Titration is a laboratory method wherein the concentration of a particular solute in a solution is determined. In an acid-base titration, a solution with a known concentration of base is slowly added to an acid or vice versa. By monitoring the progress of the reaction, a titration curve can be produced. A titration curve is the graph of the pH as a function of the volume of the added titrant. The equivalence point of a titration curve is the point at which the quantities of an acid and base that have been brought together are stoichiometrically equivalent. The titration curve may be used to determine the Ka of the weak acid or the Kb of the weak base being titrated.

A. Preparation of Standard Titration Curves In a 100 mL beaker, 25 mL of 0.1 M HOAc was

measured and transferred. The initial pH was determined with the use of a pH meter. The titrant, 0.1 M NaOH, was added by 1 mL increments. The solution was stirred then the pH was recorded afterwards. This was done after each addition of the said titrant. After an abrupt change in the pH, the volume of the increments added was changed to 0.5 mL. This was done until the pH was at 5-6 past neutral.

The procedure was repeated but with: (b) 0.1 M HCl as initial solution and 0.1 M NaOH as the titrant and (c) 0.1 M NH4OH as the initial solution and 0.1 M HCl as the titrant.

B. Analysis of Unknown The same procedure in Part A was done

but with 0.1 M HCl titrant used for the unknown base and 0.1 M NaOH for the unknown acid.

I. Titration of 25 ml 0.1 M HOAc, (which is a weak acid) and 0.1 M NaOH (which is a strong base).

Before Titration: Initial pH of 0.1 M HOAc

HOAc + H2O H3O + + OAc-

Initial 0.1 M 0 0Change -x +x +xEquilibrium 0.1-x x x

Ka= [H3O + ] [OAc-] = 1.8 x 10 -5 (pH will depend on [H3O + ] = x)

[HOAc]x2/ (0.1-x)= 1.8 x 10 -5 x= 1.3 x 10 -3 pH= -log (1.3 x 10 -3 ) = 2.89

Using the Henderson-Hasselbach equation, (pH= pKa + log [base]/[acid]), it is possible to predict the change in pH as the titrant is added.

 + 5 ml NaOH= 30 ml (solution) (before equivalence point)

0.1 M NaOH = 0.1 mol NaOH/ 1L= 0.1 mmol/ml5 ml NaOH x 0.1 mmol/ml = 0.5 mmol25 ml HOAc x 0.1 mmol/ml= 2.5 mmol

HOAc + OH- OAc- + H2OInitial 2.5 mmol 0.5 mmol 0Change -0.5 mmol -0.5

mmol+0.5 mmol

Equilibrium 2 mmol 0 0.5 mmol

pH= pKa + log ( [base]/[acid]) -log (1.8 x 10 -5 ) + log (2 mmol/30 ml)/ (0.5 mmol/30)pH= 4.14

+ 25 ml NaOH= 50 ml (solution) (Equivalence point)  25 ml NaOH x 0.1 mmol/ml= 2.5 mmol

HOAc + OH- OAc- + H2OInitial 2.5 mmol 2.5 mmol 0Change -2.5 mmol -2.5

mmol+2.5 mmol

Equilibrium 0 0 2.5 mmol

OAc- (weak base) will undergo hydrolysis

OAc- + H2O HOAc + OH-

Assume [OH] is small and [NaOAc]= 2.5 mmol/50 ml= 0.05 M

Kb= [OH] [HOAc] / [OAc]= Kw/Ka= 1 x10 -14 / 1.8 x 10 -5 =5.6 x10 -10

Kw/Ka= 1 x10 -14 / 1.8 x 10 -5 = x2/ 0.05x= 5.27 x 10 -6

pOH= -log (5.27 x 10 -6) pOH=5.28 pH= 14-5.28= 8.72

As 0.1 M NaOH is added in increments of 1 ml, the pH of the solution shows little change because of the buffer solution being set up. Due to its buffer capacity, which is the amount of acid or base (in this case 0.1 M NaOH) the buffer can neutralize before the pH will change to an appreciable degree. However, the change in pH becomes more pronounced as it nears the equivalence point. The pH of the solution depends on the concentration of HOAc that has not been neutralized. The goal of titration is to reach the equivalence point, which is the point where the added solute (NaOH) reacts completely with the substance present (HOAc). In this titration, equivalence point is attained when 25 ml of 0.1 M NaOH is added to 25 ml of 0.1 M HOAc. Thus, the concentration of the acid and base is the same. Since HOAc is a weak acid which reacted to NaOH which is a strong base, the solution becomes slightly basic at the equivalence point, with a pH of 8.72. The pH at the equivalence point is always above 7 when a weak acid-strong base titration happens, because the salt formed has an anion that is a weak base (OAc-), which will undergo hydrolysis(pH upon computation is 8.72). After the equivalence point, the pH is determined by the concentration of OH- from the excess NaOH.

