Lab 1-Acetic Acid in Vinegar Faris(1)

Introduction

ABSTRACTConcentration of solution is the amount of solute in a given amount of solvent which can be expressed by two specific terms namely molarity and percent by mass. The aim of this experiment is to determine the molarity of a solution and percent by mass of acetic acid in vinegar. The first experiment is done by titration of Potassium hydrogen phthalate (KHP) with sodium hyroxide (NaOH) solution for standardization. The titration runs for twice. Next, the vinegar (acidic solution) was titrated with standardize sodium hydroxide (standard base). This experiment is also done for twice. According to the result obtained, titration of NaOH with both KHP and vinegar shows higher pH value as the volume of NaOH added increase. At some point, the curve of the graph shows rapid change of pH value. This is because at that point, the volume of NaOH is at sufficient enough to completely neutralize the acidic solution. The hydrogen ion concentration thus decreases and contributes to higher pH reading. As a conclusion, the equivalence point can be occurs when the moles of acid in the solution equals to the moles of base added in the titration and the concentration for acidic solution thus can be determined.INTRODUCTIONConcentration of any solution is depends on the amount of solute in the specific amount of solvent. The higher the amount of the solute, the solution will become more concentrated while the lower the amount of the solute, the lower the concentration of the solution. More concentrated solution will contains a large quantity of solute in a given amount of solvent. But for less concentrated solution or dilute solutions, the solute contain are little. Both terms molarity and percent by mass specifically can express the concentration since molarity is the number of moles of solute per liter of solution and percent by mass is the mass in grams of solute per grams of solution time percent.

The molecular formula for acetic acid is CH3COOH. Acetic acid is dilute acidic solution since it contains less amount of acidic solute. The concentration of this weak acid in a vinegar solution can be determined by performing a titration process with standard base solution like standard NaOH. The concentration of acetic acid can be expressed in the term of molarity and also in the term of percent by mass through this titration method.A titration method is the method in which the small increments of solutions of known concentration are added to a specific amount of solution of unknown concentration until the stoichiometry for that reaction can be achieved. By knowing the quantity of one solution required to complete the titration, calculation for the other unknown solution concentration can be conduct. Through this way too, the added quantity of one solution that can achieved the exact amount necessary for stoichiometry reaction with another reactant will be used for further calculation. This point is known as equivalence point and can be used to determine the concentration for acidic solution.

OBJECTIVES To determine the concentration of acetic acid in vinegar in term of molarity of a solution and percent by mass.

To standardize sodium hydroxide solution by applying the titration method. To distinguish the equivalence point during the acid-base reaction.THEORYThe equivalence point actually occurs when the moles of acid are equal with the mole of base in the solution. For this experiment, the equivalence point thus can be determined clearly after the titration of stoichiometry amount of 1 mole sodium hydroxide as strong base required to neutralize 1 mole of the acetic acid which is weak acid. The stoichiometric equation for the reaction is shows below: NaOH(aq) + CH3COOH(aq) NaCH3CO2(aq) + H2O

When that point had been reached, the pH value for the solution will undergo a sudden change. By reading a pH value through the pH meter, the acidity or basicity of a solution can also be determined according to pH scale. For the solution with pH lower than 7, it is acidic solution. While pH equal to 7 is neutral and pH greater than 7 is a basic solution.

During the titration process as sodium hydroxide (NaOH) is incrementally added to the acid solution, some of the hydrogen ion will be neutralized and the pH of solution will gradually increase. Before the solution of acid going to become more basic as sodium hydroxide is incrementally added, most of the hydrogen ion concentration [H30+] will be removed from the solution. The sufficient amount of sodium hydroxide added then will neutralize the acidic solution and cause the sudden sharp increase in pH as show in Figure 1.1.

Specifically, before the acid going to become a basic solution as added with standardize base solution, it will first be in the neutral state. The sudden sharp increase in pH is the equivalence point where neutralization takes places. Since the pH value of the solution are depends on the hydrogen ion, pH = -log10[H3O+], the removal of the hydrogen ion during neutralization will cause the pH values to increase rapidly.Figure 1.1

Titration sample of vinegar that contains acetic acid with the standardized sodium hydroxide solution will be performed through this experiment. To prepare standard sodium hydroxide solution, a primary standard acid solution is initially prepared to be mixed with. Primary standard acid or bases have several common characteristic which make it suitable to use in standardize process such as: available in at least 99.9 purity

high molar mass so that error in weighing / measurement can be minimized stable when in contact with heat or bring to high temperature high solubility in solventCommon primary acid that widely uses is such as potassium hydrogen phthalate, KHC8H4O4 while sodium carbonate, Na2CO3, is the most commonly used base. To conduct standardization process, titration of the solution with a primary standard must be run first. In this experiment, NaOH solution will be titrated with potassium hydrogen phthalate (KHP). KHC8H4O4(aq) + NaOH(aq) KNaC8H4O4(aq) + H2O(l)

