51

Neutralization

Titration-I

UNIT 8 NEUTRALIZATION TITRATION-I

Structure

8.1 Introduction Objectives

8.2 Basic Concepts of Titrimetry, Primary and Secondary Standards

8.3 Titration Curves Titration of A Strong Acid Versus Strong Base

Titration of Weak Acid Versus Strong Base

Titration of a Weak Base Versus Strong Acid

Titration of Weak Acid Versus Weak Base

Titration of Sodium Carbonate Versus Strong Acid

Titration of Polyprotic Acid Versus Strong Base

8.4 Theory of Indicators Ostwald’s Theory

Modern Quinoid Theory

8.5 Colour Change Range of an Indicator

8.6 Selection of Indicator and Indicator Error 8.7 Summary

8.8 Terminal Questions

8.9 Answers

8.1 INTRODUCTION

The term titrimetric analysis refers to quantitative chemical analysis carried out by

determining the volume of a solution of accurately known concentration, which is

required to react quantitatively with the solution of the substance to be determined.

The solution of accurately known strength is called the standard solution. The weight

of the substance to be determined is calculated from the volume of the standard

solution used and the known laws of stoichiometry. The standard solution is usually

added from a burette. The process of adding the standard solution until the reaction is

just complete is known as titration, and the substance to be determined is titrated.

The point at which this occurs is called the equivalence point or the theoretical (or

stoichiometric) end-point. The end point is judged usually, by the addition of an

auxiliary reagent, known as indicator.

The neutralization titrations include the titration of free bases, or those formed from

the salts of weak acids by hydrolysis with standard acids (acidimetry), and the

titration of acids or those formed by the hydrolysis of salts of weak bases, with

standard base (alkalimetry). These reactions involve the combination of hydrogen and

hydroxide ions to form water. This chapter describes the various types of acid–base

neutralization titrations, including the titration of strong acids or bases and weak acids

or bases. Through a description of the theory of indicators, the selection of a suitable

indicator for detecting the completion of a particular titration has been discussed.

Objectives

After studying the unit, you should be able to:

• State and explain the concept of titrimetry

• explain the nature of neutralization titration curves

• understand primary and secondary standards

• describe the theory of indicators

• select the appropriate indicator for a particular titration

• calculate the pH of acid, base and buffer solution.

52

Estimations Based On

Kinetic and Acid-Base

Equilibria Studies

8.2 BASIC CONCEPT OF TITRIMETRY, PRIMARY AND

SECONDARY STANDARDS

The acid – base titration involves a neutralization reaction in which an acid is reacted

with an equivalent amount of base. The titrant is always a strong acid or a strong base.

The object of neutralization, say, an alkaline solution with a standard solution of an

acid in the determination of the amount of acid which is exactly equivalent chemically

to the amount of base present. The point at which this is reached is equivalence point

or theoretical end point; an aqueous solution of the corresponding salt results. If both

the acid and base are strong electrolytes, the resultant solution will be neutral and have

a pH of 7; but if either the acid or base is a weak electrolyte, the salt will be

hydrolyzed to a certain degree, and the solution at the equivalence point will be either

slightly alkaline or slightly acidic. The exact pH of the solution at the end point can

readily be calculated from the ionization constant of the weak acid or the weak base

and the concentration of the solution. For any actual titration the correct end point will

be characterized by a definite value of the hydrogen – ion concentration of the

solution, the value depending upon the nature of the acid and the base and the

concentration of the solution.

Primary Standard

A primary standard is a substance which satisfies the following requirements:

1. It must be easy to obtain, to purify, to dry (preferably at 110-120oC) and to

preserve in pure state.

2. The substance should remain unaltered during weighing i.e., it should not be

hygroscopic, or oxidized by the air, or affected by carbon dioxide.

3. The substance should be capable of being tested for impurities by qualitative

and other tests of known sensitivity.

4. It should have a high equivalent so that the weighing errors may be negligible.

5. The substance should be readily soluble under the experimental conditions.

6. The reaction with the standard solution should be stoichiometric and practically instantaneous.

In practice, it is difficult to obtain a primary standard, and a compromise between the

above ideal requirements is necessary. The commonly employed primary standards

include sodium carbonate, sodium tetraborate, potassium hydrogen phthalate,

constant-boiling-point hydrochloric acid, potassium hydrogen iodate, and benzoic

acid.

Secondary Standard

A substance, which fulfills the requirement that it can be weighed accurately to

provide a known amount of reactant but which is not a pure substance, is called

secondary standard. It may be used for standardizations, and whose content of the

active substance has been found by the comparison against a primary standard.

SAQ 1

a) Mention any two requirements of a primary standard.

…………………………………………………………………………………………...

…………………………………………………………………………………………...

b) How can secondary standards be used for standardizations?

…………………………………………………………………………………………...

…………………………………………………………………………………………...

53

Neutralization

Titration-I

8.3 TITRATION CURVES

A titration curve is constructed by plotting pH of the solution during titration as

ordinates and the amount of acid or base added as abscissa. These curves are useful to

indicate equivalence point graphically. The change in pH in the neighborhood of the

equivalence is of greatest importance, as it enables us to select an indicator, which will give the smallest error. The nature of titration curve depends on the ionization

constants of acid and base employed in titration i.e., their strength. The principles of

acid–base equilibria are important for the construction and interpretation of titration

curves in neutralization titrations.

