TABLE OF CONTENT

No.TitlePage

1Abstract2

2Introduction2

3Objectives3

4Theory3

5Apparatus4

6Procedure4

7Results5 - 13

8Calculations14 - 16

9Discussion17

10Conclusion18

11Recommendations18

12References19

13Appendices19

ABSTRACT

Acetic acid, CH3COOH is an organic compound that is in form of colorless solution and classified as a weak acid. Acetic acid is the main component of vinegar apart from water. In this experiment, the molarity of a solution and the percent by mass of acetic acid in vinegar is determined by using titration with a standardized sodium hydroxide, NaOH solution. The experiment is divided into two parts which are standardizing the NaOH solution is the first part and the second part is proceeded with the determining the molarity of a solution and the percent by mass of acetic acid in vinegar. In standardizing the NaOH solution, 250 mL of distilled water is used to dilute approximately 6 g of NaOH solid in order to prepare 0.6 M NaOH solution. This NaOH solution is then titrated with potassium hydrogen phthalate, KHC8H4O4 or KHP solution which has been prepared by diluting 1.5 g of KHP granules in 30mL of distilled water. The experiment is then preceded to the second part of the experiment which standardized NaOH solution is titrated the with 10 mL vinegar that has been diluted with 100 mL of distilled water. Both titration for part 1 and 2 are repeated thrice to get more accurate results. Based on results, it can be conclude that the greater the mass of solute in the acid solution, the more concentrated the solution becomes thus, the higher the molarity and more volume of NaOH needed to neutralize the acid. The experiment is completed and successfully conducted.

INTRODUCTION

Concentration of solution is the amount of solute in a given amount of solvent. A concentrated solution contains a relatively large quantity of solute in a given amount of solvent. Dilute solutions contains a relatively little solute in a given amount of solvent. There are two specific terms toexpress concentration, namely molarity and percent by mass:

Molarity is the number of moles of solute per litre ofsolution.

Molarity (M) = _moles of solute__ Litre of solution(Equation 21)Percent by mass is the mass in grams of solute per 100 grams of solutionPercent solute = _grams of solute_ X 100% Grams of solution (Equation 21)Vinegar is a dilute solution of acetic acid. The molecular formula for acetic acid is CH3COOH. Both molarity and percent by mass of acetic acid in the vinegar solution can be determined by performing a titration. A titration is a process in which small increments of a solution of known concentration are added to a specific volume of a solution of unknown concentration until the stoichiometry for that reaction is attained. Knowing the quantity of the known solution required to complete thetitration enables calculation of the unknown solution concentration. The purpose of titration is to determine the equivalence point of the reaction. The equivalence point is reach when theadded quantity of one reactant isthe exact amount necessary for stoichiometric reaction with anotherreactant.

OBJECTIVES

This experiment is conducted to determine the molarity of a solution and the percent by mass of acetic acid in vinegar by titration with a standardized sodium hydroxide solution. Furthermore, the students will know how to handle materials, apparatus and the correct ways to conduct the experiment as well as being careful to avoid experimental error. Moreover, this experiment also teach the students on the correct steps to calculate percentage of substance, molarity of solution, and other calculations involve throughout the experiment.

THEORY

In the titration process, a burette is used to dispense a small, quantifiable increment of solution of known concentration. In this experiment, the equivalence point occurs when the moles of acid in the solution equals the moles of base added in the titration. For example, the stoichiometric amount of 1 mole of the strong base, sodium hydroxide (NaOH), is necessary to neutralize 1 mole of the weak acid, acetic acid (CH3COOH).

The sudden change in the solution pH shows that the titration has reached the equivalence point. pH is an aqueous solution is related to its hydrogen ion concentration. pH scale is a method of expressing the acidity or basicity of a solution. Solution with pH 7 are basic. The titration is initiated by inserting a pH electrode into a beaker containing the acid solution. As NaOH, is incrementally added to the acid solution, some of the hydrogen ion will be neutralized. As the hydrogen ions concentration decreases, the pH of the solution will gradually increase. When sufficient NaOH is added to completely neutralize the acid, the next drop of NaOH added will cause a sudden sharp increase in pH. The volume of based required to completely neutralized the acid is determined at the equivalence point of titration.