Once the sodium hydroxide solution has been standardized it will be titrated with 10.00mL

aliquots of vinegar (acidic solution) CH3COOH(aq) + HaOH(aq) NaCH3COOH(aq) + H2O

By knowing the standardized NaOH concentration, the molarity and percent by mass of acetic acid in the vinegar solution can be determined through calculation and relating.

Sample calculation for standardizing a based with KHP

Dissolve 1.523 gram in 20ml distilled water titrated with NaOH. Determine molarity of the NaOH solution.

1. Calculate the moles of KHP used in titration.

1.523 g KHC8H4O4 X 1 mol KHC8H4O4 = 0.007458 mol KHC8H4O4 204.2 g KHC8H4O42. Calculate the mole of NaOH required neutralizing the mole of KHP

0.007458 mol KHP x 1 mol NaOH = 0.007458 mol NaOH

1 mol KHP

3. Calculate the molarity of the NaOH solution.

15.30mL NaOH x 1L = 0.01530L NaOH

1000mL M = mol NaOH = 0.007458 mole NaOH = 0.4875 M NaOH

L of solution 0.01530 L solution

Sample calculations for determining the acetic acid concentration in vinegar by titration with standard base. A 10.00 mL aliquot of vinegar requires 16.95 mL of the 0.4875 M standaedized NaOH solution to reach the equivalence point of the titration. Calculate the molarity and the percent by mass of CH3COOH in the solution. Assume the density of the vinegar solution is 1.00 g/mL.

1. Calculate mol of NaOH that reacted

16.95mL NaOH x 1 L = 0.01695 L NaOH 1000 mL

0.01695 L NaOH x 0.4875 mol NaOH = 0.008263 mol NaOH

1 L NaOH solution2. Calculate the moles of CH3COOH neutralized by the moles of NaOH

0.008263 mol NaOH x 1 mol CH3COOH = 0.008263 mole CH3COOH

1 mol NaOH

3. Calculate the molarity of the CH3COOH solution

10 mL CH3COOH x 1 L = 0.010 L CH3COOH solution

1000 mL

M = mol CH3COOH = 0.008263 mol CH3COOH = 0.8263 M CH3COOH


4. Calculate the mass of acetic acid in the solution


1000 mL

0.010L CH3COOHx 0.8263 mol CH3COOH x 60.06 g CH3COOH = 0.4963g CH3COOH

1 L solution 1 mol CH3COOH

5. Calculate the mass of acetic acid solution

10 mL CH3COOH solution x 1 g CH3COOH solution = 10.00 g CH3COOH solution

1 mL CH3COOH solution 6. Calculate the percent by mass of acetic acid in the solution

Percent mass CH3COOH = g CH3COOH x 100%

g CH3COOH Percent mass CH3COOH = 0.4963g CH3COOH x 100% = 1.963% CH3COOH

10.00 g CH3COOHPROCEDURE

STANDARDIZATION OF SODIUM HYDROXIDE (NaOH) SOLUTION

1. A beaker was used to prepare 150 mL and approximately 0.6M sodium hydroxide solution from solid NaoH.

2. The calculation was checked first with the laboratory instructor prior to preparing the solution and the approved data is recorded.

3. Weighted a 250 mL beaker then recorded the data as nearest to 0.001g. 4. Next, KHP was put into three different beakers by amount of 1.5 g respectively. Mass of the beaker together with KHP was recorded to the nearest 0.001g. 5. The difference mass of KHP with the beaker use was also calculated and the data was taken. 6. 30 mL of distilled water was added to the beaker.

7. The solution was stirred well with stirrer so that the KHP was dissolved completely.

8. Then, the KHP solution was titrated with NaOH and the pH value was recorded at intervals according to the addition of 1 mL NaOH solution.9. Steps 2 and 3 were repeated to conduct second and third standardization of the NaOH solution with KHP.10. The graph of pH value versus volume of NaOH added was plotted. 11. The volume of NaOH required to completely neutralizing the KHP solution in each titration was determined from the graph at the pH = 7.12. After that, the molarity and the average morality of sodium hydroxide for Titration 1, Titration 2 and Titration 3 were calculated.

