), and loss of carbon dioxide must be prevented as
far as possible by keeping the tip of the burette immersed in
59
4. Determination of Boric Acid
Boric acid acts as a weak monoprotic acid (Ka= 6.4x10-10), it
cannot therefore titrated accurately with standard alkali.
However, by the addition of certain organic polyhydroxy
compounds, such as glycerol, mannitol, sorbitol, or glucose, it
acts as a much strong monobasic and can be directly titrated with
sodium hydroxide; using phenolphthalein as indicator.
NaOH + H3BO3 = NaBO2 + 2 H2O
The effect of polyhydroxy compounds has been explained
on the basis of the formation of 1:1 and l:2 -mole ratio complexes
between the hydrated borate ion and 1,2- or 1,3-diols:
60
5- Determination of Borax
When borax is dissolved in water, it is hydrolysed into:
Na2B4O7 + 7 H2O = 4 H3BO3 + 2 NaOH
If the aqueous solution is titrated with standard hydrochloric
acid using methyl orange as indicator, it is the NaOH that is
actually titrated; boric acid being of no effect on die indicator,
and net reaction is:
Na2B 4O7 5 H2O + 2 HC1 = 4 H3BO3 + 2 NaCl
Na2B 4O7 10 H2O = 2 HCl
The residual solution can be titrated for the remaining boric
acid with standard sodium hydroxide after adding glycerol and
using phenolphthalein as indicator. The reaction would be:
Na2B4O7 + 5 H2O + 2 HC1 = 4 H3BO3 + 2 NaCl
Na2B4O7.10 H2O = 4 NaOH
In other words, when a pure sample of borax containing no free
acid is titrated, the volume of standard alkali used would be exactly
double the volume of standard acid.
If the borax solution is treated directly with glycerol and titrated
against standard sodium hydroxide, the solution could then be
61
regarded as boric acid which is half neutralised.
Glycerol
Na2B4O7 + 5 H2O ----→ 2 NaH2BO3 + 2 H3BO3
Na2B4O7.10 H2O = 2 NaOH
6- Determination of Mixture of Boric Acid and Borax
Solutions of alkali borates may be titrated with standard acid
(e.g. HC1) using methyl orange as indicator. They react towards this
indicator as if they were solutions of alkali hydroxides. They behave as
dinormal bases when titrated with acids.
Na2B4O7.10 H2O + 2 HC1 = 4 H3BO3 + 2 NaCl + 5 H2O
While the liberated boric acid consumes 4 molecules of NaOH
when titrated with alkali (e.g. NaOH), using phenolphthalein as
indicator in presence of glycerol (More details in the practical part).
7- Determination of Nitrogen by Kjeldahl's Method
Nitrogen is found in a wide variety of substances. Examples
include amino acids, proteins, synthetic drugs, fertilizers, explosives
soils, potable water supplies, and dyes. The most common method for
determining organic nitrogen is the Kjeldahl method, [n this method,
the sample is decomposed in hot, concentrated sulphuric acid to
62
convert the bound nitrogen to ammonium ion. The resulting solution is
then cooled, diluted and made basic. The librated ammonia is
distilled, collected in an acidic solution, and determined by
neutralization titration.
44221
2SOH
cba HSONH c OH bCO aNHC 42
-233
OH44 SO c CNHHSONH c
-
HCl d ClNH c HCl d)(c CNH 43
The digestion is speeded up by adding potassium sulphate
to increase the boiling point and by a catalyst such as a selenium,
mercury, or copper salt. The amount of the nitrogen containing
compound is calculated from the weight of nitrogen analyzed by
multiplying it by the gravimetric factor.
8- Determination of Amino Acids
Amino acids are amphoteric substances that contain both acidic
and basic group (i.e., they can act as acids or bases). The acid
group is a carboxylic acid group (-COOH), and the basic group is an
amine group (-NH2). In aqueous solutions, these substances tend to
undergo internal proton transfer from the carboxylic acid group
to the ammo group because the RNH2 is a stronger base than
RCO2−. The result via a Zwitter ion.
63
R CH
NH3
CO2
Since they are amphoteric, these substances can be titrated in
aqueous solutions. We can consider die conjugate acid of the
Zwitter ion as a diprotic acid which ionizes stepwise.
R CH
NH3
CO2HKa1
R CH
NH3
CO2
Ka2
R CH
NH2
CO2
congugate acid of Zwitter ion
Zwitter Congugate baseof Zwitter ion
Kal and Ka2 values are known for amino acids. The hydrogen ion
concentration of the Zwitter ion is calculated in the same way as
for amphoteric salt.