8.3.1 Titration of A Strong Acid Versus Strong Base

In the case of a strong acid versus strong base, both the titrant and 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 result of neutralization is conversion of the HCl to a neutral

solution of NaCl.

The calculation of the titration curves involves computation of the pH for the

concentration of the particular species at the various stages of the titration. The pH during neutralization processes will be calculated as follows:

1. Up to the equivalence point the pH of the solution is determined by the amount

of the strong acid remaining present.

2. At the equivalence point pH is 7.

3. After passing the equivalence point the pH value is defined by the excess of the

base.

Before any NaOH is added to 1.0 M HCl solution, its pH is zero. As the titration

proceeds, part of H+ is removed from the solution as water. So the concentration of H

+

gradually decreases but this decrease is not significant, probably, due to the reason that the strong acids are good buffers at low pH. The results of calculation of pH during the

titration of 100 cm3 of HCl with NaOH of equal concentration are presented in Table

8.1. Fig. 8.1 is the graphical representation of these data.

Fig. 8.1: Neutralisation curves of 100 cm

3 of HCl with NaOH of same concentration

(calculated)

54

Estimations Based On

Kinetic and Acid-Base

Equilibria Studies

After 50 cm3 addition of NaOH solution, 50 cm

3 of un-neutralized acid will be present

in a total volume of 150 cm3.

[H+] = 50 x 1/150 M = 0.33 M , or pH = 0.48

After 75 cm3 addition of NaOH, [H

+] = 25 × 1/175 M = 0.143 M, or pH = 0.84

After 90 cm3addition of NaOH, [H+] = 10 × 1/190 M = 0.0526 M, or pH = 1.3

After 99 cm3addition of NaOH, [H

+] = 1 × 1/199 M = 0.00503 M, or pH = 2.3

After 99.9 cm3 addition of NaOH, [H+] = 0.1 × 1/199.9 M = 0.0005 M, or pH = 3.3

After 100 cm3 addition of base, the pH will change sharply to 7, i.e., the theoretical

end point provided carbon dioxide is absent; the resulting solution is simply one of

sodium chloride.

Where the pH for the over titrations of 0.10 and 1.0 cm3 corresponding to pH 10.7 and

11.7, respectively can be calculated as given below:

After 100.1 cm3 addition of NaOH, [OH─] = 0.1/200.1 M = 0.0005 M,

pOH = 3.3, and pH = 10.7

With 101 cm3 of base, [OH

─] = 1/201 = 0.005 M, pOH = 2.3, and pH = 11.7

These results show that as the titration proceeds, initially the pH rises slowly, but

between the addition of 99.9 and 100.1 cm3 of alkali, the pH of the solution rises from

3.3 to 10.7, 4.3 to 9.7 and 5.3 to 8.7 in case of 1 M, 0.1 M and 0.001 M solutions

respectively. Further addition of base does not cause a significant change in pH. The

results of titration are presented in Table 8.1 for 1M, 0.1 M and 0.01 M solutions of

acid and base, respectively.

Table 8.1: pH during Titration of 100 cm3 of HCl with NaOH of Equal

Concentration

NaOH added

cm3

1.0 M solution

pH

0.1 M solution

pH

0.01 M solution

pH

0.0 0.0 1.0 2.0

50.0 0.5 1.5 2.5

75.0 0.8 1.8 2.8

90.0 1.3 2.3 3.3

98.0 2.0 3.0 4.0

99.0 2.3 3.3 4.3

99.5 2.6 3.6 4.6

99.8 3.0 4.0 5.0

99.9 3.3 4.3 5.3

100.0 7.0 7.0 7.0

100.1 10.7 9.7 8.7

100.2 11.0 10.0 9.0

100.5 11.4 10.4 9.4

101.0 11.7 10.7 9.7

102.0 12.0 11.0 10.0

110.0 12.7 11.7 10.7

125.0 13.0 12.0 11.0

150.0 13.3 12.3 11.3

55

Neutralization

Titration-I

In the quantitative analysis we are especially interested in the change of pH near the

equivalence point. This part is accordingly shown on a large scale in Fig. 8.2, on

which are also indicated the colour change intervals of some of the common

indicators. The magnitude of the break will depend on both the concentration of the

acid and the concentration of the base. The reverse titration will be the mirror image of

these titrations.

Fig. 8.2: Neutralisation curves of 100 cm3 of HCl with NaOH of same concentration in

vicinity of equivalence point (calculated)

8.3.2 Titration of Weak Acid Versus Strong Base

We illustrate this case by the titration curve of 0.1 M acetic acid with 0.1 M sodium

hydroxide as shown in Fig. 8.3. The neutralization reaction is:

CH3COOH + Na+ + OH − →

H2O + Na+ + CH3COO–

The acetic acid, which is only few percent ionized, depending on the concentration, is

neutralized to water and an equivalent amount of the salt, sodium acetate. Before the

titration is started, we have 0.1 M CH3COOH. As soon as the titration is started some

of the CH3COOH is converted to CH3COONa, and a buffer system is set up. As the

titration proceeds, the pH slowly increases as the ratio [CH3COO–] / [CH3COOH]

changes. At the mid point of titration, [CH3COO–] = [CH3COOH], and the pH is equal

to pKa. At the equivalence point we have a solution of CH3COONa. Since this is

Bronsted base (it hydrolyzes), the pH at the equivalence point will be alkaline. The

pH will depend on the concentration of CH3COONa. The greater the concentration,

the higher the pH. As excess of NaOH is added beyond the equivalence point, the

ionization of base CH3COO– is suppressed to a negligible amount, and the pH is

determined only by the concentration of excess OH − . Therefore, the titration curve

beyond the equivalence point follows that for the titration of a strong acid.