In the first set of titration, a procedure called standardization is conducted to determine the concentration of the NaOH solution, which is difficult to prepare an accurate concentration. In this experiment, titration of a vinegar sample with a standardized NaOH solution will be performed. To standardize the NaOH solution, a primary standard acid solution is initially prepared. In general, primary standard solutions are produce by dissolving a weighed quantity oa pure acid or base in a known volume of a solution. Primary standard acid or bases have several common characteristics : They must be available in at least 99.9 purity They must have high molar mass to minimize error in weighing They must be stable upon heating They must soluble in the solvent of interest

Potassium hydrogen phthalate, KHP and oxalic acid are common primary standard acids. Sodium carbonate is the commonly used base. Most acids and bases are mostly available in primary standard form. To standardized one of these acidic or basic solutions, titration of the solution with a primary standard should be performed. In this experiment, NaOH solution will be titrated with KHP. Once the NaOH solution has been standardized it will be titrated with 10.00 mL aliquots of vinegar. Knowing the standardized NaOH concentration, we can determine the molarity and percent by mass of acetic acid in the vinegar solution.

APPARATUS AND MATERIALS

pH meterNaOH solutionanalytical balance vinegarburette clamp potassium hydrogen phthalate (KHP)beaker (100, 250 mL) droppersfilter funnel glass rod pipette 10mLspatula distilled water burette

PROCEDUREPART A : STANDARDIZATION OF SODIUM HYDROXIDE SOLUTION1. 250mL of approximately 0.6M sodium hydroxide solution was prepared from NaOH solid. The solution was prepared in a beaker; the calculation was checked with the laboratory instructor to prepare the solution.2. A beaker was placed on the balance and tare. 1.5g of KHP was added to the beaker. The mass of KHP was recorded to the nearest 0.001g. 30mL of distilled water was added to the beaker and the solution was stirred until the KHP dissolved completely.3. This solution was titrated with NaOH and the pH with 1mL additions of NaOH solution was recorded.4. Steps 1 to 3 was repeated and two more solutions for NaOH standardization was prepared.5. The graph oh pH versus NaOH was plotted and the volume of of NaOH required to neutralize the KHP solution in each titration was determined from the graph.6. The molarity of sodium hydroxide for titrations 1, 2, and 3 was calculated.7. The average molarity of the sodium hydroxide solution was calculated.

PART B : MOLARITY OF ACETIC ACID AND MASS PERCENT IN VINEGAR

1. 10mL of vinegar was transferred to a clean, dry 250mL beaker using a 10mL volumetric pipette. 75 to 100mL of water was added into the beaker to cover the pH electrode tip during titration.2. 1mL of NaOH was added to the vinegar solution and the pH was recorded.3. The above steps was repeated twice more.4. The graph of pH versus NaOH volume added was plotted and the volume of NaOH required to neutralize the vinegar in each titration was determined from the graph. 5. The molarity of acetic acid in vinegar for titrations 1, 2, and 3 was calculated.6. The average molarity of acetic acid for each titration was calculated.7. The percent by mass of acetic acid in vinegar for titrations 1, 2, and 3 was calculated.8. The percent by mass of acetic acid in vinegar was calculated.RESULTSPART A : STANDARDIZATION OF SODIUM HYDROXIDE SOLUTIONMass of potassium acid phthalate used : 1.500gTitration 1 :

Based on graph, molarity of NaOH solution for titration 1 = 0.3498 MVolume of NaOH (mL)PHVolume of NaOH (mL)pH

15.10136.22

25.28146.30

35.38156.38

45.48166.50

55.56176.62

65.66186.76

75.74196.96

85.82207.28

95.90218.96

105.982212.66

116.062312.92

126.142413.10

Titration 2 :Based on graph, molarity of NaOH solution for titration 2 = 0.5665 MVolume of NaOH (mL)PH

15.17

25.34

35.49

45.62

55.76

65.90

76.01

86.16

96.27

106.45

116.65

126.98

138.07

1413.02

Titration 3:

Titration 3 :

Based on graph, molarity of NaOH solution for titration 3 = 0.4329 MVolume of NaOH (mL)PH

15.18

25.30

35.42

45.58

55.70

65.87

75.96

86.12

96.24

106.40

116.48

126.60

136.72

146.84

156.96

167.20

178.08

1812.98

1913.08

2013.30

Overall results :Titration 1

Titration 2Titration 3

Mass of KHP (g)1.500

1.5041.503

Volume of NaOH used to neutralize the KHP solution (mL)21.0013.0017.00

PART B : MOLARITY OF ACETIC ACID AND MASS PERCENT IN VINEGAR Mass of acetic acid solution : 10.00gTitration 1 :Graph of pH versus volume of NaOH for titration 1