[H+] = 2.Ka1Ka
When the Zwitter ion of an amino acid is titrated with strong
acid, a buffer region is first established, consisting of Zwitter ion
(salt) and the conjugate acid. Halfway to the equivalence
point, the f> H is determined by the conjugate acid (Ka1). When
the Zwittter ion is titrated with a strong base, a buffer region of
64
conjugate base (the salt) and Zwitteaon (now the acid) is
established. Halfway to the equivalence point, pH=pKa2 (as with
carbonate and bicarbonate) and at the equivalence point,
the pH is determined by the conjugate base (whose Kb= Ka2 /
Kw).
9- Biphasic Titrations
This type of titrations concerned with water-soluble salts the
acid of which is insoluble in water, but soluble in an immiscible
solvent. For example, the alkaline metal salicylates, and benzoates
are determined by this method.
Taking sodium salicylate as an example, its determination by the
two phase titration method depends upon tlie reaction expressed by
the following equation:
C6H4(OH)COONa + HCl = C6H4(OH)COOH + NaCl
Since the salicylic acid liberated during the titration is a
sufficiently strong acid to give a pH which will be "acid" to any usual
indicator, this acid has to be removed as it is liberated. The ratio of
solubility of salicylic acid in ether to that in water is 250:1, and
consequently when equal volumes of ether and water and if one
gram of acid are used, only 1/250 gm of the acid will remain in the
aqueous layer. If the volume of ether is double or triple that of water,
65
then the salicylic acid remaining in the aqueous phase will be quite
negligible. This fact is the basis for biphasic titrations.
C6H4(OH)COONa + HC1 = C6H4(OH)COOH + NaCl
HCl = C6H4(OH)COONa
Notes:
1 - In viewing the colour change of the bromophenol blue
indicator during the titration, place white paper halfway round the
separator and do not use artificial light.
2- At the end point, a bluish-green colour indicates under titration
and a yellowish-green colour indicates over titration.
3 - The same procedure can be applied to the determination of
sodium benzoate.
10- Double Indicator Titration
These are direct titrations in which a mixture of two
monobasic acids are titrated in such a way that the quantity of
each acid present is known. This is accomplished by using two
indicators. With equal initial concentrations of the two acids, it is
possible to determine each separately with an accuracy of
less than 1% if the difference in the ionisation constants of the
two acids is at least 104. If there is 100 times as much of one acid as
of the other, there must be a difference of 106 in the ionisation
66
constants. Thus, it is possible to titrate hydrochloric acid in
presence of acetic acid (ka= 1.8x10-5).
The hydrogen ion of the hydrochloric acid suppresses the
ionisation of the other weak acid by common ion effect, so that
the sodium hydroxide added neutralises the hydrochloric acid first.
When this reaction is complete, the pH rises to that of the other
weaker acid solution which is then titrated. Fig. 6 shows a curve
for a mixture of hydrochloric acid and acetic acid titrated with
standard sodium hydroxide.
The pH change at the first equivalence point is not very great,
and the end point determined by a colour indicator would not
be very reliable. However, a more exact location of the
equivalence point is made possible by pH-meter (potentiometric
measurements).
Titration curve for 50 mL of a mixture of HC1 and HOAc with O.l M NaOH.
67
11- Residual (or Back) Titrations
These consist in the addition of a known excess of standard
solution to a weighed amount of the sample and after the
reaction is complete, the residual quantity of the added standard is
determined.
In general this method is used for:
a) Volatile substances, such as ammonia, some of which would be
lost during the titration, e.g. determination of ammonium
chloride by the indirect method.
68
b) Insoluble substances, such as calcium oxide and calcium
carbonate, which require excess of the standard solution to
effect a quantitative reaction.
c) Substances which require heating with the standard reagent
during the determination in which decomposition or loss of
the reactants or products would occur in the process e.g. of
back titration is the determination of mixture of CaO and
CaCO3.
12- Determination of Mixture of Calcium Oxide and Calcium
Carbonate
The estimation is based on:
a) Cao suspended in water is alkaline to phenolphthalein.
b) Phenolphthalein loses its colour when (H+) is lower than that
required to attack the precipitate of CaCO3.
c) CaO is more soluble in sucrose solution than in water, the
complex calcium saccharate being equally alkaline as CaO. It
can be titrated with standard acid (e.g. HCl).
d) Add a known excess of standard HCl sufficient to dissolve all
the carbonate and oxide. Boil-off CO2 cool and titrate the
excess acid with standard NaOH using phenolphthalein
indicator.
e) Subtract the volume of standard HCl used in step (c) from
that consumed by the mixture in step (d), the difference is
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equivalent to CaCO3 in the sample.
