TABLE OF CONTENT

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1Table of content1

2Summary2

3Introduction3

4Objectives3

5Theory4

6Apparatus6

7Procedures8

8Results9

99Sample of calculations14

10Discussions18

11Conclusions19

12Recommendations20

13References20

14Appendices21

SUMMARY

Acetic acid is classified as weak acid and is the main component of vinegar apart from water. The molarity of a solution and the percent by mass of acetic acid can be determined in this experiment by using the process of titration with sodium hydroxide solution, NaOH. The experiment is conducted by standardizing NaOH first and is preceded by determining the molarity of acetic acid and percent of vinegar. The first part which is the process of standardizing NaOH solution is first conducted by diluting 6g of NaOH solids with 250 ml of distilled water to produce 0.6 M of NaOH solution. This NaOH solution is then titrated with potassium hydrogen phthalate, KHP which had been diluted with 1.5g of KHP solids and 30 ml of distilled water. The second part which is the step to determine the molarity of acetic acid and percent of vinegar is conducted by titrating 10 ml of vinegar with 100 ml of water with the previously prepared NaOH solution. Both of these two processes are repeated twice. Based on the result of this experiment, it can be said that the greater the mass of solute in the acid solution, the more concentrated the solution become thus, the higher the molarity and more volume of NaOH needed to neutralized the acid.

INTRODUCTION

Concentration is the amount of solute in a given amount of solvent. A relatively large amount of solute in a given amount of solvent is known as concentrated solution whereas a relatively small amount of solute in a given amount of solvent is known as dilute solution. The specific terms that are used to refer to concentration are either molarity or percent by mass. Both molarity and percent by mass can be calculated by using the formula below.

Molarity (M) = moles of solute litre of solution (Equation 1 1)

Percent by mass = (grams of solute grams of solution) x 100% (Equation 1 2)

Vinegar with the molecular formula of CH3COOH is a dilute solution of acetic acid. The concentration of acetic acid in the vinegar can be determined through process of titration. A titration is a technique where a solution of known concentration is used to determine the concentration of an unknown solution. Typically the titrant (the known solution) is added from a burette to a known quantity of the analyte (the unknown solution) until the reaction is complete which is when the stoichiometry of that reaction is attained. The purpose of titration is to determine the equivalent point of reaction.

OBJECTIVESThe aim of the experiment is to determine the molarity of a solution and the percent by mass of acetic acid in vinegar by titration with the standardized sodium hydroxide solution.

THEORYA burette is always used to dispense a small, quantifiable increment solution of known concentration in the titration process. The volume dispensed from the burette should be estimated to the nearest 0.001 mL. When the moles of acid in the solution equals the moles of base added in the titration, the equivalent point occurs. For example, the stoichiometric amount of 1 mole of strong base, sodium hydroxide (NaOH) is necessary to neutralize 1 mole of weak acid, acetic acid (CH3CO2H), as indicated in equation 1 3:

NaOH (aq) + CH3CO2H (aq) NaCH3CO2 (aq) + H2O (l) (Equation 1 3)

When the titration has reached the equivalence point there would be a sudden change in the pH of the solution. pH in an aqueous solution is related to its hydrogen in concentration. Symbolically, the hydrogen ion concentration is written as [H3O+]. pH is defined as the negative of the logarithm of the hydrogen ion concentration.

pH = log10 [H3O+] (Equation 1 4)

The method of expressing the acidity or basicity of a solution is by using pH scale. Solutions with pH < 7 are acidic, pH = 7 are neutral, and pH > 7 are basic For example,, a solution having an H3O+ concentration of 2.35 x 102 M would have a pH of 1.629 and is acidic. pH electrodes will be used in this experiment in order to read the pH of the solution. pH electrode will be inserted into a beaker containing acid solution that will be added with NaOH. The pH reading will increase as the hydrogen ions will be neutralized when NaOH solution is added. The solution is said to be neutralized when sufficient amount of NaOH is added to remove most of H3O+ in the solution. In this experiment, titration of a vinegar sample with standardized sodium hydroxide solution will be performed. A primary standard acid solution is initially prepared. Primary standard solutions are produce by dissolving a weighed quantity of pure acid or base in a known volume of solution. Primary standard acid or bases have several common characteristics:

They must be available in at least 99.9 purity They must have a high molar mass to minimize error in weighing They must be stable upon heating They must be soluble in the solvent of interest