Based on the graph ;Molarity of acetic acid in vinegar solution for titration 1 = 1.26 MPercent of acetic acid in vinegar solution for titration 1 = 7.57%

Volume of NaOH (mL)PHVolume of NaOH (mL)pH

13.10194.80

23.36204.86

33.63214.94

43.70225.01

53.84235.12

63.93245.20

74.00255.26

84.09265.44

94.16275.59

104.23285.86

114.30296.90

124.373011.01

134.423111.53

144.493211.72

154.563311.88

164.613411.97

174.683512.03

184.74

Titration 2 :

Based on the graph ;Molarity of acetic acid in vinegar solution for titration 2 = 1.2465 MPercent of acetic acid in vinegar solution for titration 2 = 7.49%

Volume of NaOH (mL)PHVolume of NaOH (mL)pH

13.25174.75

23.44184.84

33.63194.92

43.75205.06

53.86215.13

63.96225.18

74.05235.31

84.13245.43

94.23255.61

104.29265.89

114.36277.05

124.43288.50

134.502911.52

144.563011.74

154.623111.88

164.713211.97

Titration 3 :

Based on the graph ;Molarity of acetic acid in vinegar solution for titration 3 = 1.206 MPercent of acetic acid in vinegar solution for titration 3 = 7.24%

Volume of NaOH (mL)PHVolume of NaOH (mL)pH

13.27174.89

23.70184.98

33.78195.04

43.82205.15

54.00215.23

64.06225.36

74.12235.58

84.17245.86

94.31257.70

104.37268.20

114.46278.90

124.522811.13

134.592911.48

144.673011.75

154.743112.00

164.823212.04

Overall result :Titration 1Titration 2Titration 3

Volume of NaOH used to neutralize the vinegar solution (mL)28.0027.7026.80

CALCULATIONSPART A : STANDARDIZATION OF SODIUM HYDROXIDE SOLUTION No. of moles of NaOH = 0.25(0.6) = 0.15 mol NaOHMolecular weight of NaOH = 23+16+1 = 40g/molMass of NaOH = No. of moles NaOH x Molecular weight of NaOH = 0.15g x 40g/molMass of NaOH = 6.0369g Titration 1 :No. of moles of KHP = _______ ____Mass of KHP _(g)_____________ = _____1.5 g______ = 0.007346 mol of KHP Molecular weight of KHP(g/mol) 204.1 g/mol1 mol of KHP = 1 mol of NaOH0.00746 mol of KHP = 0.00746 mol of NaOHMolarity of NaOH (M) = _moles of NaOH_(mol) = 0.007346 mol = 0.3498 M of NaOH Litre of solution(L) 0.021 LTitration 2 :No. of moles of KHP = _______ ____Mass of KHP _(g)_____________ = _____1.504 g______ = 0.007365 mol of KHP Molecular weight of KHP(g/mol) 204.1 g/mol1 mol of KHP = 1 mol of NaOH0.007365 mol of KHP = 0.007365 mol of NaOHMolarity of NaOH (M) = _moles of NaOH_(mol) = 0.007365 mol = 0.5665 M of NaOH Litre of solution(L) 0.013 LTitration 3 :No. of moles of KHP = _______ ____Mass of KHP _(g)_____________ = _____1.503 g______ = 0.007360 mol of KHP Molecular weight of KHP(g/mol) 204.1 g/mol1 mol of KHP = 1 mol of NaOH0.007360 mol of KHP = 0.007360 mol of NaOHMolarity of NaOH (M) = _moles of NaOH_(mol) = 0.007360 mol = 0.4329 M of NaOH Litre of solution(L) 0.017 LAverage molarity of NaOH = 0.3498M+0.5665M+0.4329M = 0.4497 M of NaOH 3

