Acid Base Titrations in Aqueous Solvents

Full text

(1)

-1-

Acid Base Titrations in Aqueous Solvents

Introduction:

All kind of titrations in various samples are performed today in process analysers and laboratories, by far the most of them are acid base titrations. Acids and bases play an important role in many processes and are controlled at a regularly basis. In most cases titration in water is quite adequate to determine the acid or base concentration. Generally for the titration of acids a strong base like sodium or potassium hydroxide is used. For the titration of bases a very strong acid like hydrochloric acid is used, the hydrochloric acid can be replaced by nitric acid or sulphuric acid if the sample contains substances reacting with chloride such as silver ions.

Titration is an absolute method so requires no calibration, on top of that multiparameter determinations are possible: with one acid as a titrant several bases can be determined. Titration is a quite versatile technique where it concerns the concentration range that can be analysed.

Variations in sample size (between 0.1 and 100 ml), in concentration of the titrant (between 0.01 M and 1 M) and that of the burette size (between 2 ml and 50 ml) can be made. Most determinations in this monograph are potentiometric titrations with a glass electrode as indication system. In section 9 an example is given with an antimony electrode and section 10 on a photometric indication.

1) Classification of Acids according to strength:

An acid dissolved in water gives the next reaction: HA + H

2O ÙH3O + + A-

Depending on the strength of an acid the equilibrium lies to the right or to the left. A very strong acid, like hydrochloric acid completely dissociates in water. A weak acid, like acetic acid only partly dissociates and most of the acetic acid is present in the form of the non dissociated acid.

For the classification of the acid strength the dissociation constant Ka is used.

[ ]

] ][

[

3 HA

A O

K a

=

H +

Example: Calculation of the part of the acid that has not been dissociated.

An Acid HA with Ka=10

-4

is dissolved in water at a concentration of 1 mol / l Reaction: HA + H

2

O Ù H

3

O

+

+ A

-

So [H

3

O+] is equal to [A

-

]

The concentration is 1 mol / l in total, so [HA] + [A

-

] = 1 mol / l

[ ]

] ][

[

3

HA A O

K

a

= H

+

] [

] 4 [

10

2

HA A

− =

The greater part of the total concentration HA is in non dissociated form HA so [HA] ~ 1 mol/l and [A

-

]

2

= 10

-4

and [A

-

] =10

-2

So about 1 % of the acid has been dissociated and 99% is not.

Definition

pKa = -log Ka

For easy operations normally the unit pKa is used for the acid strength.

pKa means the -log of the Ka value.

Example: Ka of Acetic Acid is: 1.75 * 10

-5

So log Ka = - 4.76 and the pKa = 4.76

(2)

2a) Classification of Acids according to pKa :

Some examples of acids, besides the listed ones many more of them can be found in literature.

Very Strong Acids pKa < 0 pKa

HClO

4

-9

HI -9

HCl -6

H

2

SO

4

-3

HNO

3

-1

Strong Acids 0 < pKa< 4.5

HSO

4-

1.9

H

3

PO

4

2.1

HF 3.1

Formic Acid 3.7

Weak Acids 4.5 < pKa <9.5

Acetic Acid 4.7

H

2

PO

4-

7.2

HCN 9.1

Very Weak Acids pKa >9.5

HPO

4-2

12.3

Phenol 10

The very strong acids all completely dissociate in water, the differences in pKa value in this group are only noticeable in non-aqueous solvents such as acetone and alcohols. In water all the very strong act the same. The figure below presents titration curves of acids with different pK values using the degree of conversion λ versus pH.

λ = 0 at start of the titration and λ = 1 in the Endpoint

Fig.1

Acids with pK numbers from 2 - 10

2.00 4.00 6.00 8.00 10.00 12.00

0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 1.10 1.20 pH

pK

10

9

8

7

6

5

4

3

2

λ

(3)

- 3-

As can be seen from the curves:

-the potential jump gets smaller with higher pK values

-acids can be titrated with a reasonable accuracy till a pK of 8 -very weak acids show no inflection point at all !

-for the weaker acids (in the range of pK 3.5 and 10.5) at the factor λ = 0.5 the pK values correspond to the pH values.