ml titrant pH

0 2.89

5 4.14

10 4.56

15 4.92

20 5.34

25 8.72

30 11.96

35 12.22

40 12.36

45 12.46

50 12.52

55 12.57

60 12.61

Theoretical Titration of 0.1 M HOAc with 0.1 M NaOH

ml titrant pH

0 3.11

1 3.4

2 3.67

3 3.85

4 3.98

5 4.09

6 4.2

7 4.3

8 4.37

9 4.45

10 4.52

11 4.6

12 4.67

13 4.73

14 4.81

15 4.88

16 4.95

17 5.03

18 5.11

19 5.22

ml titrant pH

20 5.32

21 5.43

22 5.58

23 5.77

24 6.09

25 6.63

26 9.88

26.5 10.22

27 10.82

27.5 10.93

28 11.21

28.5 11.32

29 11.39

29.5 11.46

30 11.52

30.5 11.57

31 11.6

Titration of 0.1 M HOAc with 0.1 M NaOH

II. Titration of 25 ml 0.1 M HCl (strong acid) with 0.1 M NaOH (strong base)

Before Titration: Initial pH of 0.1 M HCl[H+] = [HCl]pH= -log (0.1M)pH= 1

+ 5ml= 30 ml (solution)(before equivalence point)Acid mmol= 0.1 mmol/ml x 25 ml= 2.5 mmolAdded Base mmol= 0.1 mmol/ml x 5 ml= 0.5 mmol

H3O + + OH- 2 H2OInitial 2.5 mmol 0.5 mmolChange -0.5 mmol -0.5

mmolEquilibrium 2mmol 0

(pH will depend on [H3O + ])

pH= -log (2mmol/30 ml)pH= 1.18

+ 25 ml= 50 ml (solution) (equivalence point)Added base mmol= 0.1 mmol/ml x 25 ml= 2.5 mmol

H3O + + OH- 2 H2OInitial 2.5 mmol 2.5 mmolChange -2.5 mmol -2.5

mmolEquilibrium 0 0

Since no excess acid or base is present, the solution is neutral. pH=7

+30 ml= 55ml (solution) (after equivalence point)Added base mmol= 0.1 mmole/ml x30 ml= 3 mmol

H3O + + OH- 2 H2OInitial 2.5 mmol 3 mmolChange -2.5 mmol -2.5

mmolEquilibrium 0 0.5 mmol

(pH depends on excess [OH])pOH= -log (0.5 mmol/55ml)pOH= 2.04 pH= 14-2.04= 11.96

In a titration of a strong acid and a strong base, the solution of the strong acid will always have a higher initial pH (initial pH of HCl is 1, whereas initial pH of HOAc is 2.89). Just as with HOAc, the initial pH depends on the concentration of HCl. Before the titration reaches the equivalence point, pH increases slowly, but experiences a steeper rise of pH as it reaches the equivalence point. Theoretically, in a strong acid-strong base titration, the pH at the equivalence point is always 7, because the cation of the strong base (Na +) and the anion of the strong acid (Cl-) do not hydrolyze. The pH of the solution after the equivalence point will be determined by the concentration of excess NaOH.

Theoretical Titration of 0.1 M HCl with 0.1 M NaOH

ml titrant pH

0 1

5 1.18

10 1.37

15 1.6

20 1.95

25 7

30 11.96

35 12.22

40 12.36

45 12.46

50 12.52

55 12.57

60 12.61

ml titrant pH

1 1.67

2 1.63

3 1.59

4 1.57

5 1.58

6 1.67

7 1.63

8 1.62

9 1.61

10 1.62

11 1.64

12 1.65

13 1.67

14 1.69

15 1.71

16 1.74

17 1.77

18 1.8

19 1.84

20 1.9

ml titrant pH21 1.9522 2.0123 2.1224 2.2425 2.4526 2.9227 6.36

27.5 9.6728 10.38

28.5 10.7729 10.91

29.5 11.0730 11.22

30.5 11.2931 11.33

31.5 11.432 11.43

32.5 11.4633 11.51

33.5 11.5134 11.52

34.5 11.5635 11.57

35.5 11.59

Titration of 0.1 M HCl with 0.1 M NaOH

Theoretical Titration of 0.1 M HCl w ith 0.1 M NaOH

0

2

4

6

8

10

12

14

0 20 40 60 80

ml NaOH

pH pH

III. Titration of 25 ml 0.1 M NH4OH (weak base) with 0.1 M HCl (strong acid).

Before Titration: Initial pH of 0.1 M NH4OH

NH4OHNH4 +

+ OH-

Initial 0.1 M 0 0Change -x +x +xEquilibrium 0.1-x x x

Kb= 1.8x 10 -5= [NH4+] [OH-]/[NH4OH] Assume: 0.1 >>x

1.8x 10 -5 = x 2

X= 1.3 x 10 -3

pOH= -log (1.3 x 10 -3) = 2.89 pH= 14-2.89= 11.11

+ 5ml= 30 ml (solution) (before equivalence point)

