UNIVERSITI TEKNOLOGI MARAFAKULTI KEJURUTERAAN KIMIA

ENGINEERING CHEMISTRY LABORATORY(CHE485)

No. Title Allocated Marks (%) Marks

1 Abstract/Summary 52 Introduction 103 Aims 54 Theory 105 Apparatus 56 Methodology/Procedure 107 Results 108 Calculations 109 Discussion 2010 Conclusion 511 Recommendations 512 Reference / Appendix 5

TOTAL MARKS 100

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Date :

NAME : NUR FARHANA BINTI JELENSTUDENT NO. : 2014824828GROUP : EH 241 2BEXPERIMENT : DETERMINATION OF ACETIC ACID IN VINEGARDATE PERFORMED : 27/03/2015SEMESTER : 2PROGRAMME / CODE : BACHELOR OF CHEMICAL ENGINEERING & PROCESSSUBMIT TO : EN. MOHD AIZAD AHMAD

Abstract Unknown acid solution is an organic compound that is in form of colourless solution and

classified as a weak acid. In this experiment, the acid ionization constant Ka, of weak acid

solutions is determined by using titration with standardization of sodium hydroxide and by

measuring the pH of the weak acids. The experiment is set up by measure 10 mL of two

unknown acid. This, two unknown acid solution then titrated with sodium, hydroxide

solution. Both titrations for two unknown acid solution are repeated to perform second trial to

get more accurate result. Another method is by measuring the pH of the weak acid by using

pH meter. The pH meter then is put into each of the unknown acid solution until the pH

reading is constant. Based on the results, it can be conclude that, the larger the Ka value, the

stronger the acid.

Introduction

Acid and bases can be described as ‘weak acid and base’ or ‘strong acid and base’. Acid

ionization constant values are usually used for acid. Weak acid interact with water to produce

concentrations of ions. The degree of ionization is described by the equilibrium constant Ka,

HA(aq) + H2O(l) ↔ H3O+(aq) + A- (aq)

Since an equilibrium exist, an equilibrium constant Ka can be written as :

Ka = [H3O+][A-] / [HA]

The Ka value is an indication of acid strength. The larger the value of Ka, the stronger the

acid. This value is characteristic of the acid and can be used to identify an unknown acid. A

similar system exist for bases is Kb.

Theory

We will be using the two methods for determining the Ka values of two unknown acid

solution. A solution of unknown acid is prepared and then titrated with standardization of

sodium hydroxide solution. Titration is a technique where a solution known concentration is

used to determine the concentration of unknown solution. Typically, the titrant is added from

a burette to a known quantity of the unknown solution until the reaction is complete.

Knowing the volume of titrant added allows the determination of the concentration of the

unknown. Usually, indicator is used as a signal end of the reaction called end point. The other

method is by measuring the pH of the unknown acid solution. It is a suitable method because

we can easily get through the pH reading of the two unknown acid solution.

Purpose

To determine the ionization constant Ka, for weak acid by using two method which is

titration with standardization of sodium hydroxide solution and the other method is by

measuring the pH of the weak acid.

Apparatus and Material

pH meter, 10 mL of two unknown acid solution, burette, standard base solution (0.10 M

NaOH), beakers, distilled water, dropper, stirrer bar, retort stands and clamps.

Procedure

1. Firstly, 10 mL of two unknown acid solution is put into each beaker. Then, the two

unknown acid solution are titrated with standardization of sodium hydroxide solution.

The pH reading is recorded for every 0.5 mL of sodium hydroxide solution needed to

neutralize the acid. The equivalence point is reached when pH reading are constant.

2. The experiment is repeated to perform second trial to get an accurate result.

3. Based on the result, a graph of pH against volume of sodium hydroxide needed to

neutralize the acid solution is plotted. From the graph, we can determine the Ka of the

two unknown acid solution.

4. Then, the experiment is proceed to a second method which is measuring the pH of

weak acid. The pH meter is put into two unknown acid solution until the pH reading

become constant. The constant pH is recorded.

