UNIVERSITI TEKNOLOGI MARA FAKULTI KEJURUTERAAN KIMIA

ENGINEERING CHEMISTRY LABORATORY (CHE485)

No. Title Allocated Marks (%) Marks

1 Abstract/Summary 5  2 Introduction 5  3 Aims 5  4 Theory 5  5 Apparatus 5  6 Methodology/Procedure 10  7 Results 10  8 Calculations 10  9 Discussion 20  10 Conclusion 5  11 Recommendations 5  12 Reference / Appendix 5  13 Supervisor’s grading 10  

TOTAL MARKS 100  

Remarks:

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Date : Date :

NAME : WAN FATHI HATIM DIYANA BT WAN NORMANSTUDENT NO. : 2010146941GROUP :GROUP 5EXPERIMENT : DETERMINATION OF THE CONCENTRATION OF ACETIC

ACID IN VINEGARDATE PERFORMED : 30 SEPTEMBER 2011SEMESTER : 2PROGRAMME / CODE : EH220-2SUBMIT TO : MADAM NOR SHARLIZA BT MOHD SAFAAI

ABSTRACT

The experiment is carried out to determine the molarity of standardizedsodium hydroxide solution and acetic acid in vinegar by titration between potassium hydrogen phthalate (KHP) and standardized NaOH solution and vinegar with standardized NaOH solution. The percent mass of acetic acid in vinegar is also determined. The titration is done first between KHP and standardized sodium hydroxide solution (NaOH). The pH for every 1mL additions of standardized NaOHin KHP solution is recorded. The titration is stopped once the pH of KHP solution indicates that it becomes alkali and we gain enough information to plot the graph of titration. The same method goes for titration of vinegar with standardized NaOH solution. The titration is done three times for each solution to minimize the error. From the experiment, the molarity of standardized NaOH in titration 1,2,3 are0.594M,0.589M and 0.586Mrespectively. The average amount of molarity of standard NaOH solution used in the experiment is 0.590M. Other than that, we were able to calculate the molarity of acetic acid for titration 1,2 and 3 which are 0.9145M,0.9263M and 0.9381 M respectively while the average molarity is 0.9263M. Furthermore, the percent by mass for each titration 1,2 and 3 are 5.49%, 5.56% and 5.63% respectively while the average percent by mass is 5.56%.

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 contain relatively little solute in a given amount of solvent. There are two specific terms to express concentration, namely molarity and percent by mass:

Molarity is the number of moles of solute per liter of solution.

Molarity(M) = moles of solute (mol) Liter of solution (L) (Equation 2-1)

Percent by mass is the mass in grams of solute per 100 grams of solution.

Percent solute (%) = grams of solute (g) x 100%

grams of solution(g)(Equation 2-2)

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 a 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 the titration 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 the added quantity of one reactant is the exact amount necessary for stoichiometric reaction with another reactant.

OBJECTIVES

To determine the molarity of standardized sodium hydroxide solution that is needed to titrate the KHP solution and molarity and percent by mass of acetic acid in vinegar by titration with standardized sodium hydroxide solution.

THEORY

In the titration process, a burette is used to dispense a small, quantifiable increment of solution of known concentration (Figure 2-1). A typical burette has the smallest calibration unit of 0.1 mL (Figure 1-1), therefore the volume dispensed from the burette should be estimated to the nearest 0.05mL.

Figure 1-1: a) A typical burette

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 strong base, sodium hydroxide (NaOH) is necessary to neutralize 1 mole of the weak acid, acetic acid (CH3COOH), as indicated in equation 2-3.

NaOH (aq) + CH3COOH(aq) NaCH3COO(aq) + H2O (l) (Equation 2-3)