56

Estimations Based On

Kinetic and Acid-Base

Equilibria Studies

Fig. 8.3: The pH titration curve of weak acid (CH3COOH) and strong base (NaOH)

The slowly rising region before the equivalence point is called the buffer region. It is

flattest at the midpoint, and so the buffer capacity is greatest at a pH corresponding

to pKa. The buffering capacity also depends on the concentrations of CH3COOH and

CH3COO–, and the total buffering capacity increases as the concentration increases. In

other words, the distance of the flat portion on either side of pKa will increase as

[CH3COOH] and [CH3COO–] increase.

For plotting the titration curve the pH values can be calculated as:

1. pH of weak acid can be calculated from the following equation:

pH = ½ pKa – ½ log [acid] … (8.1)

Strictly, [H+] = 0.5 ( (Ka2

+ 4Ka [acid]0.5) – Ka )

2. Up to equivalence point pH of the solution is determined by the dissociation

exponent of the weak acid and by the ratio of the concentration of free acid

(HA) and titrated acid (A − = salt) (buffer solution):

[ ][ ]acid

saltlogpKpH a += … (8.2)

3. The pH at the equivalence point is greater than 7 due to the alkaline hydrolysis

of the resulting salt:

pH = ½ pKw + ½ pKa + ½ log [salt]

or, pH = 7 + ½ pKa – ½ pc … (8.3)

Concentration of the salt is c mol dm − 3 to be consistent we use here pc = -log c.

4. After passing the equivalence point the excess of the base determines the pH of

the solution as if the hydrolyzing salts were not present at all.

The initial pH of 0.1 M acetic acid solution is computed from Eq. (8.1); the

57

Neutralization

Titration-I

dissociation of acid is relatively so small that it may be neglected in expressing the

concentration of acetic acid.

pH = ½ (-log 1.82 × 10 − 5) – ½ log 0.1

or, pH = 2.87 when 50 cm3 of 0.1 M alkali has been added,

[salt] = 50 × 0.1/150 = 3.33 × 10 − 2

and [acid] = 50 × 0.1/150 = 3.33 × 10 − 2

pH can be computed from Eq. (8.2)

pH = –log 1.82 x 10 − 5 + log 2

2

1033.3

1033.3−

−

×

×

or, pH = 4.74

The pH of solution at the equivalence point can be calculated using Eq. (8.3).

pH = 7 –1/2 log 1.82 × 10 − 5 + 1/2 log 5 × 10 − 2

= 7 + 2.37 – 1/2 (1.3) = 8.72

The pH values at other points on the titration curve can be similarly calculated. After

the equivalence point has been passed, the solution contains excess of OH − ions

which will repress the hydrolysis of the salt; the pH may be assumed, with sufficient

accuracy for our purpose, to be that due to the excess of base present, so that in this

region the titration curve will almost coincide with that for 0.1 M hydrochloric acid.

The results of titration are presented in Table 8.2. The results for the titration of 100

cm3 of a weaker acid (Ka = 1 × 10 − 7

) with 0.1 M sodium hydroxide at the laboratory

temperature are also included.

Table 8.2: Neutralization of 100 cm3 of 0.1 M acetic acid (Ka = 1.82 ×××× 10

− 5) and

of 100 cm3 of 0.1 M – HA (Ka = 1 ×××× 10

− 7) with 0.1 M sodium hydroxide.

0.1M-HA(Ka = 1 ×××× 10− 7

) 0.1 M NaOH used

cm3

0.1 M acetic acid

pH pH

0.0 2.9 4.0

10.0 3.8 6.0

25.0 4.3 6.5

50.0 4.7 7.0

90.0 5.7 8.0

99.0 6.7 9.0

99.5 7.0 9.3

99.8 7.4 9.7

99.9 7.7 9.8

100.0 8.7 9.9

100.2 10.0 10.0

100.5 10.4 10.4

101.0 10.7 10.7

110.0 11.7 11.7

125.0 12.0 12.0

150.0 12.3 12.3

200.0 12.5 12.5

58

Estimations Based On

Kinetic and Acid-Base

Equilibria Studies

For 0.1 M acetic acid and 0.1 M sodium hydroxide, it is evident from the titration

curve that neither methyl orange nor methyl red can be used as indicators. The

equivalence point is at pH 8.7, and it is necessary to use an indicator with a pH range

of slightly alkaline side, such as phenolpthalein, thymolpthalein, or thymol blue (pH

range, as base, 8.0-9.6). For the acid with Ka = 1 × 10 − 7 the equivalence point is at pH