2b) Classification of Bases according to pK

b

:

A base dissolved in water gives the next reaction :

B + H2O Ù OH - + HB +

Strong bases are for instance NaOH, KOH, and tetra methyl ammonium hydroxide.

Analogue to acids a strong base reacts completely and a weak base partly dissociates.

The strength of a base is given by the K

b

value, following the formulae:

] [

] ][

[ B

HB Kb= OH +

For easy operations the next equation is used: Definition

pKb = -log Kb

Some examples of bases

Weak bases 4.5 < pKb < 9.5 Very weak bases pKb > 9.5

Ammonia 4.7 Caffeine 13.4

Pyridine 8.8 Acetamid 13.5

Aniline 9.4 Urea 13.8

Equation: pKa+pKb=pKw

Water dissociates for a very small part in [H

3

O

+

] and [OH

-

] , following the reaction:

2 H

2

O Ù H

3

O

+

+ OH

-

, the K of this reaction Kw = 10

-14

and the pKw = 14 Ammonia reacts with water: NH

3

+ H20 Ù NH4

+

+ OH

-

The (NH

4

)

+

reacts with water NH

4+

+ H20 Ù NH3 + H

3

O

+

Because the ratio between [H

3

O

+

] and [OH

-

] is known by the pKw, the pKa and pKb can easily be calculated if one of both is known because pKa+pKb = pKw =14.0 So Ammonia (NH

3

) is a weak base with pKb = 4.7 Therefore the Ammonium (NH

4

)

+

is a weak acid with pKa = 9.3

In literature many tables of the pKa of acids and pKb of bases can be found, sometimes the constant is represented just as pK and the reader has to find out himself if is the stated pK is the pKa or the pKb. Unfortunately some tables do not mention the pKb of bases but just record the pKa of the corresponding acid. Because the sum of pKa and pKb is 14, calculation of pKb of the corresponding base is easy.

3a) Polybasic Acids:

Some acids react with water in steps; a well-known example is Phosphoric Acid (Fig. 2a) During the titration H

3

PO

4

reacts with hydroxide in three steps.

H

3

PO

4

+ OH

-

Î H

2

PO

4-

+ H

2

O with pKa1 = 2.1 H

2

PO

4-

+ OH

-

Î HPO

4-2

+ H

2

O with pKa2 = 7.2 HPO

4-2

+ OH

-

Î PO

4-3

+ H

2

O with pKa3 = 12.3

A titration could theoretically result in three endpoints. The last EP however cannot be detected

because of the high pKa number (fig 2a).

(4)

Fig. 2a Fig. 2b

3b Polyacid bases:

The Na

3

PO

4

sodium salt can be titrated with HCl ( fig. 2b) , so again a titration in 3 steps that ends in H

3

PO

4

. The pKb values can be calculated from the pKa numbers of phosphoric acid, using the expression pKa + pKb = pKw=14. The corresponding pKb numbers are respectively 1.7, 6.8 and 11.9. Again the last EP cannot be detected because the base is too weak.

4 Separation of a Mixture of Acids first idea

Mixtures of two acids can be determined in one titration if there is a significant difference in pKa value. The strongest acid is titrated first, resulting in the first endpoint, the second endpoint can represent the weaker acid. If the difference in pK is too small no separate endpoints will be found, only the sum of the acids can be determined. In this section some examples are given

4a) HCl and HNO

3

The determination of hydrochloric acid and nitric acid is showed in fig. 3. The pKa numbers are respectively -6 for the hydrochloric and - 1 for the nitric acid. As mentioned earlier these negative values have no meaning in water and both acids act similar. The result is that no seperation occurs. The sum of both acids of coarse can easily be determined. In this particular case the HCl, containing chloride can be determined by a titration with AgNO

3

. The difference between the results in the titration with NaOH and the result in the titration with AgNO3 represents the HNO

3

.

