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.

5

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

6

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.

8

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

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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–

21

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)

24

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.

26

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

69

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):

73

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.

75

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

76

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:

79

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

81

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+