Potassium hydrogen phthalate, KHC8H4O4, and oxalic acid, (COOH)2, are common primary standard acids. Sodium carbonate, Na2CO3, is the most commonly used base. Most acids and bases (e.g. HCL, CH3COOH, NaOH, and KOH) are mostly available in primary standard form. Titration of the solution with a primary standard should be performed in order to standardize one of these acidic or basic solutions. In this experiment, NaOH solution will be titrated with potassium hydrogen phathalate (KHP). The reaction equation for this is:

KHC8H4O4 (aq) + NaOH (aq) KNaC8H4O4 (aq) + H2O (l) (Equation 1 5)

Once the sodium solution has been standardized it will be titrated with 10.00 mL aliquots of vinegar. The reaction equation for vinegar with NaOH is:CH3COOH (aq) + NaOH (aq) NaCHCOO (aq) + H2O (l) (Equation 1 6)

Knowing the standardized NaOH concentration and using Equation 1 6, the molarity and percent by mass of acetic acid in the vinegar solution can be determined.

APPARATUS

3) pH meter to read the ph of the solution 1) Retort stand to hold burette

4) Burette - to place the NaOH solution for titration 2) Beaker to place the solutions and chemicals that are used

7) Stirrer to stir the acid solution that is added with NaOH during titration processbeaker.

5) Weighing machine to weigh the chemicals in solid form or materials used

Chemicals: Sodium hydroxide solid (NaOH), potassium hydrogen phthalate granules (KHC8H4O4), vinegar, distilled water6) Measuring cylinder - to measure and transfer the right amount of solutions needed from its actual container into the beaker.

PROCEDURES

Standardization of Sodium Hydroxide Solution1) 250 mL of approximately 0.6 M NaOH solution was prepared in a beaker from NaOH solid2) A 250 mL of beaker was weighed and the mass was recorded to the nearest 0.001g. 1.5g of KHP was then added to the beaker. The beaker that contained KHP was then weighed and the mass was again recorded to the nearest 0.001g. 300 mL of distilled water was added to the beaker. The solution was stirred until the KHP solution had completely dissolved.3) This solution was then titrated with the previously prepared NaOH. The pH of the solution was read and recorded with every 1 mL addition of NaOH.4) Steps 1- 3 was repeated to perform another trial to standardize the NaOH solution5) The graph of pH versus NaOH was plotted. The volume of NaOH required to neutralize the KHP solution in each titration was determined from the plots of the graph.6) The molarity of NaOH for titration 1 & 2 was calculated7) The average molarity of NaOH for titration 1 & 2 was also calculated. The resulting NaOH concentration was used in part B of the experiment.Molarity of Acetic Acid and Percent of Vinegar1) 10 mL of vinegar was transfered to a clean, dry 250 mL of beaker. Sufficient water was added (100mL) to cover the pH electrode tip during titration.2) For every 1 mL titration of NaOH into vinegar, the pH value was read and recorded3) The steps above were repeated.4) The graph of pH vs volume of NaOH added was plotted and the volume of NaOH required to neutralize the vinegar in each titration was determined from these plots. The data was recorded.5) The molarity and percent by mass of acetic acid in vinegar for titration 1 and 2 was calculated.6) The average molarity and average percent by mass of acetic acid for each titration was also calculated.