The sudden change in the solution pH shows that the titration has reached the equivalence point. pH in an aqueous solution is related to its hydrogen ion 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 = - log 10[ H3O+] (Equation 2-4)

pH scale is a method of expressing the acidity or basicity of a solution. Solutions with pH<7 are acidic, pH= 7 are neutral, pH >7 are basic. For example, a solution having an [ H3O+] concentration of 2.35 x 10-2 M would have a pH of 1.629 and is acidic. pH electrodes will be used in this experiment. The titration is initiated by inserting a pH electrode into a beaker containing the acid solution (pH within 3-5). As sodium hydroxide, NaOH, is incrementally added to the acid solution, some of the hydrogen ions will be neutralized. As the hydrogen ion concentration decreases, the pH of the solution will gradually increase. When sufficient NaOH is added to completely neutralize the acid (most of the [ H3O+] ions are removed from the solution), 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 this experiment, titration of vinegar sample with standardized sodium hydroxide solution will be performed. To standardize the sodium hydroxide solution, a primary standard acid solution is initially prepared. In general, 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 (KHP) and oxalic acid (COOH)2 are common primary standard acids. Sodium carbonate Na2CO3 is the most commonly used base. Most acids and bases (e.gHCl, CH3COOH, NaOH and KOH) are mostly available in primary standard form. To standardize one of these acidic or basic solutions, titrations of the solution with a primary standard should be done. In this experiment, NaOH solution will be titrated with potassium hydrogen phthalate (KHP). The reaction equation for this is:

KHC8H4O4 (aq) + NaOH (aq) KNaC8H4O4 (aq) + H2O (l) (Equation 2-5)

Once the NaOH 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) NaCH3COO (aq) + H2O (l) (Equation 2-6)

Knowing the standardized NaOH concentration and using equation 2-6, we can determine the molarity and percent by mass of acetic acid in the vinegar solution.

APPARATUS/MATERIAL

Apparatus

- 500 mL volumetric flask- 50 mL burette- 100 mL beaker- Rod stirrer

- Magnetic stirrer- pH electrodes- 250 mL beaker- 250 mL measuring cylinder

Material

- Solid sodium hydroxide - 0.6M standardized sodium hydroxide

solution

- potassium hydrogen phthalate (KHP)- Distilled water- Vinegar

PROCEDURES

2.4.1 Standardization of sodium hydroxide solution.

1. 250mL of approximately 0.6M sodium hydroxide solution is prepared from NaOH solid. The solution is prepared in a beaker. The calculation should be checked with the lab instructor prior to prepare the solution. The calculation is recorded.

2. A beaker is placed and tared on the balance. 1.5 grams of KHP is added to the beaker. The mass of KHP is recorded to the nearest 0.001g. 30mL of distilled water is added to the beaker. The solution is stirred until the KHP has dissolved completely.

3. The solution is titrated with NaOH and the pH is recorded with every 1mL additions of NaOH solution.

4. Steps 1 to 3 are repeated and two solutions are prepared for NaOH standardization.

5. The graph of pH versus NaOH is plotted. From the plots, the volume of NaOH required to neutralize the KHP solution in each titration are determined.

6. The molarity of sodium hydroxide for titrations 1,2 and 3 are calculated.

7. The average molarity of the sodium hydroxide solution is calculated. The resulting sodium hydroxide concentration will be used in part B of the experiment.

2.4.2 Molarity of acetic acid and mass percent in vinegar.

1. 10.00mL of vinegar is transferred to a clean, dry 250mL beaker using a 10mL volumetric pipette. Sufficient water between 75mL to 100mL is added to cover the pH electrode tip during the titration.

2. 1mL of NaOH is added to the vinegar solution and the pH is recorded.

3. The above steps are repeated twice more.

4. The graph of pH versus NaOH volume added is plotted and from the plots determines the volume of NaOH required to neutralize the vinegar in each titration. The data is recorded.

5. The molarity of acetic acid in vinegar for titrations 1,2 and 3 are calculated.

6. The average molarity of acetic acid for each titration are calculated.

7. The percent by mass of acetic acid in vinegar for titrations 1, 2 and 3 are calculated.

8. The percent by mass of acetic acid in vinegar are calculated.

RESULTS

2.5.1 Standardization of sodium hydroxide (NaOH) solution.

Table 1.1: The mass of KHP for each titration

Titration 1 Titration 2 Titration 3

Mass of KHP (grams)

1.5058 1.5037 1.5073

Table 1.2 : The pH of KHP for every 1mL of NaOH for each titration.