= 10.0, but here the rate of change of pH in the neighborhood of the stoichiometric

point is very less pronounced, owing to considerable hydrolysis. Phenolphthalein will

commence to change colour after 92 cm3 of alkali have been added, and this change

will occur to the equivalence point; thus the end point will not be sharp and the

titration error will be appreciable. With thymolphthalein, however, the colour change

covers the pH range 9.3-10.5; this indicator may be used, the end-point will be more

sharp than for phenolphthalein, but nevertheless somewhat gradual, and the titration

error will be about 0.2 per cent. Acids having Ka < 1 × 10 − 7 cannot be satisfactorily

titrated in 0.1 M solution with simple indicator. In general it may be stated that weak

acids (Ka > 5 ×10 − 6) should be titrated with phenolphthalein, thymolphthalein, or

thymol blue as indicators.

Fig. 8.4: Titration of 50 cm3 of 0.1 M-H3PO4 with 0.1 M-KOH

Fig. 8.4 shows the titration curves for 50 cm3 of 0.1 M solutions of weak acids of

different Ka values titrated with 0.1 M KOH. The sharpness of the end point decreases

as Ka decreases. As in Fig. 8.4 sharpness will also decrease as the concentration

decreases. Generally for macro titrations (ca. 0.1 M), acids with Ka values of 10 − 6 can

be titrated accurately with a visual indicator; and with suitable colour comparisons,

those with Ka values approaching 10 − 8 can be titrated with reasonable accuracy. A pH

meter can be used to obtain better precision for the very weak acids by plotting the

titration curve. Weaker acids can be titrated in nonaqueous solvents that do not

possess the acidity or basicity of water.

59

Neutralization

Titration-I

8.3.3 Titration of a Weak Base Versus Strong Acid

The titration of a weak base with a strong acid is similar to the above case, but the

titration curves are reverse of those for a weak acid versus strong base. This

neutralization titration can be illustrated by the titration curve for 100 cm3 of 0.1 M

ammonia with 0.1 M hydrochloric acid as shown in Fig. 8.5. The neutralization

reaction is,

NH3 + H+ + Cl − → NH4+ + Cl −

Fig. 8.5: The pH titration curve of 100 cm3 0.1M ammonia (NH4OH) with 0.1M

hydrochloric acid (HCl)

At the beginning of the titration, we have 0.1 M NH3 and the pH is calculated for a

weak base. As soon as some acid is added, some of the NH3 is converted into NH4+

and the buffer region is formed. End point is judged by inflection point of the titration

curve but the jump in pH is smaller than in case of strong base - strong acid titration.

The pH values obtained in the course of titration can be calculated as given below:

1. The pH of the solution of a weak base:

pH = pKw – ½ pKb + 1/2 log [base] … (8.4)

Strictly, [H+] = Kw / (0.5 ( (

2

bK + 4Kb [base])0.5

– Kb )

2. The pH upto the equivalence point:

[ ][ ]salt

baselogpK pKpH bw +−= … (8.5)

3. The pH of the equivalence point is lower than 7, due to the hydrolysis of

resulting salt:

pH = 1/2 pKw – 1/2 pKb –1/2 log [salt]

or, pH = 7 – 1/2 pKb + 1/2 pc … (8.6)

C and pc have the same meaning as used in Eq. (8.3).

4. After the attainment of equivalence point, the solution contains the excess of H+

ions, hydrolysis of the salt will be repressed, and the subsequent pH change may

be assumed with sufficient accuracy for our purpose, to those due to the excess

of acid present.

60

Estimations Based On

Kinetic and Acid-Base

Equilibria Studies

Example

Titration of 100 cm3 of 0.1 M aqueous ammonia (Kb = 1.8 × 10 − 5) with 0.1 M HCl at

the ambient temperature. The pH of the solution at the equivalence point is given by

the Eq. (8.6):

pH = 7 – 1/2 pKb + 1/2 pc

= 7 – 2.37 + ½ (1.3)

= 5.28

For the titration of weak base and strong acid, those indicators are generally used

which change their colour in the acidic pH range. It is clear from the Fig. 8.4 that

neither thymolphthalein nor phenolpthalein can be employed in the titration of 0.1M

aqueous ammonia. The equivalence point is at pH 5.3, and it is necessary to use an

indicator with a pH range on the slightly acid side (3 – 6.5), such as methyl orange,

methyl red, bromophenol blue, or bromocresol green. The last named indicators are

applicable for the titration of all weak bases (Kb > 5 × 10 − 6) with strong acids

For the weak base (Kb = 1 × 10 − 7), bromophenol blue or methyl orange may be used;

no sharp colour change will be obtained with bromo-cresol green or with methyl red,

and the titration error will be considerable.

8.3.4 Titration of Weak Acid Versus Weak Base

This can be illustrated by titration of 100 cm3 of 0.1 M acetic acid (Ka = 1.8 × 10 − 5

)

with 0.1 M aqueous ammonia (Kb = 1.8 × 10 − 5). The pH at the equivalence point is

given by:

pH = 1/2 pKw + ½ pKa – ½ pKb … (8.7)

= 7.0 + 2.37 – 2.37 = 7.0

The neutralization curve upto the equivalence point is almost identical with that using

0.1 M sodium hydroxide as the base; beyond this point the titration is virtually the

addition of 0.1 M aqueous ammonia solution to 0.1 M ammonium acetate solution and

Eq. (8.5) is applicable to the calculation of the pH. The titration curve for the

neutralization of 100 cm3 0.1 M acetic acid with 0.1 M aqueous ammonia at the

laboratory temperature is shown in Fig. 8.6. The main feature of the curve is that the

change of pH near equivalence point and, indeed, during the whole of the

neutralization curve is very gradual.