TO DETERMINE THE MOLARITY OF ACETIC ACID AND MASS PERCENT IN VINEGAR1. 10.00 mL of vinegar was transferred to a clean, safe and dry 250 mL beaker. 10 mL volumetric pipette was used in this step.

2. Sufficient volume of water (around 75-100 mL) was added to cover the pH electrode tip during the titration process. Make sure the water is also clean and free from any contaminants.3. Steps above were repeated to perform a second and third titration of vinegar with the standard NaOH solution.4. The graph of pH value versus volume of NaOH added was plotted. 5. From the graph, the volume of NaOH required to neutralize vinegar in each titration (equivalence point) was determined and recorded.

6. The molarity and the average molarity of acetic acid in vinegar for Titration 1, Titration 2 and Titration 3 were calculated.7. The percent by mass and the average percent by mass of acetic acid in vinegar for Titration 1, Titration 2 and Titration 3 were calculated.APPARATUS Potassium Hydrogen Phthalate (KHP) Magnetic Stirrer

Standard NaOH solution / Solid NaOH Ring Stand Vinegar

Burette

Volumetric Pipette 250 mL Beaker pH meter Volumetric FlaskRESULTSA) Standardization of sodium hydroxide solution

Titration 1Volume of NaOH (mL)pHVolume of NaOH (mL)pH

03.5296.04

13.83106.06

24.191112.26

34.501212.27

44.651312.71

54.811412.92

65.011512.97

75.181613.01

85.53


03.4494.83

13.60105.02

23.91115.25

34.07125.67

44.24136.32

54.391412.28

64.401512.69

74.541612.70

84.69


03.5095.05

13.71105.27

23.92115.60

34.10126.22

44.251312.29

54.421412.70

64.551512.71

74.711612.93

84.87

Titration 1Titration 2Titration 3

Mass of KHP (g)1.50641.52431.5180

Volume of NaOH to neutralize the KHP solution (mL)10.0013.1012.10

B) Molarity of acetic acid and mass percent in vinegar


03.05114.62

13.07124.80

23.25135.00

33.51145.20

43.69155.57

53.861611.17

63.981712.11

74.141812.35

84.241912.57

94.362012.66

104.49

Titration 2

Volume of NaOH (mL)pHVolume of NaOH (mL)pH

03.10114.50

13.14124.76

23.20135.01

33.50145.10

43.64155.60

53.811611.08

63.951712.09

74.131812.40

84.201912.60

94.352012.74

104.39

Titration 3

Volume of NaOH (mL)pHVolume of NaOH (mL)pH

03.12115.15

13.50125.28

23.84135.45

34.05145.68

44.22156.05

54.361610.65

64.481711.65

74.601811.92

84.711912.04

94.802012.13

104.95

Titration 1Titration 2Titration 3

Volume of NaOH to neutralize the vinegar solution (mL)15.1015.1015.10

CALCULATIONA) Standardization of sodium hydroxide solution

Titration 11. The moles of KHP used in titration :1.5064 g KHC8H4O4 X 1 mol KHC8H4O4 = 0.007377 mol KHC8H4O4 204.2 g KHC8H4O42. The mole of NaOH required to neutralizing the mole of KHP : 0.007377 mol KHP x 1 mol NaOH = 0.007377 mol NaOH

1 mol KHP

3. The molarity of the NaOH solution :10.0000 mL NaOH x 1L = 0.010000 L NaOH

1000mL

M = mol NaOH = 0.007377 mol NaOH = 0.7377 M NaOH

L of solution 0.010000 L solutionTitration 21. The moles of KHP used in titration : 1.5243g KHC8H4O4 X 1 mol KHC8H4O4 = 0.007465 mol KHC8H4O4 204.2 g KHC8H4O42. The mole of NaOH required to neutralizing the mole of KHP :0.007465 mol KHP x 1 mol NaOH = 0.007465 mol NaOH

1 mol KHP3. The molarity of the NaOH solution :13.1000 mL NaOH x 1L = 0.013100 L NaOH

1000mL

M = mol NaOH = 0.007465 mol NaOH = 0.5698 M NaOH

L of solution 0.013100 L solutionTitration 3.