Volume of NaOH(mL)

Titration 1(pH)

Titration 2(pH)

Titration 3(pH)

Initial 4.20 3.95 3.931 4.32 4.15 4.132 4.40 4.34 4.323 4.61 4.51 4.494 4.70 4.64 4.635 4.86 4.78 4.776 5.08 4.90 4.897 5.12 5.02 5.038 5.25 5.15 5.179 5.40 5.30 5.31

10 5.59 5.47 5.4911 5.80 5.69 5.7212 6.15 5.99 6.0913 11.03 10.69 11.0214 12.19 11.99 12.0715 12.39 12.82 12.32

2.52 Molarity of acetic acid and mass percent in vinegar.

Table 1.3: The pH of the vinegar for every 1 mL of NaOH for each titration.

Volume of NaOH(mL)

Titration 1(pH)

Titration 2(pH)

Titration 3(pH)

Initial 2.50 2.85 2.881 3.05 3.44 3.392 3.44 3.73 3.733 3.76 3.95 3.964 3.96 4.12 4.125 4.12 4.26 4.256 4.25 4.38 4.377 4.40 4.50 4.498 4.51 4.63 4.619 4.63 4.74 4.72

10 4.75 4.86 4.8411 4.90 5.00 4.9612 5.04 5.15 5.1113 5.22 5.35 5.2814 5.43 5.60 5.5415 5.81 6.18 5.9116 10.38 11.23 9.0317 11.53 11.73 11.4518 11.79 11.92 11.7519 11.53 12.04 11.8920 12.03 12.12 12.06

CALCULATIONS

2.5.1 Standardization of sodium hydroxide (NaOH) solution

0 2 4 6 8 10 12 14 160

2

4

6

8

10

12

14

12.4 mL NaOH at the equivalence

point

Titration 1 of KHP with NaOH solution

Volume of NaOH (mL)

pH o

f KHP

Equivalence point

Graph 1.1: Titration 1 of KHP with NaOH solution

0 2 4 6 8 10 12 14 160

2

4

6

8

10

12

14

12.5 mL NaOH at the equivalence point

Titration 2 of KHP with NaOH solution

Volume of NaOH (mL)

pH o

f KP

Equivalence point

Graph 1.2: Titration 2 of KHP with NaOH solution

0 2 4 6 8 10 12 14 160

2

4

6

8

10

12

14

12.6 mL NaOH at the equivalence point

Titration 3 of KHP with NaOH solution

Volume of NaOH (mL)

pH o

f KHP

Equivalence point

Graph 1.3: Titration 3 of KHP with NaOH solution

The molarity of Sodium Hydroxide solution (NaOH) for each titration can be determined by using:

KHC8H4O4 (aq) + NaOH (aq) KNaC8H4O4 (aq) + H2O (l)Mole of KHP (mol) = Mass of KHP (g)Molecular mass (g)Mole of NaOH = Mole of KHP Molarity of NaOH = mole of NaOH (mol)x 1000

Litres of NaOHsolution(L)

Titrations Calculations

1

Mole of KHP (mol) = 1.5058 g KHP 204.2 g KHP = 7.37x10-3mol of KHPMole of KHP = Mole of NaOH = 7.37x10-3mol of NaOHMolarity of NaOH = 7.37x10 -3 molx1000 12.4 mL of NaOH = 0.594 M

2 Mole of KHP (mol) = 1.5037 g KHP 204.2 g KHP = 7.36x10-3mol of KHPMole of KHP = Mole of NaOH = 7.36x10-3mol of NaOHMolarity of NaOH = 7.36x10 -3 molx1000

12.5mL of NaOH = 0.589 M

3

Mole of KHP (mol) = 1.5073 g KHP 204.2 g KHP = 7.38x10-3mol of KHPMole of KHP = Mole of NaOH = 7.38x10-3mol of NaOHMolarity of NaOH = 7.38x10 -3 molx1000 12.6 mL of NaOH = 0.586 M

The average molarity of NaOH for each titration can be determined by using:

Average Molarity = Molarity of Titration 1 + Molarity of Titration 2 + Molarity of Titration 3 Number of titration