Fig. 8.6: Titration curve of weak base and weak acid

61

Neutralization

Titration-I

In this sort of titration, no sudden change in pH, and hence no sharp end point can be

found with any simple indicator. A mixed indicator which exhibits a sharp colour

change over a very limited pH range, may sometimes be found which is suitable. Thus

for acetic acid – ammonia solution titrations, neutral red methylene blue indicator may

be used, but on the whole, it is best to avoid the use of indicators in titrations involving

both a weak acid and a weak base.

8.3.5 Titration of Sodium Carbonate Versus Strong Acid

Sodium carbonate is a Bronsted base that is as a primary standard for the standardization of

strong acids. It hydrolyses in two steps:

CO32 − + H2O � HCO3

¯ + OH¯ 2w1 10x1.2KH −

===2

1a

b K

KK .. (8.8)

HCO3¯ + H2O � CO2 + H2O + OH − 8w

2 10x3.2KH −===

1

2a

b K

KK ... (8.9)

where 1a

K and 2a

K refer to the Ka values of H2CO3.

HCO3¯ is the conjugate acid of CO3

2 − and H2CO3 is the conjugate acid of HCO3¯ and

Kb values are calculated for salts of weak acids and bases (i.e., from KaKb = Kw).

A titration curve for Na2CO3 with HCl is shown in Fig. 8.7 (solid line). Even though

1bK is considerably larger than the 10 − 6 required for a sharp end point, the pH break

is decreased by the formation of CO2 beyond the first equivalence point. The second

end point is not very sharp either, because 2bK is smaller than 10 − 6. Fortunately, this

end point can be sharpened because the CO2 produced from the neutralization of

HCO3− is volatile and can be boiled out of the solution. This is described below.

Fig. 8.7: Titration of 100 cm3 of 0.1 M-Na2CO3 with 0.1 M-HCl

62

Estimations Based On

Kinetic and Acid-Base

Equilibria Studies

At the start of the titration, the pH is determined by the hydrolysis of the Bronsted

base −2

3CO . After the titration is begun, part of the −2

3CO is converted to HCO3¯ ; and

−

3

-2

3 /HCOCO buffer region is established. At the first equivalence point, there remains

a solution of HCO3¯, and

21 aa ][H KK≈+ . Beyond the first equivalence point, the

HCO3¯ is partially converted to H2CO3 (CO2) and a partial buffer region is established,

the pH being established by [HCO3¯]/[CO2]. The pH at the second equivalence point is

determined by the concentration of the weak acid CO2.

Phenolphthalein is used to detect the first equivalence point, and methyl orange is used

to detect the second one. Neither the point, however, is very sharp. In actual practice,

the phenolphthalein end point is used only to get an approximation of where the

second end point will occur; phenolphthalein is colourless beyond the first end point

and does not interfere. The second equivalence point, which is used for accurate

titrations, is normally not very accurate with methyl orange indicator because the

gradual changes in colour of the methyl orange. This is caused by the gradual decrease

in pH due to the HCO3¯ /CO2 buffer system beyond the first end point.

If beyond the first end point the solution is boiled after each addition of HCl to remove

the CO2 from the solution, the buffer system of −

3HCO /CO2 would be removed,

leaving only HCO3¯ in the solution. This is both a weak acid and a weak base whose

pH (≈ 8.3) is independent of concentration (21 aa ][H KK=

+ or21 bb ][OH KK=

− .

Then the pH would remain essentially constant until the equivalence point when we

are left with a neutral solution of water and NaCl (pH = 7.0).

8.3.6 Titration of Polyprotic Acid Versus Strong Base

Diprotic acids can be titrated stepwise, if 1aK ≥ 10

4 x

2aK , the solution behaves like a

mixture of two acids with constants 1aK and

2aK respectively. Thus for sulphurous

acid (H2SO3), 1aK = 1.7 × 10 − 2

and 2aK = 1.0 × 10 − 7

, it is evident that there will be a

sharp change of pH near the first equivalence point, but for the second stage the

change will be less pronounced, yet first sufficient for the use of , say thymolphthalein

as indicator. For carbonic acid (H2CO3), 1aK = 4.3 × 10 − 7

and 2aK = 5.6 × 10 − 11

,

only the first stage will be just discernible in the neutralization curve, the second stage

is far too weak to exhibit any point of inflexion and there is no suitable indicator for

direct titration.

Now the titration curve for diprotic acid (H2A) versus Na2CO3 is to be discussed.

During titration upto first equivalance point, a solution of HA − / H2A buffer region is

established. At the first equivalance point, a solution of HA¯ exists, and

21 aa ][H KK≈+ or

21 aa Kp2

1Kp

2

1pH += . Beyond this, a A2 − / HA − buffer exists;

and finally the second equivalance point, the pH is determined from the hydrolysis of

A2 − .