CaO + 2HC1 = CaCl2 + H2O
2 HCl = CaO
Notes:
1- Alcohol is added to prevent formation of lumps of CaO on
addition of the sugar solution.
2- Sucrose may get acidic on standing owing to bacterial
fermentation. For this, it must be neutralised just before use.
13- Determination of Barium Chloride
The alkaline earth metal (e.g. Ba2+) is precipitated as its
insoluble carbonate by the addition of a known excess of standard
Na2CO3. The solution is then boiled, cooled to about 0°C and the
excess sodium carbonate is back titrated with standard HCI using
phenolphthalein as indicator; multiplying the volume of acid by 2.
The volume used in precipitating the metal ion is obtained
by difference.
Na2CO3 + BaCl3 = BaCO3 + 2NaCl
Notes:
1. Solution must be dilute and heated to 70°C to prevent
formation of Ba(HCO3)2.
2. Boiling renders the precipitated BaCO3 dense.
70
3. It is possible to neutralise the excess of Na2CO3 with HCI
without decomposing the precipitated BaCO3 because
ph.ph. loses its colour when (H+) is lower than that required to
attack BaCO3.
4- Ca, Ba & Sr in neutral solutions of their salts may be similarly
determined.
14- Determination of Acetylsalicylic Acid (Aspirin)
Esters of the type in which the hydroxyl group is esterified,
such as acetylsalicylic acid, readily dissolve in dilute sodium
hydroxide solution and are completely hydrolysed by boiling or
by heating for a few minutes on a water bath with an excess of
base liberating the sodium salts of acetic acid and salicylic
acid. The residual base can then be back titrated with
standard acid, using phenolphthalein as indicator.
CH3.CO.O.C6H4.COOH + 2 NaOH = CH3.COONa + C6H4(OH)COONa
71
Lecture 7 (2 hrs) ……/…./…….. (Acid-base)
Titration in non aqueous solvents, Brönsted definitions of acids and
bases, Solvent and solvent properties, Relative of acidity and basicity,
Factors affecting levelling effects of solvents¸ The choice of solvents for
non – aqueous medium, Detection of end point in non–aqueous
titration, Application of Non- aqueous Acid – base titration
Titration in Non – aqueous solvents
Non– aqueous titrimety found wide application in the field of
pharmaceutical analytical chemistry, especially after the
introduction of many new complex organic medicinal agents
which are too weakly basic or too weakly acidic (pK more then
8) to be titratable in aqueous solution. With non–aqueous
solvents, the difficulty of weak reactivity is overcome; solvents
are used to enhance to basic (or acidic) characteristic of a
compound so as to make it suitable for non– aqueous titration
with standard acid (or standard base). Furthermore, the solvent
enhances the strength of the titrant, so that a more complete
neutralisation reaction and sharper end point than is possible in
aqueous media are obtained.
The simplicity, speed, precision and accuracy of non–
aqueous methods are equivalent to those of the classical
procedures in aqueous media; and the same laboratory
72
equipment may be employed with the exception that extra
precautions should be taken where moisture and carbon
dioxide are to be excluded; and temperature is to be
corrected.
Brönsted definitions of acids and bases:
According to Brönsted, an acid is defined as any substance in
ionic or molecular form, which produces or donates protons,
H+; while a base is any substance which accepts or acquires
protons. When an acid ionises or transfers protons, it produces a
base; and when a base accepts protons, an acid is formed;
Acid H+ + base
Protons
The base produced by such a process is called the conjugate
base of the acid; it may be an anion, or a neutral particle. The
acid in this instance can also be considered to be the
conjugate acid of the base taking part in the reaction.
Solvent molecules are involved, either as acid or as base.
Thus, when ammonia (a base) is dissolved in water, an acid
base reaction takes place in which the solvent (water) acts as
an acid (proton donor):
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NH3 + H2O NH4+ + OH–
Base1 acid1 acid2 base2
Here, the hydroxyl ion is the conjugate base of the acid water,
and the ammonium ion is the conjugate acid of the base
ammonia. On the other hand, when gaseous hydrogen
chloride is dissolved in water, the solvent (water) acts as a base
(proton acceptor):
HCl + H2O H3O+ + Cl–
Acid1 base1 acid2 base2
The chloride ion is the conjugate base of the acid. HCl, and
hydronium ion, H3O+, is the conjugate acid of the base H2O.