= 0.594 M + 0.589 M + 0.586 M 3

= 0.590 M

2.5.2 Molarity of acetic acid and mass percent in vinegar

0 5 10 15 20 250

2

4

6

8

10

12

14

15.5mL of NaOH at the equivalence

point

Titration 1 of acetic acid with NaOH so-lution

Volume of NaOH (mL)

pH o

f ace

tic a

cid Equivalence point

Graph 1.4 : Titration 1 of acetic acid with NaOH solution

0 5 10 15 20 250

2

4

6

8

10

12

14

15.7 mL NaOH at the equivalence point

Titration 2 of acetic acid with NaOH so-lution

Volume of NaOH solution (mL)

pH o

f ace

tic a

cid

Equivalence point

Graph 1.5: Titration 2 of acetic acid with NaOH solution

0 5 10 15 20 250

2

4

6

8

10

12

14

15.9mL NaOH at the equivalenc e point

Titration 3 of acetic acid with NaOH solu-tion

Volume of NaOH (mL)

pH o

f ace

tic a

cid Equivalence point

Graph 1.6: Titration 3 of acetic acid with NaOH solution

The molarity and percent mass of acetic acid in vinegar for each titration can be determined by using:

NaOH (aq) + CH3COOH(aq) NaCH3COO(aq) + H2O (l)Mole of NaOH = Molarity of NaOH x Litres of NaOH solution 1000Mole of acetic acid = Mole of NaOHMolarity of acetic acid = mole of acetic acid x 1000Litres of solution of vinegarMass of acetic acid = mole of acetic acid x molecular mass of acetic acidMass of acetic acid solution = Volume of acetic acid solution x 1g acetic acid solution 1 mL acetic acid solutionPercent by mass of acetic acid (%) = Mass of acetic acid(g) x 100% Mass of acetic acid solution (g)

Titrations Calculations1 Mole of NaOH = 0.590 M NaOH x 15.5mL

1000 = 9.145 x 10-3molNaOH

Mole of acetic acid = 9.145 x 10-3mol Molarity acetic acid = 9.145 x 10 -3 mol x 1000 10mL of solution = 0.9145 MMass of acetic acid = 9.145 x 10-3mol x 60.06 g acetic acid = 0.549 g of acetic acidMass of solution= 10mL x 1g solution 1mL of solution = 10.0 g of solutionPercent by mass(%) = 0.549 g x 100% 10.0 g

= 5.49 % of acetic acid2 Mole of NaOH = 0.590 M NaOH x 15.7mL

1000 = 9.263 x 10-3molNaOH

Mole of acetic acid = 9.263 x 10-3mol Molarity acetic acid = 9.263 x 10 -3 mol x 1000 10mL of solution = 0.9263 MMass of acetic acid = 9.263 x 10-3mol x 60.06 g acetic acid = 0.556 g of acetic acidMass of solution= 10mL x 1g solution 1mL of solution = 10.0 g of solutionPercent by mass(%) = 0.556 g x 100% 10.0 g

= 5.56 % of acetic acid

3

Mole of NaOH = 0.590 M NaOH x 15.9mL 1000 = 9.381 x 10-3molNaOH

Mole of acetic acid = 9.381 x 10-3mol Molarity acetic acid = 9.381 x 10 -3 mol x 1000 10mL of solution = 0.9381 MMass of acetic acid = 9.381 x 10-3mol x 60.06 g acetic acid = 0.563 g of acetic acidMass of solution= 10mL x 1g solution 1mL of solution = 10.0 g of solutionPercent by mass(%) = 0.563 g x 100% 10.0 g = 5.63 % of acetic acid

Average molarity = Molarity of titration 1 + Molarity of titration 2 + Molarity of titration 3 3

= 0.9145M + 0.9263 M + 0.9381M 3 = 0.9263 MAverage percent by mass (%)=5.49 % +5.56 %+ 5.63% 3 = 5.56 %

DISCUSSION

At the end of the experiment, we had managed to calculate the molarity of NaOH used to titrate the weak acid which is KHP. The molarity of NaOH is determined to know the standardization of the NaOH solution. We also managed to determine the molarity and percent mass of acetic acid in vinegar by using the molarity of standardization of NaOH solution in the first experiment. All of the calculations are vital to fulfill the objectives of our experiment.