Fig.3

EP1

EP2

2 3 4 5 6 7 8 9 10 11 12

0 1 2 3 4 5 6 7 8 9 10

Na

3

PO

4

with 0.2 M HCl

pH

V [ml]

EP1

EP2

2 3 4 5 6 7 8 9 10 11 12

0 1 2 3 4 5 6 7 8 9

H3PO4 with 0.2 M NaOH

pH

V [ml]

E P 1

1 2 3 4 5 6 7 8 9 10

0 .0 2 .5 5.0 7.5 10.0 1 2.5

H C l + H N O3 w ith 0 .2 M N aO H

p H

V [m l]

HCl (pK

a

–6) and HNO

3

(pK

a

–1)

-No separation -Sum can be

determined

(5)

- 5- 4b) HCl and Phosphoric Acid

Solutions of hydrochloric acid are dissociated for 100 %, solutions of phosphoric acid are dissociated for about 25 %. These values as such, indicate that a titration of a mix of both acids will give problems in detecting the first endpoint. And indeed no separation occurs.

The first endpoint detected in the mix represents the sum in the HCl and first endpoint of H

3

PO

4.

Fortunately the endpoint from the reaction H

2

PO

4-

+ OH

-

Î HPO

4-2

+ H

2

O can be used for the determination of the phosphoric acid. Because the same consumption of hydroxide is used for the reaction H

3

PO

4

+ OH

-

Î H

2

PO

4-

+ H

2

O the determination of HCl is possible too, just by subtracting the difference of the detected endpoints from the first endpoint.

Fig. 4

4c) Hydrochloric and Acetic Acid

Both acids can be determined, the separation even occurs if the ratios hydrochloric acid and acetic acid are varied from 1-10 to 10-1. In figure 5 curves are shown of the ratios 1-10, 1-1 and 10-1. The endpoint in this mixture of hydrochloric acid is about pH 3.5. HCl alone gives an endpoint at about pH 7 (see fig. 4). The inflection of the hydrochloric acid in the mix is much smaller because of the presence of the inflection of the acetic acid.

Fig.5

2 3 4 5 6 7 8 9 10 11 12

0. 1. 2. 3. 4. 6. 7. 8. 9. 10. 12.

A) H3PO3

B)

HCl

C)

HCl+ H3PO3

pH

V [ml]

A B C

Titrant 0.2 M NaOH HCl ( pK

a

–6 )

H

3

PO

4

( pK

a

2.1) H

2

PO

4-

( pK

a

7.2)

-No separation H

3

PO

4

and HCl

-Separation H

3

PO

4

and H

2

PO

4-

EP2

EP1

EP2

EP1 EP2

EP1

2 3 4 5 6 7 8 9 10 11

0 1 2 3 4 5 6 7 8 9 10 11 12

H Cl + Acetic Acid with 0.2 M N aO H

pH

V [ml]

HCl (pK

a

-6) Acetic Acid (pK

a

4.7)

-Separation HCl and Acetic acid

A) HCl -Acetic Acid 1 - 10 B) ´´ 1 - 1 C) ´´ 10 - 1

A B C

(6)

4d) Hydrochloric Acid + Formic Acid

The titration results in a poor inflection for the hydrochloric acid. Accuracy will be far from optimal, furthermore if the performance of the electrode declines false endpoints will be found. To check inflections, the differentiated ERC curve as represented in fig.6b can be viewed. Large potential breaks in the normal curve, like the second endpoint, result in sharp peaks and the top reaches high ERC values above 70. The first endpoint is a poor inflection: the top reaches a value of 7.

Into account has to be taken that this peak is on a sloping baseline, after correction the ERC value drops to 3. This figure is definitely too low for quantitative measurements. Discrete values of 5 are the minimum, and advised is 10 for a reasonable accuracy. So for mixed acids of this composition only the total of acids can be determined.

Fig. 6a Fig. 6b

4e) Phosphoric acid and Acetic Acid

The EP2 from the acetic acid is the smallest inflection. EP3 and EP1 have larger inflections, it is advised to use these endpoints for the calculation of the acetic acid. The phosphoric acid can be calculated from the EP1. The difference between EP3 and EP1 represents the sum of the consumption of the H

2

PO

4-

and the acetic acid. As for phosphoric acid solutions the consumption of the H

3

PO

4

and the H

2

PO

4-

consumption are basically the same, the difference is the acetic acid.