RESULTSStandardization of sodium hydroxide solution

Results for titration1 and 2Titration 1Titration 2Titration 3

Volume (mL)pHpHpH

04.084.114.08

14.374.294.29

24.584.404.46

34.664.634.65

44.834.774.94

54.944.895.01

65.085.065.14

75.195.175.18

85.345.245.28

95.535.495.41

105.685.735.73

116.126.116.02

1210.9311.066.78

1310.9611.8711.52

1412.0612.1312.10

1512.1812.2412.19

1612.2012.3312.26

1712.2612.3712.30

11.18 mLGraph 1-1 Graph of pH versus volume of NaOH for titration 1

11.18 mLGraph 1-2 Graph of pH versus volume of NaOH for titration 2

12.04 mLGraph 1-3 Graph of pH versus volume of NaOH for titration 3

Titration 1Titration 2Titration 3

Mass of beaker (g)96.68399.35396.262

Mass of beaker + KHP (g)98.186100.85597.762

Mass of KHP1.5031.5021.500

Volume of NaOH to neutralize the KHP solution (mL)11.1811.1812.04

Molarity of NaOH solution0.6583 M0.6583 M0.6105 M

Average molarity of NaOH

0.6424

Molarity of Acetic Acid and Percent of VinegarTitration 1Titration 2

Volume (mL)pHpH

02.732.7

23.293.29

43.563.62

63.833.82

83.943.96

104.114.11

124.254.22

144.354.33

164.484.46

184.604.60

204.724.72

224.854.84

245.045.02

255.155.12

265.295.25

275.455.42

285.755.68

296.426.85

3010.7910.79

3111.2011.24

3311.5411.53

3511.7011.69

3711.8211.79

3911.9111.87

4111.9711.93

4312.0312.03

4512.0512.06

4712.0612.09

29.13 mLGraph 2-1 Graph of pH versus volume of NaOH for titration 1

29.04 mLGraph 2-2 Graph of pH versus volume of NaOH for titration 2

Titration 1Titration 2

Volume of NaOH to neutralize vinegar (mL)29.1329.04

Molarity of NaOH solution 1.871 M1.866M

Average Molarity of NaOH solution1.8685 M

Percent by mass of acetic acid in vinegar (%)11.24 11.21

Average percent by mass of acetic acid in vinegar (%)11.225

SAMPLE OF CALCULATIONStandardization of sodium hydroxide, NaOH solution

i) Calculation for preparing 150 mL of approximately 0.6 M NaOH solution

Molarity (M)= moles of soluteLitre of solution0.6= moles of solute (250 X 103)L

Moles of solute= (0.6) (0.25)= 0.15 mol

No. of moles= mass Molecular weight of NaOH

Mass= (no. of moles) (MW NaOH)= (0.15) (22.99 + 16.00 + 1.01g)= 6 g

ii) Volume of NaOH needed to neutralize the KHP (by using the data of titration 1)

Perform interpolation:7 10.93= Vol 126.12 10.93 11 12

Vol = 11.18 mL of NaOH

iii) Molarity of NaOH

Moles of KHP:1.503 g KHC8H4O4X 1 mol KHC8H4O4= 0.00736 mol KHC8H4O4 204.2 g KHC8H4O4Moles of NaOH required neutralizing moles of KHP:0.00736 mol KHC8H4O4X 1 mol NaOH= 0.00736 mol NaOH1 molKHC8H4O4

Molarity of NaOH solution:11.18 mL NaOHX1L= 0.01118 L NaOH 1000 mLM = mol of NaOH = 0.00736 mol NaOH L of solution 0.01118 L NaOH = 0.6583 M

iv) Average molarity of NaOHMave = (M1 + M2 + M3) / 3 = (0.6583 + 0.6583 + 0.6105) / 3 = 0.6424 M NaOHMolarity of acetic acid and mass percent in vinegar

i) Volume of NaOH needed to neutralize the vinegar (by using the data of titration 1)Perform interpolation:7 10.79 = Vol 306.42 10.79 29 30

Vol= 29.13 mL of NaOH

ii) Molarity of acetic acid in vinegarMoles of NaOH that reacted:29.13 mL of NaOHX1L= 0.02913 L NaOH 1000 mL0.02913 L NaOHX 0.6424 mol NaOH= 0.01871 mol NaOH 1L NaOH solution

Moles of CH3COOH neutralized by moles of NaOH:0.01871 mol NaOHX 1 mol CH3COOH= 0.01871 mol CH3COOHMol NaOH

Molarity of CH3COOH:10 mL CH3COOHX 1L =0.010 L CH3COOH 1000 mL

M = mol of CH3COOH = 0.01871 mol CH3COOH L of solution 0.010 L CH3COOH = 1.871 CH3COOH

iii) Average molarity of acetic acid in vinegar

Mave = (M1 + M2) /2 = (1.871 + 1.866) /2 = 1.869 M CH3COOH

iv) Percent by mass of acetic acid in vinegar

Mass of acetic acid in the solution:10 mL CH3COOHX 1L =0.010 L CH3COOH 1000 mL0.010 L CH3COOHX 1.871 mol CH3COOHX60.06 g CH3COOH1 L solution 1 mol CH3COOH= 1.124 g CH3COOH