Base mmol before titration= 0.1 mmol/ml x 25 ml= 2.5 mmolAdded acid mmol= 0.1mmol/ml x 5 ml= 0.5 mmol

NH4OH + H3O+ NH4 + + H2OInitial 2.5 mmol 0.5

mmol0

Change -0.5 mmol -0.5 mmol

+0.5 mmol

Equilibrium 2 mmol 0 0.5 mmol

pOH=pKb + log [acid]/[base] 4.75 + log9.25 + log (0.5mmol/30 ml)/ (2mmol/30 ml)=4.15pH= 14-4.15= 9.85

+25 ml= 50 ml (solution) (equivalence point)

Added acid mmol= 0.1 mmol/ml x 25= 2.5 mmoles

NH4OH + H3O+ NH4 + + H2OInitial 2.5 mmol 2.5

mmol0

Change -2.5 mmol -2.5 mmol

+2.5 mmol

Equilibrium 0 0 2.5 mmol

Hydrolysis of NH4 +

NH4 + + H2O NH3 + H3O+

Initial 0.05 M 0 0Change -x +x +xEquilibrium 0 x x

Ka= [NH4OH] [H3O+] /[ NH4

+] =5.6 x 10 -10

5.7 x 10 -10 = [H3O+]2 /(2.5mmol/50 ml)

5.6 x 10 -10 = x2/0.05-xx=5.29 x 10 -6

pH= -log (5.29 x 10 -6)pH= 5.27

+30 ml=55 ml (solution) (after equivalence point)

Added acid mmol= 0.1mmol/ml x 30 ml= 3 mmol

NH4OH + H3O+ NH4 + + H2OInitial 2.5 mmol 3 mmol 0Change -2.5 mmol -2.5

mmol+2.5 mmol

Equilibrium 0 0.5 mmol 2.5 mmol

pH= -log[H3O+]= -log (0.5 mmol/55ml)= 2.04

The pattern of the titration of a weak base-strong acid is quite similar with the titration of a weak acid-strong base, except that the pH of the solution decreases as the titration continues. When titration reaches the equivalence point, the cation of the weak base (NH4

+) will undergo hydrolysis, thus making the pH less than 7 (upon calculation, pH at equivalence point is 5.28). Similarly, the pH after the equivalence point will depend on the excess HCl present.

Theoretical Titration of 0.1 M NH4OH with 0.1 M HC

ml titrant pH

0 11.11

5 9.85

10 9.43

15 9.07

20 8.65

25 5.28

30 2.04

35 1.78

40 1.64

45 1.54

50 1.48

55 1.43

60 1.39

ml titrant pH0 9.91 9.642 9.493 9.284 9.125 8.986 8.787 8.568 8.269 7.7510 5.73

10.5 3.0811 2.66

11.5 2.4812 2.36

12.5 2.2813 2.21

13.5 2.1614 2.11

14.5 2.0715 2.04

15.5 2.0116 1.98

Titration of 0.1 M NH4OH with 0.1 M HC

Theoretical Titration of 0.1 M NH4OH w ith 0.1 M HCl

0

2

4

6

8

10

12

0 20 40 60 80

ml HCl

pH pH

Experimental Titration Curve of NH4OH w ith HCl

0

2

4

6

8

10

12

0 5 10 15 20

ml HCl

pH pH

Titration of Unknown with 0.1 M NaOH

Upon comparison with the theoretical titration curves, the unknown can be classified as a weak acid, due to the less steep change in pH as compared to the theoretical titration curve of a strong acid.

Concentration of unknown acid

[Concentration of NaOH] (Vol. NaOH)= [C. of unknown acid](Vol.at eq. point)

(0.1 M) (25 ml) = x [ (25 ml NaOH) + (27 ml unknown acid)] x= 0.05 M

When comparing all the theoretical titration curves with their respective experimental values, there is little difference in terms of the pattern the plot follows. The difference is accounted for by the different initial pH values before titration as compared to the theoretical pH values, which affects the equivalence point during titration.

In these titration curves, determining the pH at each step is essential. Before the titration reaches the equivalence point, the pH of the solution depends on the amount of the substance not yet neutralized by the titrant. After reaching the equivalence point, pH levels depend on the excess amount of the titrant in the solution. Determining the pH at equivalence point depends of the substances used in titration. For a strong acid-strong base titration, there is no need to compute for the pH level is always seven, because no cations or anions undergo hydrolysis, which will have an effect on its pH level. If one of the components is a weak acid or a weak base, the hydrolysis of the cation/anion needs to be taken into account

Sources:

Brown, Theodore L., Bursten, Bruce E., Lemay, H. Eugene Jr. (2000). Chemistry: The Central Science. 8th ed.

  Silberberg, Martin S. (2007). Principles of

General Chemistry. McGraw-Hill Intl. ed.