For triprotic acid (H3PO4), the ionization is given below:

H3PO4 � H+ + H2PO4

− 1aK = 7.5 × 10 − 3

= ]PO[H

]PO[H ][H

43

-

42

+

... (8.10)

H2PO4− � H

+ +

−2

4HPO 2aK = 6.2 × 10 − 8

= ]PO[H

][HPO ][H

42

-2

4

−

+

… (8.11)

63

Neutralization

Titration-I

−2

4HPO � H+ + −3

4PO 3aK = 5 × 10 − 13 =

]PO[H

][PO ][H2

4

-3

4

−

+

… (8.12)

The overall ionization constant is the product of the individual ionization constants:

H3PO4 � 3H+ + −3

4PO

]PO[H

][PO ][H 10x32.2

43

-3

4

322

aaaa 321

+

−=== KKKK … (8.13)

Orthophosphoric acid will behave like a mixture of three monoprotic acids. The pH of

the first equivalance point for 0.1 M H3PO4 with 0.1 M NaOH is given approximately

by:

4.6pKpK21 a2

1a2

1 =+ … (8.14)

and that of the second equivalance point by:

7.9pKpK32 a2

1a2

1 =+ … (8.15)

In the very weak third stage, the curve is flat and no indicator is available, the third

equivalance point may be computed approximately from the following equation:

pcpKpKpH 21

a21

w21

3−+= … (8.16)

= 7 + 6.15 – ½ (1.6) = 12.35

(Here the terms have their usual meaning)

The experimental neutralization curve of 50 cm3 of 0.1 M H3PO4 with KOH and

suitable indicators are shown in Fig. 8.4.

SAQ 2

a) Calculate the pH at 0, 10, 90, 100 and 110 % of the titration for the titration of

50 cm3 of 0.5 M HCl with 0.5 M NaOH.

…………………………………………………………………………………………...

…………………………………………………………………………………………...

…………………………………………………………………………………………...

…………………………………………………………………………………………...

b) Calculate the pH at 0, 25 and 50 cm3 titrant in the titration of 50 cm

3 of 1 M

acetic acid with 1 M NaOH.

…………………………………………………………………………………………...

…………………………………………………………………………………………...

…………………………………………………………………………………………...

…………………………………………………………………………………………...

c) Calculate the concentration of OH − and the pH of a solution that is 0.2 M in

aqueous NH3 and 0.1 M in NH4Cl.

…………………………………………………………………………………………...

…………………………………………………………………………………………...

…………………………………………………………………………………………...

64

Estimations Based On

Kinetic and Acid-Base

Equilibria Studies

d) What is the buffer region?

…………………………………………………………………………………………...

…………………………………………………………………………………………...

…………………………………………………………………………………………...

8.4 THEORY OF INDICATORS

Indicators are the organic substances, the presence of very small amount of which

indicates the termination of a chemical reaction by a change of colour. Indicators are

of various types, e.g., acid-base indicators, redox indicators, adsorption indicator, etc.

Acid-base indicators are the organic substances, which have one colour in acid

solution while different colour in alkaline solution. The following theories have been

put forward to explain the colour change of the acid base indicator.

8.4.1 Ostwald’s Theory

According to the Ostwald theory indicators are such weak acids ( HIn) or bases

(InOH) whose colours are different from that of the indicator-ion formed by their

dissociation. The equilibria in the aqueous solution may be written as:

HIn � H+ + In

¯

InOH � OH¯

+ In+

Unionized ionized

colour colour

If the indicator is a free amine or substituted amine the equilibrium is:

In + H2O � OH¯ + HIn+

Indicator-acids HIn dissociate in aqueous solution as follows:

HIn � H+

+ In¯

Applying the law of mass action to this dissociation

a

-

[HIn]

][In ][HK=

+

… (8.17)

from which

]In[

]HIn[]H[ a −

+= K … (8.18)

The actual colour of the indicator, which depends upon the ratio of the concentration

of the ionized and unionized forms, is thus directly related to the hydrogen–ion

concentration. Eq. (8.18) may be written as

apK]HIn[

]In[logpH +=

−

... (8.19)

In this equation [HIn] represents the concentration of the undissociated indicator-

molecule whose colour is called ‘acid colour’ while [In − ] denotes the concentration of

the indicator-anions, the colour of which is called ‘alkaline colour’. Ka is the

dissociation constant of the indicator- acid.

The indicator base may be characterized similarly to the indicator acid

InOH � In+ + OH −

65

Neutralization

Titration-I

b

]InOH[

]OH[]In[K=

−+

… (8.20)

taking the ionic product of water into consideration

bw

]InOH[]H[

]In[K

K=

+

+

… (8.21)

[ ][ ]InOH

In]H[

b

w

+

+=

K

K … (8.22)

]In[

]InOH[logpKpKpH bw +

+−= … (8.23)

where Kw represents the ionic product of water, Kb denotes the dissociation constant of

the indicator base, the colour of which is the alkaline colour ; the acid colour is due to

the In+

ions.