Mass of acetic acid solution:10 mL CH3COOH X 1 g CH3COOH solution = 10 g CH3COOH solution 1 mLCH3COOH solution

Percent by mass of acetic acid in the solution =gCH3COOH X 100 %g CH3COOH solution =1.124X 100% 10.0 = 11.24 %

v) Average percent by mass of acetic acid in vinegar:%ave = (%1 + %2) / 2 = (11.24 + 11.21%) / 2 = 11.225 %

DISCUSSIONThe objectives of this experiment which are to determine the molarity of a solution and the percent by mass of acetic acid in vinegar by titration with the standardized sodium hydroxide solution had been determined. In the first part of this experiment, sodium hydroxide is standardized first before the second part of this experiment is proceeded which the molarity as well as the percent by mass of acetic acid is determined.On the first section of this experiment, sodium hydroxide, NaOH which acts as a base is standardized by the process of titration with potassium hydrogen phthalate, KHC8H4O4 or jotted as KHP which is a primary standard acid. The change of the pH value during titration is determined by using pH meter for every 1 mL addition of NaOH to ensure that more accurate result is obtained. Based on the results and graphs, when the pH level is 7, which at this point the solution is neutral, the average amount of NaOH needed to neutralize the KHP solution is 11. 47 mL.The second section of this experiment which is to determine the molarity and the percent by mass of the acetic acid in vinegar is conducted also by using titration between NaOH with vinegar. The vinegar needed to be diluted first in order to prevent a very small titre, which will then cause the results to be less accurate. Both the molarity and percent by mass was calculated using the average volume of NaOH from the first section of this experiment which is 0.6424 mL as well as the plotted graph based on the results from the second part of the experiment. From this experiment, the average molarity of acetic acid in vinegar was 1.869 and its average percent by mass is 11.225%.Vinegar is roughly 3%-5% of acetic acid by volume. However, in this experiment, the calculated average percent by mass is 11.225%. Although the amount can still be considered as relatively small but it exceeded the normal range of acetic acid in vinegar. This may due to the errors while conducting the experiment. The experiment was conducted by using burette. However, there was air bubble trapped at the tip of the burette during the experiment, causing the volume reading of the NaOH to be slightly incorrect, hence, affect the related calculation.

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.CONCLUSION

The results from the first section of the first titration of this experiment show that the volume of NaOH needed to neutralize 1.503 g of KHP is 11.18 mL and the molarity of the solution is 0.6583 M. For the second titration of the experiment, 1.502 g of KHP needed 11.18 mL of NaOH to neutralize the solution and the molarity is 0.6583 M. While for the third titration, 1.5 g of KHP and 12.04 mL NaOH is required to neutralize the solution and the molarity of NaOH solution for titration 3 is 0.6105 M. Thus, the average molarity of NaOH solution is 0.6424 M.

As for the second section of this experiment, the first titration required 29.13 mL of NaOH to neutralize the solution. The molarity of the solution is 1.871 M and the percent by mass of acetic acid in the vinegar is 11. 24%. Meanwhile, the molarity and the percent by mass for the second titration are 1.866 M and 11.21 % respectively. The volume of NaOH required to neutralize the solution is 29.04 mL. The average molarity of NaOH solution is 1.8685M and the average percent by mass of acetic acid in vinegar is 11. 225%.

As for the conclusion, it can be said that the acetic acid in the vinegar is a dilute solution as the percent by mass and molarity are relatively small.

RECOMMENDATION

1. It is better to use both indicator and pH meter. This is because the indicator such as phenolphthalein will change the colour when the solution had reached neutral condition rather than using a pH meter alone as using a pH meter is fairly tedious and it will turns out unnecessary. By using both indicator and pH meter, the result would be more accurat2. The swirling of the solution should be constant while adding the NaOH in order to ensure that the NaOH is totally dispersed.3. Ensure that the position of eye is directly perpendicular to the meniscus when reading the volume of solution to avoid inaccuracy.4. Ensure that the tip of the burette is filled with NaOH so that no air bubbles are present in the tip.

REFERENCES

Chemistry Laboratory CHE 485, Laboratory Manual Experiment 1 Determination of the Acetic Acid Content of Vinegar, http://wwwchem.uwimona.edu.jm/lab_manuals/c10expt1.html Acetic Acid http://en.wikipedia.org/wiki/Acetic_acid

APPENDICES

6