Result

First method : titration with standardization of sodium hydroxide solution

Unknown acid 1

Titration 1 Titration 2

Volume of sodium hydroxide

to neutralize the unknown

acid, mL

11.50 mL 12.00 mL

pH values 12.24 12.24

Unknown acid 2

Titration 1 Titration 2

Volume of sodium hydroxide

to neutralize the unknown

acid, mL

12.50 mL 12.50 mL

pH value 12.24 12.20

Second method : measuring the pH

Unknown acid 1

Volume of unknown acid , mL 10 mL

pH value 4.17

Unknown acid 2

Volume of unknown acid , mL 10 mL

pH value 3.57

Calculations

i) First method

Unknown acid 1

First trial

Half equilibrium point = 4.20

pH = - log PKa

pH = - log Ka

4.2 = - log Ka

Ka = 10-4.2

= 6.309 x 10-5

Second trial

Half equilibrium point = 4.18

pH = - log PKa

pH = - log Ka

4.18 = - log Ka

Ka = 10-4.18

= 6.6069 x 10-5

Average Ka of uknown acid 1 = 6.45 x 10-5

Calculate molarity :

First trial

VNaOH = 11.50 mL , MNaOH = 0.1 M

Vunknown acid 1 = 10 mL , Munknown acid 1 = ?

nNaOH = MV1000 , nNaOH =

(0.1M )(11.50mL)1000

, nNaOH = 1.15 x 10-3 mol

Assume ;

1 mol NaOH ≡ I mol unknown acid

1.15 x 10-3 mol NaOH ≡ 1.15 x 10-3 mol unknown acid

From the assumption, the molarity of unknown acid 1 is ;

M = 1000nV

, M = 1000(1.15 x10−3)

10 , M = 0.115 M

Second trial

VNaOH = 12.00 mL , MNaOH = 0.1 M

Vunknown acid 1 = 10 mL , Munknown acid 1 = ?

nNaOH = MV1000 , nNaOH =

(0.1M )(12.00mL)1000

, nNaOH = 1.2 x 10-3 mol

Assume ;

1 mol NaOH ≡ I mol unknown acid

1.2 x 10-3 mol NaOH ≡ 1.2 x 10-3 mol unknown acid

From the assumption, the molarity of unknown acid 1 is ;

M = 1000nV

, M = 1000(1. 2x 10−3)

10 , M = 0.12 M

Average molarity of unknown acid 1 = 0.12 M

Unknown acid 2

First trial

pH = 7

Half equilibrium point = 4.75

pH = - log PKa

pH = - log Ka

4.75 = - log Ka

Ka = 10-4.75

= 1.778 x 10-5

Second trial

Half equilibrium point = 4.73

pH = - log PKa

pH = - log Ka

4.73 = - log Ka

Ka = 10-4.73

= 1.862 x 10-5

Average Ka of uknown acid 1 = 1.8200 x 10-5

Calculate molarity :

First trial

VNaOH = 12.50 mL , MNaOH = 0.1 M

Vunknown acid 1 = 10 mL , Munknown acid 1 = ?

nNaOH = MV1000 , nNaOH =

(0.1M )(12 .50mL)1000

, nNaOH = 1.25 x 10-3 mol

Assume ;

1 mol NaOH ≡ I mol unknown acid

1.25 x 10-3 mol NaOH ≡ 1.25 x 10-3 mol unknown acid

From the assumption, the molarity of unknown acid 1 is ;

M = 1000nV

, M = 1000(1.2 5x 10−3)

10 , M = 0.125 M

Second trial

VNaOH = 12.50 mL , MNaOH = 0.1 M

Vunknown acid 1 = 10 mL , Munknown acid 1 = ?

nNaOH = MV1000 , nNaOH =

(0.1M )(12.5 0mL )1000

, nNaOH = 1.25 x 10-3 mol

Assume ;

1 mol NaOH ≡ I mol unknown acid

1.25 x 10-3 mol NaOH ≡ 1.25 x 10-3 mol unknown acid

From the assumption, the molarity of unknown acid 1 is ;

M = 1000nV

, M = 1000(1. 25 x10−3)

10 , M = 0.125 M

Average molarity of unknown acid 1 = 0.125 M

ii) Second method

Unknown acid 1