PART B : MOLARITY OF ACETIC ACID AND MASS OERCENT IN VINEGAR Molarity of standard NaOH : 0.45 MTitration 1 :No. of moles of NaOH = 0.028 L NaOH x 0.45 mol NaOH = 0.0126 mol NaOH 1 L NaOH solutionNo. of moles of CH3COOH = 0.0126 mol NaOH x 1 mol CH3COOH = 0.0126 mol CH3COOH 1mol NaOH Molarity of CH3COOH = 0.0126mol CH3COOH = 1.26 M of CH3COOH0.01 L solution Mass of the acetic acid in the solution :0.01 L CH3COOH x 1.26 mol CH3COOH x 60.06 g of CH3COOH = 0.7568 g of CH3COOH 1 L NaOH solution 1 mol CH3COOHMass of acetic acid solution :10 mL CH3COOH solution x 1 g CH3COOH solution = 10.00 g of CH3COOH solution 1mL CH3COOH solution Percent by mass of acetic acid in the solution :Percent mass CH3COOH = g CH3COOH x 100% = 0.7568g CH3COOH x 100% = 7.57% CH3COOH g CH3COOH solution 10.00 g CH3COOHTitration 2 :No. of moles of NaOH = 0.0277 L NaOH x 0.45 mol NaOH = 0.012465 mol NaOH 1 L NaOH solutionNo. of moles of CH3COOH = 0.012465 mol NaOH x 1 mol CH3COOH = 0.012465 mol CH3COOH 1mol NaOH Molarity of CH3COOH = 0.012465 mol CH3COOH = 1.2465 M of CH3COOH 0.01L solution Mass of the acetic acid in the solution :0.01 L CH3COOH x 1.2465 mol CH3COOH x 60.06 g of CH3COOH = 0.74986 g of CH3COOH 1 L NaOH solution 1 mol CH3COOHMass of acetic acid solution :10 mL CH3COOH solution x 1 g CH3COOH solution = 10.00 g of CH3COOH solution 1mL CH3COOH solution Percent by mass of acetic acid in the solution :Percent mass CH3COOH = g CH3COOH x 100% = 0.7486 g CH3COOH x 100% = 7.49% CH3COOH g CH3COOH solution 10.00 g CH3COOHTitration 3 :No. of moles of NaOH = 0.0268 L NaOH x 0.45 mol NaOH = 0.01206 mol NaOH 1 L NaOH solutionNo. of moles of CH3COOH = 0.01206 mol NaOH x 1 mol CH3COOH = 0.01206 mol CH3COOH 1mol NaOH Molarity of CH3COOH = 0.01206 mol CH3COOH = 1.206 M of CH3COOH 0.01L solution Mass of the acetic acid in the solution :0.01 L CH3COOH x 1.206 mol CH3COOH x 60.06 g of CH3COOH = 0.7243 g of CH3COOH 1 L NaOH solution 1 mol CH3COOHMass of acetic acid solution :10 mL CH3COOH solution x 1 g CH3COOH solution = 10.00 g of CH3COOH solution 1mL CH3COOH solution Percent by mass of acetic acid in the solution :Percent mass CH3COOH = g CH3COOH x 100% = 0.7243 g CH3COOH x 100% = 7.24% CH3COOH g CH3COOH solution 10.00 g CH3COOH

Average molarity of acetic acid = 1.260 M+1.2465 M+1.206 M = 1.2375 M 3Average percent by mass of acetic acid in vinegar = 7.57% + 7.49% + 7.24% = 7.43% 3

Discussion The main objective of this experiment is to determine the concentration as well as the percentage by mass of acetic acid in vinegar. This experiment was conducted by using titration method of acetic acid with a standardized sodium hydroxide solution. The acid and base titration used the Arrhenius theory. This theory stated that acid are substance which produce hydrogen ions in solution and bases are substance which produce hydroxide ion in solution. In preparing the NaOH solution, we prepared the sodium hydroxide solution by adding NaOH solid with distilled water to be used in the experiment. In order to make sure that the NaOH solid dissolved completely, we have to shake the flask thoroughly. Some chemicals can be purchased ina pure form and remain pure over a long period of time. However, sodium hydroxide absorbs moisture from the air and often appears wet, thus it is easily contaminated. Thus if a solution of NaOH is prepared by weighing the NaOH solid, the concentration of the solution may not be precisely the intended concentration. KHP on the other hand, has a lesser tendency to absorb water from the air and when dried will remain dry for a reasonable period of time. KHP is a primary standard. This means that carefully prepared solutions of known concentration of KHP may be used to determine, by titration, the concentration of another solution such as NaOH.In part A, we have to conduct the standardization of NaOH solution by using titration. In this part, we need to prepare the KHP solution by adding distilled water to the KHP. Since the KHP need to dissolve completely, so we have to swirl the flask thoroughly. Based on graph and calculated result, it is shown that the average volume of NaOH needed to neutralize the primary standard acid is 17.00 mL . The solution starts to neutralize from pH 5.15 up to 13.24. this is because some of the hydrogen ions are gradually neutralized with the increment volume of NaOH. Thus, a sudden sharp increase in pH occurred as sufficient volume of NaOH is added into the acid solution. Furthermore, the pH at the endpoint of a weak acid-strong base titration is always greater than 7 because strong base allows hydrogen ions in weak acid to neutralize more easily. In part B, we have to analyse of a vinegar solution. Vinegar solution is a weak acid and it contains acetic acid. Hence, when reacted with NaOH whish is a strong base it will produce a basic solution. NaOH will ionize completely in the water to form high concentration of hydroxide ion. Meanwhile, acetic acid is a weak acid. It will dissociate partially in the water to form low concentration of hydrogen ions. The molarity of acetic acid and the mass percent in vinegar are calculated by using the average volume of NaOH resulted from the first part of experiment and with the help of graph plotted based on result from the second part of exeperiment. The average molarity of acetic acid in vinegar is 1.2375 M and the average percent by mass of acetic acid in vinegar is 7.43 % . Based on the graph plotted, there is a sharp increase in pH, from approximately pH 5.5 to pH 11.5 at the equivalence point. This sharp increase is due to the excess of around one drop of NaOH added from the burette. So, it is important to dilute the vinegar in order to avoid a very small yitre, which would reduce the accuracy of the experiment.The significance of percent by mass and molarity of solution in this experiment is that it tells whether the solution is either diluted or concentrated solution. Hence, the acetic acid in the vinegar is a dilute solution as its percent by mass and molarity are relatively small.