Fig.7 a Fig.7 b

EP1

EP2

2 3 4 5 6 7 8 9 10 11 12

0 1 2 3 4 5 6 7 8 9 10 11 12

HC l + Formic Acid

pH

V [ml]

EP1 EP2

0 10 20 30 40 50 60 70 80

0 1 2 3 4 5 6 7 8 9 10 11

HC l+Formic Acid

ERC

V [ml]

HCl (pK

a

-6) Formic Acid (pK

a

3.7) -Too poor separation

EP1

EP2 EP3

2 3 4 5 6 7 8 9 1 1 1

0 5 1 1 2 2

Phosphoric+ Acetic Acid with 0.2 M NaOH

pH

V [ml]

EP1 EP2 EP3

0 10 20 30 40 50 60

0 5 1 1 2 2

Phosphoric + Acetic Acid

ERC

V [ml]

pKa H

3

PO

4

2.1 H

2

PO

4-

7.2 Acetic

Acid 4.7

Poor

seperation

(7)

- 7- 4 f) H3PO4 + Formic acid:

No separation of the H

3

PO

4

and the formic acid, the first detected endpoint represents the sum of both acids. Because the H

2

PO

4-

step is well separated, calculation of the phosphoric as well as the formic is possible. Note in the differentiated curve the small ‘hill’ around 5 ml. The peak height is so low that the endpoint from the H3PO4 is not detected.

Fig. 8a Fig. 8b

5 Separation of a mixed bases

Analogue to acids, the simultaneous determination of bases can be done if there is a significant difference in pKb of the bases. Is the difference too small then only the sum of both bases can be determined. The stronger base, having the lowest pKb value is titrated first the one with the highest pKb is titrated last.

5a Amino ethoxy ethanol and hydroxyl amine.

Amino ethoxy ethanol pKb: 5.4 Hydroxylamine: pKb: 8.0

The amines are well separated, so good quantitative determinations are possible. See the differentiated curves in fig. 9b.

Fig. 9 a Fig. 9b

.

EP1 EP2

2 3 4 5 6 7 8 9 10 11 12

0.0 2.5 5.0 7.5 10.0 12.5 15.0 17.5

H3PO4+Formic Acid

pH

V [ml]

EP1 EP2

0 5 10 15 20 25 30 35 40 45 50 55 60

0.0 2.5 5.0 7.5 10.0 12.5 15.0 17.5

H3PO4+Formic Acid

ERC

V [ml]

pKa H

3

PO

4

2.1 H

2

PO

4-

7.2 Formic Acid 3.7

No separation H

3

PO

4

and Formic Acid

A: Amino ethoxy ethanol B : Hydroxyl amine C : Mix

A

B C

0 10 20 30 60 50 60 70

0. 2. 5. 7. 10. 12. 15.

Amines with 0.1 M H2SO4

ERC

V [ml]

A

B C

C

2 3 4 5 6 7 8 9 1 1

0. 2. 5. 7. 10. 12. 15. 17.

Amines with 0.1 M H2SO4

A) Amino Ethoxy Ethanol B) Hydroxylamine

C) Mix

pH

V [ml]

A

B C

(8)

5b) Carbonate, Bicarbonate and Hydroxide

The Carbonic acid H

2

CO

3

can dissociate in two steps, first in the formation of Bicarbonate (HCO

3-

) with a pKa of 6.3 and the second to Carbonate (CO3

–2

) with pKa 9.4. The H

2

CO

3

is not stable in water and falls apart in CO

2

and H

2

O. The carbonate and bicarbonate are stable bases and can be titrated with acid. Using the expression pKa + pKb = pKw =14 the pKb can be calculated.

So the pKb of Carbonate is 4.6 and the pKb of bicarbonate is 7.7

In samples with Carbonate and Hydroxide in about equal concentrations 3 EPs exist (Curve1 fig.10)

EP1: NaOH + HCl Î NaCl + H

2

O EP2: Na

2

CO

3

+ HCl Î NaHCO

3

+ NaCl EP3: NaHCO

3

+ HCl Î CO

2

+ H

2

O +NaCl

The first EP cannot be used for quantitative measurements, but all concentrations can be calculated from the last two remaining endpoints because the consumption for the Na

2

CO

3

is equal to the consumption of NaHCO

3 .

With low concentrations carbonate (curve2 in fig.10) two endpoints are detected.

EP1: sum of NaOH + Na

2

CO

3

and EP2: NaHCO

3

Fig. 10

6) Rules for seperation of acids and bases based on the pK

For acids rules of thumb

The negative values for the pKa of very strong acids have a meaning only in organic solvents.