There are two parts of experiments. For the first experiment, we want to determine the molarity of NaOH used to make the KHP to reach its equivalence point. The equivalence point is reached when the added quantity of one reactant is the exact amount necessary for stoichiometric reaction with another reactant which means the pH of KHP is 7. The equivalence point can be determined from the graph of pH versus volume of NaOH. The equation that will results from the titration between KHP and NaOH solution is:

KHC8H4O4 (aq) + NaOH (aq) KNaC8H4O4 (aq) + H2O (l)

We notice that in the chemical equation, the number of moles of KHP is the same as the number of moles of NaOH. From the number of moles of NaOH, we manually can calculate the concentration of NaOH. In this experiment, the titrations are done thrice. This is to minimize the error that we will get and to calculate the average amount of concentration of NaOH. The three determinations should agree within 1.0%. If not, the standardization should be repeated. Furthermore, notice that the water that being used in the experiment is distilled water. If we use tape water, the presence of carbon dioxide will affect the equation above.

In the next experiment, we want to determine the molarity of acetic acid and percent by mass of the acetic acid in vinegar. To achieve that, we titrated acetic acid solution (vinegar) with standardized NaOH to obtain the equivalence point. The equation that will results from the titration between acetic acid and NaOH solution is:

CH3COOH (aq) + NaOH (aq) NaCH3COO (aq) + H2O (l)

Notice that, the mole of acetic acid and NaOH is the same based on the equation above. Same as the first experiment, the molarity of acetic acid can be obtained manually by determining the mol of NaOH solution for each titration. As for the percent by mass of the acetic acid, the mass of acetic acid and mass of acetic solution need to be determined.

There are several techniques wrongly used in titration. For example, the opening of outlet of burette at high speed would change the exact amount of NaOH required to neutralize the KHP. This wrongly technique could result to an inaccurate reading measurements. Thus, affects the experiment results.

CONCLUSION

As a conclusion, the experiment that we have done was successful. At the end of the experiment, we had managed to obtain the amount of molarity of standard NaOH solution used in the experiment using appropriate calculation. We got 0.594M for titration 1, 0.589M for titration 2, and 0.586M for titration 3 and the average amount of molarity of standard NaOH solution used in the experiment is 0.590M. Other than that, we were able to calculate the molarity of acetic acid for titration 1,2 and 3 which are 0.9145M,0.9263M and 0.9381 M respectively while the average molarity is 0.9263M. Furthermore, the percent by mass for each titration 1,2 and 3 are 5.49%, 5.56% and 5.63% respectively while the average percent by mass is 5.56%.Thus, it can be conclude that all the objectives of the experiment had been reached.

RECOMMENDATIONS

Before starting the titration, the burette should be first cleaned and after that be rinsed with a little amount of NaOH solution. Run the NaOH solution through the burette tip as to remove any unwanted impurities.

In the titration process, it is wiser to repeat the titration several times and record all the readings before calculate the average reading, as this can increase the precision and the accuracy in determining the volume of the NaOH solution titrated.

While measuring the meniscus level of the solutions in the measuring cylinder and burette, parallax error should be avoided as this might affect the readings to be taken and recorded.

In titrating the NaOH solution using the burette, a miscalculated drop could have triggered the pH reading. Hence, it is better if for each adding of NaOH, only a small amount of the solution is titrated.

A part from that, safety issues need to be prioritized. Wearing lab coats, safety goggles, gloves and appropriate attire is a compulsory for all the students.

REFERENCES

Chang, R. (2009). Chemistry (10th Ed) New York : McGraw-Hill Companies, Inc.

(Page:311 )

SengChyeEng, Lim Boon Tik, Lau Kah Pew (2007). Chemistry for Matriculation 2 (3rd. Ed)Kuala

Lumpur : Oriental Academic Publication.

(Pages 21,22,24)

Bruce Averill, Patricia Eldredge.(2007) Chemistry (2nd Ed.),1301 SansomeSt.,San Francisco.

(Page : 47)

Norman E. Griswold, H.A. Neiding, James N. Spencer, Conrad L. Stanitski (2002). Laboratory

Handbook for General Chemistry ( 2nd. Ed. ) Canada : Brooks/Cole,

(Page 12)

EngNguan Chong, Lim Yean Ching, Lim EngWah. (2008) SPM Focus U Chemistry (2nd

Ed.)PenerbitanPelangiSdn. Bhd. ( pages 76 and 77)

Collins Cobuild :Advance Dictionary : HeinleCengage Learning (2010) (pages 914 and 23)

APPENDICES