CONCLUSIONThe results in part A of this experiment showed that when the mass oh KHP is 1.500g, volume of NaOH required to neutralize the acid is 21.00 mL and the molarity of NaOH solution for titration 1 is 0.35 M. While for titration 2, when the mass of KHP is 1.504g, volume of NaOH required to neutralize the acid is 13.00 mL and the molarity of NaOH solution for titration 2 is 0.57 M. for titration 3, when the mass of KHP is 1.503g, volume of NaOH required to neutralize the acid is 17.00 mL and the molarity of NaOH solution for titration 3 is 0.43 M. So, the average molarity ofsodium hydroxide required to neutralize the acid is 0.45 M. For part B, the molarity of acetic acid in vinegar solution for titration is 1.26 M, the percent of acetic acid in vinegar solution for titration 1 is 7.57 % and the volume of NaOH required to neutralize the solution is 28.00 mL . For titration 2, molarity of acetic acid in vinegar solution is 1.2465 M, percent of acetic acid in vinegar is 7.49 % and the volume of NaOH required to neutralize the solution is 27.7 mL . And for titration 3, molarity of acetic acid in vinegar solution is 1.206 M, percent of acetic acid in vinegar is 7.24 % and volume of NaOH required to neutralize the solution is 26.8 mL . So the average molarity of acetic acid in vinegar solution is 1.2375 M, average percent of acetic acid in vinegar is 7.43 % and the average volume of NaOH required to neutralize the solution is 27.5 mL.Thus, it can be concluded that the greater the mass of solute in the acid solution, the more concentrated the solution becomes. Hence, the higher the molarity and more volume of NaOH needed to neutralize the acid.

RECOMMENDATIONThere are few recommendations and precautions that have to be considered during conducting the experiment in order to get an accurate value and readings of data.Firstly, ensure that the tip of burette is filled with NaOH so that no air bubbles are present in the tip because air bubbles can cause error in the measurement of the concentration because the actual volume of unknown will be less.Secondly, it is better to use an indicator such as phenolphthalein as it color will change at a pH of ~9.0 rather than using pH meter as using a pH meter is fairly tedious and it will turns out unnecessary.Ensure that the position of eye is directly perpendicular to the meniscus when reading the volume of solution to avoid inaccuracy. Besides that, the swirling of the solution should be constant while adding the NaOH in order to ensure that the NaOH is totally dispersed.It is better to carry out three accurate titration so that the experimental error is reduced by calculating the average value.Lastly, the reaction between acid and base should be rapid and without any side reactions. Thus, if there are any interfering substances, they should be removed.

REFERENCES1. Rabiatul Adawiyah Abdol Aziz ; Laboratory Manual . Universiti Teknologi Mara : Engineering Chemistry Laboratory.2. Tan,Y.T., Ashy Kumren. (2011) Chemistry for Matriculation. Selangor, Malaysia : Oxford Fajar.3. www.chemguide.co.uk/physical/acidbaseeqia/theories.html.4. http://alpha.chem.umb.edu/chemistry/ch313/Experiment%204%20vinegar.pdf

APPENDICES

Sample figure of equipments used in this experiment.

pH meter

Analytical balance

1