For the rule of thumb in water, use for all of the very strong acids in these calculations the value pK = 0 .As can be seen in the foregoing examples successful seperation based on the difference of pK values is achieved easier with weaker acids.

Strong acids: Difference in pKa at least pKa > 4 Weak acids: Difference in pKa at least pKa > 2

For bases rules of thumb:

Strong bases: Difference in pKb at least pKb >4 Weak bases: Difference in pKb at least pKb >2

E P 2 E P 1

E P 3 E P 2

E P 1

1 2 3 4 5 6 7 8 9 10 11 12 13

0.0 0.7 1.4 2.1 2.8 3.5 4.2 4.9 5.6 6.3 7.0

H ydro xide and C arbonate w ith 0.2 M H C l

pH

V [m l]

2 1) High CO

3=

and high OH

-

1

2) Low CO

3=

and high OH

-

(9)

- 9- 7) Titration of very weak acids

For Ammonia (NH

3

) the pKb is 4.7, using the formulae pKa+pKb=14 the pKa for the corresponding acid the Ammonium ion (NH4)+ is 9.3. The (NH4)+ ion can be titrated with sodium hydroxide, because the pKa value is above 8 the accuracy will be limited. As a result the curve shows a small inflection. The results of different quantities of a standard solution of ammonium chloride result in a relative standard deviation of 0.5 % . Note that the endpoint itself varies with the concentration.

Fig. 11 Different concentrations (NH

4

)

+

Standard solution of ammoniumchloride, 1-10 ml of standard + 50 ml water Titrant 10 ml burette 0.2 M NaOH, DET ( MPD = 4, drift 0.1 pH /min)

No. ml of sample EP in ml EP in pH NH4Cl in

g/l

1 1 1.039 10.40 11.12

2 2 2.052 10.56 10.98

3 4 4.135 10.73 11.06

4 6 6.161 10.81 10.99

5 8 8.214 10.88 10.99

6 10 10.273 10.88 11.02

Average 10.71 11.03

Relative Standard deviation 0.5 %

8) Examples of the formation of a second compound

Sometimes chemistry can help to solve a titration problem, some acids or bases form on addition of an auxiliary solution a compound that can be titrated easier.

8a) Ammonia and Formaldehyde

An example is the reaction of ammonia with formaldehyde, adding an excess of formaldehyde converts ammonia to the hexamethylene tetramine.

Reaction: 4 NH

3

+

6 HCOHÎ (CH

2

)

6

N

4

+ 6 H

2

O

The hexamethylene tetramine ( pKb = 9.1) is a much weaker base then ammonia (pKb = 4.7) . So for the titration of ammonia with acid, the addition of formaldehyde would not improve the titration. But if we add formaldehyde to ammonium chloride, the weak acid ammonium (pKa 9.3) is converted to a to the stronger hexamethylene tetramine ion (pKa 4.9). This improves the curves remarkable. To check the improvement of the formaldehyde addition some titrations were performed with NaOH 0.2 M and standard samples of HCl, NH4Cl and a mix of both (with the

E P 1 E P 1

E P 1

E P 1 E P 1 E P 1

8 9 1 0 1 1

0 2 4 6 8 1 0 1 2

N H 4 C l w ith 0 .2 M N a O H

p H

V [m l]

(10)

addition of formaldehyde 5 ml 37 % pre neutralised to pH 7). When the determination of the ammonium is done in two determinations, the overshoot of titrant in the HCl titration has to be added to the consumption in the ammonium titration.

1 NH

4

Cl 2 HCl

3 mix in one titration

4) HCl + NH

4

Cl mix in two titrations, titration of the HCl 5) HCl + NH

4

Cl mix in two titrations, second titration

of the NH

4

Cl after addition of formaldehyde

Fig. 12 a Fig. 12b

No formaldehyde addition Formaldehyde addition

Sample HCl det. NH

4

Cl det. Sample HCl (4) NH

4

Cl (5).