The dissociation of water, in this view, is simply an acid– base
reaction:
H2O + H2O H3O+ + OH–
Acid1 base1 acid2 base2
According to Brönsted concept, a neutralisation reaction can
be expressed as the difference between two half reaction:
(acid1) HAc H+ + Ac– (base1)
(acid2) H2O H+ + OH– (base2)
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HAc + OH– H2O + Ac–
Acid1 base2 acid2 base1
Not only water, but also other solvents that have acidic or basic
properties can be used. For example, when HCl is dissolved in
ethyl alcohol or in glacial acetic acid, the same takes place:
HCl + C2H5OH C2H5OH2+ + Cl–
HCl + HC2H3O2 H2C2H3O2+ + Cl–
The basic strength of the solvent determines the extent to
which these reactions proceed. Thus, while water is a
sufficiently strong base to cause complete ionisation of
hydrochloric acid, yet glacial acetic acid – a weaker base
then water – cause only partial ionisation of the hydrogen
chloride molecules. In other words, hydrogen chloride is not a
strong acid if dissolved in glacial acetic acid. As a result the
classification of acids or bases as strong or weak, is dependent
on the degree of basicity or acidity of the solvent. If, on the
other hand, the solvent is inert, i.e. has no appreciable acidic or
basic properties, such as benzene and carbon tetrachloride,
no ionisation can occur when an acid or a base is dissolved in
these solvents. The acidic property of the solute becomes
apparent only when a soluble base is added to the solution,
when an equilibrium is set up between the acid and base.
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Solvent and solvent properties
Solvents may generally be divided into four classes with respect
to their acid – base properties;
1 – Amphiprotic solvents are these which show both acidic
and basic properties and undergo self – dissociation or
autoprotolysis. One molecule acting as an acid with a second
molecule acting as a base. Water is the most common
example of this class, although other solvents are capable of
similar dissociation reactions.
H2O + H2O H3O+ + OH–
CH3OH + CH3OH CH3OH2+ + CH3O–
CH3COOH + CH3COOH CH3COOH2+ + CH3COO–
NH3 + NH3 NH4+ + NH2
–
2 H2N–CH2CH2–NH2 H2N–CH2CH2–NH3+ + H2N–CH2CH2NH–
Methanol and ethanol show acid – base properties similar to
those of water; glacial acetic acid shows stronger acid
properties than water; ammonia and ethylenediamine show
basic properties stronger than that of water.
2 – Aprotic solvents or inert solvents, show no obvious acidic or
basic properties, have no dissociable proton, and show no
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tendency to donate or accept a proton. Typical examples of
this group are benzene, chloroform and carbon tetrachloride
3 – Protophillic solvents: – this class is more basic than water
and exemplified by many organic oxygen compounds such as
ethers (including dioxan) ketones, such as acetone, esters, such
as ethyl acetate and the amines (including liquid ammonia).
They are basic substances and react with an acidic solution
with the formation of a solvated proton together with the
cojugated base of the acid.
HB + S HS+ + B–
Acid Solvent Solvated proton Conjugated base
Acids vary in their inherent tendency to donate protons to
basic substances, their inherent strength varying directly with
the ease with which they can fulfil this function. Similarly, an
inherently strong base is one which readily accepts a proton,
whilst a weak base shows little affinity for protons.
If an acid is dissolved in a weakly basic solvent, the acidic
strength of the solution will vary with the inherent acidic
strength of the solute: a strong acid, such as perchloric acid will
produce a strongly acidic solution, whereas a weak acid such
as propionic acid will form a weakly acidic solution. If a strong
77
basic solvent is used, the effect of the basic medium on an
inherently strong acid will be small. But a weak acid will tend to
donate protons more readily, thus its acidic strength is
enhanced. In fact, all acids tend to become indistinguishable
in strength from one another in strongly basic solvents. This
phenomenon is called the “levelling effect”. For example. 0.1 N
solutions of acetic, benzoic, formic, thiocyanic, nitric,
hydrocyanic, hydrochloric, hydrobromic, hydroiodic and
perchloric acids all have the same acid strength in liquid
ammonia, although their strengths differ widely in water.
4 – Protogenic solvents: – this class includes substances which
are more acidic than water and is exemplified mainly by
sulphuric acid and anhydrous acetic acid. They exert a
“levelling effect” on bases; arguments similar to those given
before for protophilic solvents and acids being applicable.