8.4.2 Modern Quinoid Theory

According to the modern quinoid theory an acid-base indicator is a dynamic

equilibrium mixture of two alternative tautomeric forms; ordinarily one form is

benzenoid while the other is quinoid. Out of these one form exists in the acidic

solution, while the other in alkaline solution. Change in pH causes the transition of

benzenoid form to quinoid form and vise versa and consequently a change in colour.

The colour changes in case of methyl orange and phenolphthalein are given below:

Methyl Orange

Na O3S N=N N(CH

3)

2+

+

Yellow Benzenoid form (in bases)

Na O3S N N=

H H

H

N (CH3)

2+

+

+

+

Red Quinoid form (in acids)

Phenolphthalein

C O HO C

O

COO

H

HOH

OH

C=O

+

++

Colourless, benzenoid form

(in acid)

Red, Quinoidform

(in alkali)

66

Estimations Based On

Kinetic and Acid-Base

Equilibria Studies

SAQ 3

Why the quinoid form of the indicator is coloured or darker than benzenoid form?

…………………………………………………………………………………………...

…………………………………………………………………………………………...

8.5 COLOUR CHANGE RANGE OF AN INDICATOR

A large number of acid-base indicators are available, which possess different colour

according to the hydrogen-ion concentration in the solution. The important

characteristics of these indicators is that the change from a predominantly acid colour

to predominantly alkaline colour is not sudden and abrupt, but takes place within a

small interval of pH (generally about two pH units) termed the colour-change

interval of the indicator . In this pH interval the indicator shows mixed colours of

different shades of the acid and alkaline colours, i.e. the colour intensity of one colour

indicators increases gradually. The acidic red colour of methyl orange and methyl red

is more sensitively perceivable beside the alkaline yellow colour, since the colour

intensity of red form is greater. The position of the colour-change interval in the pH

scale varies widely with different indicators. For most acid-base titrations we can

therefore select an indicator, which exhibits a distinct colour change at a pH close to

that obtained at the equivalence point. Table 8.3 summarizes a selected list of

indicators suitable for neutralization titration.

Table 8.3: The pH Transition Ranges and Colours of Some Indicators

Indicator pH range Colour in

acid solution

Colour in

alkaline solution pK����In

Brilliant cresylblue (acid) 0.0-1.0 Red-orange Blue -

Cresol red (acid) 0.2-1.8 Red Yellow -

Thymol blue (acid) 1.8-2.8 Red Yellow 1.7

m-Cresol purple 1.2-2.8 Red Yellow -

Bromo-phenol blue 3.0-4.6 Yellow Blue 4.1

Methyl Yellow 2.9-4.0 Red Yellow 3.3

Ethyl Orange 3.0-4.5 Red Orange -

Methyl orange 3.1-4.4 Red Orange 3.7

Congo red 3.0-5.0 Blue Red -

Bromo-cresol green 3.8-5.4 Yellow Blue 4.7

Methyl red 4.2-6.3 Red Yellow 5.0

Ethyl red 4.5-6.5 Red Orange -

Propyl red 4.6-6.6 Red Yellow -

Chlorophenol red 4.8-6.4 Yellow Red 6.1

4-Nitrophenol 5.6-7.6 Colourless Yellow 7.1

Bromo-cresol purple 5.2-6.8 Yellow Purple 6.1

Bromo-phenol red 5.2-6.8 Yellow Red -

Bromo-thymol blue 6.0-7.6 Yellow Blue 7.1

Neutral red 6.8-8.0 Red Orange -

Phenol red 6.8-8.4 Yellow Red 7.8

Cresol-red (base) 7.2-8.8 Yellow Red 8.2

m-Cresol purple 7.6-9.2 Yellow Purple -

Thymol blue (base) 8.0-9.6 Yellow Blue 8.9

o-Cresol-phthalein 8.2-9.8 Colourless Red -

Phenol-phthalein 8.3-10.0 Colourless Red 9.6

Thymolphthalein 8.3-10.5 Colourless Blue 9.3

Alizarin yellow R 10.1-12.0 Yellow Orange red -

Brilliant crystal blue (base) 10.8-12.0 Blue Yellow -

Tropaeolin O 11.1-12.7 Yellow Orange -

67

Neutralization

Titration-I

SAQ 4

What is the usual pH range for the colour change of an indicator at the end point?

…………………………………………………………………………………………...

…………………………………………………………………………………………...

…………………………………………………………………………………………...

…………………………………………………………………………………………...

8.6 SELECTION OF INDICATOR AND INDICATOR

ERROR

The primary consideration in choosing an indicator for a given titration is that the

indicator should change, that is the end point should occur, within the required

increment, ∆v of the equivalence point. As a general rule it may be stated that for a

titration to be feasible there should be change of approximately two units of pH at or

near the stochiometric point produced by the addition of small volume of the reagent.

The pH range at either side of the equivalence point (0.1-1 cm3) may be calculated,

and the difference will indicate whether the change is large enough to permit a sharp

end point to be observed. Alternatively, the pH change on the both sides of the

equivalence point is noted from the neutralization curve determined by potentiometric

titration. If the pH change is satisfactory an indicator should be selected that changes

its colour at or near the equivalence point. Next, the molar absorptivity of the indicator

in its two forms should be known in order that the amount of indicator required to give

a readily observable change can be determined, and this amount should be small to

avoid consumption of titrant. Finally the personal preference to the operator can be

consulted as a selection of colours, etc., most studied to his vision. Yet, again it should

be added that the indicator reaction should be fast.