1 NH4

Cl 6.169 ml HCl + NH

4

Cl 5.472 ml 6.177 ml

2

5.483 ml HCl + NH

4

Cl 5.471 ml 6.160 ml

3 HCl + NH4

Cl 5.473 ml 6.189 ml HCl + NH

4

Cl 5.473 ml 6.149 ml

average 5.472 ml 6.162 ml

Relative

Stand. Dev 0.02 % 0.23 %

Comparing the standard deviations it is obvious that the deviation of the NH

4

Cl is relatively high compared to the HCl deviation. This example follows the rule that deviations will increase with increasing pK value.

8b) Boric acid and Mannitol

Fig.13 On the addition of the sugar

Mannitol the Boric Acid with a pKa of 9.2 can be converted to the boric ester.

Because this ester has a pK of 6 titration results will improve remarkably.

EP2

EP1 EP1

EP1

2 3 4 5 6 7 8 9 10 11 12

0.0 2.5 5.0 7.5 10.0 12.5

HCl, NH4Cl,HCl+NH4Cl

pH

V [ml]

EP1 EP1

2 3 4 5 6 7 8 9 10

-0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6.5

HCl, 5 ml Formaldehyde 37 %, NH4Cl

pH

V [ml]

1 2 3

4 5

EP1 EP1

4.68 6.24 7.80 9.36 10.92

0 1 2 3 4 5 6 7 8 9 10 11 12

Boric acid Boric Acid and Mannitol

pH

V [ml]

(1) (2)

1

2

(11)

- 11-

9) Hydrogen Fluoride with an Antimony Electrode

For potentiometric acid base titrations the glass electrode is used almost exclusively as indication system. As the glass is attacked by fluoride solutions below the pH of 4, the the alternative of an Antimony Electrode can be used here.

Fig. 14

10) Acid Base Titrations using Color Indicators.

One of the oldest forms of titration is the addition of color indicators for Endpoint indication.

A color indicator is an organic dye that has different colors depending on the pH of the solution.

An example is methyl orange, above pH 4.4 the color is yellow and below pH 3.1 the color is red.

The region between pH 3.1 and 4.4 is called, the pH interval, between these two pH values the color change is from red to orange to yellow. For an acid having its endpoint at about pH 3.7 methylorange can be used. The titration can be performed visually, the addition of titrant is continued until the original red color changes to orange.

Fig. 15

.

pH indicator pH interval Color change

Methylorange 3.1 - 4.4 red - yellow

Methyl red 4.2 - 6.3 red - yellow

p- Nitrophenol 5.6 - 7.6 colorless- yellow

Phenolphthalein 8.2 - 9.8 colorless -red violet

E P 1

4 5 6 7 8 9 1 0

0 1 2 3 4 5 6 7

H F w i t h 0 . 2 M N a O H ( A n t i m o n y e l e c t r o d e )

p H

V [ m l ]

ml 3

5 pH

EP red

yellow

(12)

Thymolphtalein 9.3 - 10.5 colorless -blue

The disadvantage of visual indication is when the titration curve has not a clear inflection, the color change of the indicator is to gradually then. Adding more indicator to the solution will introduce a second problem because the indicator itself is also an acid or base and so is consuming acid or base. When two acids or bases are present in the sample, for each determination a different indicator has to be used.

In fig. 16a Sodium carbonate is titrated a fotometric indication using the indicators phenolphthalein (PP) and methylorange (MO.) is used . At the beginning of the titration the solution is red/violet and due to the color of the PP and the transmission is low. Adding acid will lower the pH and the PP will become colorless. At this stage the resulting solution gets yellow due to the yellow color of the MO and this results in a high Transmission. Continuing the titration the MO will change its color from yellow to red and the Transmission will drop again. The raise and fall in the transmission results in the detection of an endpoint.

As a comparison fig. 16 b gives the determination with a glass electrode for detection of the endpoint.

Fig. 16 a Fig. 16b

Jaap Guijt 20 February 2004

EP2 EP1

2 3 4 5 6 7 8 9 10 11

0 1 2 3 4 5 6 7 8 9 10 11 12

Na2CO3 with glass electrode,

pH

V [ml]

EP2 EP1

55 60 65 70 75 80 85 90 95 100

0 1 2 3 4 5 6 7 8

Na2CO3, fotometric. 540 nm

phenolphthaleine + methylorange addition

% T

V [ml]

Figure

Updating...

References

Updating...