B + CH3COOH BH+ + CH3COO–
For example, the aliphatic amines and alkylanilines, all of
which are weak bases in water, behave as strong bases in
acetic acid, and many substances which are extremely weak
bases in water – such as urea, the oximes etc. – show
pronounced basic properties in acetic acid.
78
Solutions of strong acids, such as perchloric acid, in acetic
acid can be used for titrating very weak bases such as oximes
and amides, which cannot be titrated in water.
Sulphuric acid is such a strong acid medium that almost all
compounds containing oxygen or nitrogen will accept a
proton from it to some degree, thus belaving as bases. Many
substances normally regarded as acids exhibits basic properties
in sulphuric acid : thus most carboxylc acids are strong bases,
forming the ion RCOOH2+. Aliphatic and armatic
nitrocompounds, sulphones and sulphonic acids also behave
as bases through as weak ones.
Detection of end point in non-aqueous titration
The end point can be determined either by
1 - Instrumental method “Potentiometrically”:
It involves the measurement of a glass potential that
responds to the concentration of the solvated proton.
2 – Visual methods:
Many acid – base coloured indicators used in aqueous
titration, are also applicable in non – aqueous medium for
example:
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Crystal violet is suitale for titration of basic substances, its
colour changes from violet to bluish green.
Thymol blue, is suitable for titration of acidic substance, its
colour changes from yellow to blue
Azo violet is also suitable for titration of acidic compounds
where its colour changes from red to blue.
Application of Non- aqueous Acid – base titration:
[1] Titration of weak basic substance:
Many weak bases can be determined successfully in acetic
acid medium using acetous perchloric acid as a standard
titrant. Which can be standardized against potassium acid
phtahlate using 0.5% acetous crystal violet.
COOK
COOH
+ HClO4
COOH
COOH
+ KClO4
Examples
1 – Aromatic amines, amides, urea and other very weak
nitrogenous base including alkaloids
2 – another impotant application is the analysis of primary,
secondry and tertiary amines,
80
[a ] The total mixture is determned by acetous
perchloric where all amine react according to the
following equations:
R – NH2 + HClO4 = R – NH3+ + ClO4
–
R2 – NH + HClO4 = R2 – NH2+ + ClO4
–
R3 – N + HClO4 = R3 – NH+ + ClO4–
[b ] The tertiary amine can be determined alone by
refluxing the mixture with acetic anhydride to allow
acetylation of both primary and secondary amine
leaving the tertiary one which can be titrated
against perchloric
H3C C
O
C
O
O
H3C
+ R-NH2 R - NH - C -CH3
O
+ CH3COOH
H3C C
O
C
O
O
H3C
+ R2-NH R2 - NH - C -CH3
O
+ CH3COOH
[c ] Another portion of the mixture is heated with
salicylaldehyde which forms Schiff’s base with
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primary amine leaving both 2ry and 3ry amine and
allow their titration versus perchloric
CHO
OH
+ NH2 - R
- H2O CH = NR
OH
Shiff's base
The concentration of tertiary amine is directly calculated from
(b), while primary amine concentration is obtained by
subtracting the volume of perchloric acid used in (b) from that
in (a), a – b = y. add the volume in (b) to (y) then subtract from
the total volume in (a) (which is corresponding to the free
amine) to obtain the volume of acid equivalent to the
secondary amine.
[II] – Titration of weak acidic compounds:
Several basic solvents have been employed to determine
acids (phenols, enols, sulphonamides ….etc) that are too weak
to be titrated in water, for examples: ethylenediamine,
dimethylformamide, pyridine, dimethyl sulphoxide and
butylamine, in addition to, alcohols and aprotic solvent
mixtures such as benzene, methanol, and acidic solvents such
82
as acetone and acetonitrile.
The presence of water in a sample to be titrated in
dimethylformaide causes high results may be due to hydrolysis
of the solvent with formation of formic and dimethyl amine
H C
O
N
CH3
CH3
+ H2O HCOOH + H - N
CH3
CH3
Weak acidic substances usually determined in such basic
solvents by titration with sodium methoxide or potassium
methoxide in benzene – methanol using thymol blue as
indicator
Eg. Sulphonamide:
All sulphonamides contain the weak acidic group – SO2NH
which can be titrated with KOCH3 in benzene – methanol,
using dimethylformamaide or butylamine as a solvent and
thymol blue or azo – violet, respectively, as indicator.
R - SO2 - N- R
H
+ CH3OK R - SO2 - N- R
K
CH3OH+