Indicator Error

The indicator error follows from the fact, that the indicator itself will consume a

certain amount of the standard solution. The amount of this consumption of standard

solution depends first of all upon the nature of indicator and its concentration; whether

it is alkaline for instance, etc. If ct. vt >> cind. vind where ct and cind are the

concentrations of titrant and the indicator solution; vt and vind are the volumes used of

titrant and indicator, respectively, then the indicator error will be negligible.

SAQ 5

What is the indicator error?

…………………………………………………………………………………………...

…………………………………………………………………………………………...

…………………………………………………………………………………………...

…………………………………………………………………………………………...

…………………………………………………………………………………………...

8.7 SUMMARY

In this unit we have described the basic concept of titrimetry. The neutralization

curves for strong acid with strong base, weak acid with strong base, weak base with

strong acid, weak acid with weak base, sodium carbonate with strong acid, and

polyprotic acid with strong base have been illustrated. The nature of titration curves

68

Estimations Based On

Kinetic and Acid-Base

Equilibria Studies

has been explained. The calculations of pH of the solution before titration, at the

equivalence point, and after the equivalence point have been illustrated. The theory of

indicators and the selection of the appropriate indicator for a particular titration have

been discussed.

8.8 TERMINAL QUESTIONS

1. Only strong acid or base is used as the titrant, why?

2. Calculate the [OH−

], pOH, pH and percent ionization for 0.2 M aqueous NH3.

3. Calculate the concentration of the species in a 0.1 M H2SO4. K2 = 1.2 × 10− 2

.

4. When the mixtures of acids (or bases) can be titrated stepwise?

5. Calculate the pH of a solution prepared by adding 25 cm3 of 0.10 M NaOH to

30 cm3 of 0.20 M acetic acid.

6. Write the criteria for choosing an indicator.

8.9 ANSWERS

Self Assessment Questions

1. a) A primary standard is a substance which satisfies the following

requirements:

• It must be easy to obtain, to purify, to dry ( preferably at 110-

120oC) and to preserve in pure state.

• The substance should remain unaltered during weighing i.e., it

should not be hygroscopic, or oxidized by the air, or affected by

carbon dioxide.

• The substance should be capable of being tested for impurities by

qualitative and other tests of known sensitivity.

b) Secondary standards may be used for standardizations by finding the

content of the active substance comparing against a primary standard.

2. a) 0.30, 0.39, 1.58, 7.00, and 12.48

b) 2.37, 4.74, and 9.07

c) [OH¯] = 3.6 x 10

− 5 M and pOH = 9.56

d) The slowly rising region before the equivalence point is called the buffer

region. It is flattest at the midpoint, and so the buffer capacity is greatest

at a pH corresponding to pKa.

3. The increased per cent conjugation in the quinoid form of the indicator results

the shifting of the λmax from a shorter to a greater wave length (i.e., from the

ultra violet region to the visible region).

4. The point at which the colour change for the indicator occurs in a titration is

called end point. Typically, colour changes occur over a range of 1.5 to 2.0.

5. Section 8.6, 2nd Para.

69

Neutralization

Titration-I

Terminal Questions

1. The magnitude of the break at the equivalance point is significant and the end

point can be obtained with greater accuracy.

2. [OH−

] = 1.9 x 10− 3

M, pOH = 2.72, pH = 11.28 and % ionization = 0.95%

ionized.

3. In first step, complete ionization of H2SO4 is complete.

−++→+ 43242 HSOOHOHSOH

0.10 M 0.10 M 0.10 M

In second step, ionization is not complete.

OHHSO 24 +−

� −+

+2

43 SOOH and ]HSO[

]SO[]OH[

4

2

432 −

−+

=K =1.2 × 10 − 2

Let x = ]HSO[ 4

−that ionizes. Therefore at equlibrium

OHHSO 24 +−

� −+

+2

43 SOOH

(0.1-x) M (0.10 + x )M xM

↓ ↓

I step II step

2

4

2

432 102.1

)10.0(

)()10.0(

]HSO[

]SO[]OH[ −

−

−+

×=−

+==

x

xxK

x cannot be ignored because K is too large

[H3O+] = [

−2

4SO ] = 0.01M. The concentrations of species in 0.1 M H2SO4 are:

[H2SO4] ≈ 0.0 M ; −

4HSO = (0.10-x) M = 0.09 M; [−2

4SO ] = 0.01 M

[H3O+] = (0.10 + x) M =0.11M

[OH−

] = Kw/ [H3O+] =1.0 × 10

− 14/0.11 = 9.1 × 10

− 14

4. There should be an appreciable difference in their strength and one acid or base

should be at least 104 times weaker than the other to titrate separately.

5. CH3COOH + NaOH � CH3COONa + H2O

m mol of CH3COOH formed = m mol of NaOH added =2.5 m mol

Unneutralized CH3COOH = 6.0 –2.5 = 3.5 m mol

pH= 4.76 + log 2.5/3.5 = 4.61

6. Choose an indicator with a pKa near the equivalence.