1. The moles of KHP used in titration :1.5180 g KHC8H4O4 X 1 mol KHC8H4O4 = 0.007434 mol KHC8H4O4 204.2 g KHC8H4O42. The mole of NaOH required to neutralizing the mole of KHP : 0.007434 mol KHP x 1 mol NaOH = 0.007434 mol NaOH

1 mol KHP 3. The molarity of the NaOH solution :12.1000 mL NaOH x 1L = 0.012100 L NaOH

1000mL

M = mol NaOH = 0.007434 mol NaOH = 0.6144 M NaOH L of solution 0.012100 L solution

Molarity for Titration 1 = 0.7377 M NaOH

Molarity for Titration 2 = 0.5698 M NaOH Molarity for Titration 3 = 0.6144 M NaOHAverage molarity for Sodium Hydroxide (NaOH) = (0.7377 + 0.5698 + 0.6144) / 3 = 0.6406 M B) Molarity of acetic acid and mass percent in vinegarTitration 11. Mol of NaOH that reacted : 15.10 mL NaOH x 1 L = 0.01510 L NaOH

1000 mL


1 L NaOH solution

2. The mole of CH3COOH neutralized by the mole of NaOH : 0.011139 mol NaOH x 1 mol CH3COOH = 0.011139 mole CH3COOH

1 mol NaOH

3. The molarity of the CH3COOH solution : 10 mL CH3COOH x 1 L = 0.010 L CH3COOH solution

1000 mL



4. The mass of acetic acid in solution : 10 mL CH3COOH x 1 L = 0.010 L CH3COOH solution

1000 mL

0.010 L CH3COOH x 1.1139 mol CH3COOH x 60.06 g CH3COOH = 0.6690 g CH3COOH


5. The mass of acetic acid solution : 10 mL CH3COOH solution x 1 g CH3COOH solution = 10.00 g CH3COOH solution

1 mL CH3COOH solution

6. The percent by mass of acetic acid in the solution : Percent mass CH3COOH = 0.6690 g CH3COOH x 100% = 6.690 % CH3COOH

10.00 g CH3COOH

Titration 2

1. Mol of NaOH that reacted :

15.10 mL NaOH x 1 L = 0.01510 L NaOH

1000 mL


1 L NaOH solution2. The mole of CH3COOH neutralized by the mole of NaOH :

0.008604 mol NaOH x 1 mol CH3COOH = 0.008604 mol CH3COOH

1 mol NaOH 3. The molarity of the CH3COOH solution :


1000 mL



4. The mass of acetic acid in solution :


1000 mL



5. The mass of acetic acid solution :


1 mL CH3COOH solution

6. The percent by mass of acetic acid in the solution :

Percent mass CH3COOH = 0.5168 g CH3COOH x 100% = 5.168 % CH3COOH

10.00 g CH3COOHTitration 31. Mol of NaOH that reacted :

15.10 mL NaOH x 1 L = 0.01510 L NaOH

1000 mL


1 L NaOH solution

2. The mole of CH3COOH neutralized by the mole of NaOH :

0.009277 mol NaOH x 1 mol CH3COOH = 0.009277 mol CH3COOH

1 mol NaOH

3. The molarity of the CH3COOH solution :


1000 mL



4. The mass of acetic acid in solution :


1000 mL



5. The mass of acetic acid solution :


1 mL CH3COOH solution 6. The percent by mass of acetic acid in the solution :

Percent mass CH3COOH = 0.5572 g CH3COOH x 100% = 5.572 % CH3COOH 10.00 g CH3COOHAverage molarity of acetic acid = 1.1139 + 0.8604 + 0.9277 = 0.9673 M CH3COOH

3Average percent by mass of acetic acid acid in vinegar = (6.690 + 5.168 + 5.572) / 3

= 5.81 % CH3COOH

DISCUSSIONThe acidity, basicity as well as neutrality of the solution can be determined regarding to the pH scale instead of litmus paper. According to this scale in which hydrogen ion is dependent, solution with pH lower than 7 is an acidic, solution with pH equal to 7 is neutral and solution with pH greater than 7 is basic. That is the higher the pH value, the solution is more basic while the lower pH value stand for more acidic solution.

For this experiment, the determination of the acetic acid concentration is achieved by applying the acid-base titration. The acetic acid in vinegar is titrating with sodium hydroxide, which is initially undergoes a standardization process with the KHP, also by titration method to determine the volume of sodium hydroxide that have been standardize. From the equivalence point where sufficient volume have been traced, the molarity of NaOH solution then can be calculated according to the moles of NaOH required to neutralize the moles of KHP.In the first experiment for Part A, the mass of KHP is weighed to be 1.5064 g with 16 drop of NaOH added for the titration. The equivalence point is achieved at 10.00 mL of NaOH droplet. The NaOH molarity at this equivalence point is recorded to be 0.7377 M. Second experiment for this part is conduct with 1.5243 g weighed KHP with also 16 mL droplet of NaOH and the equivalence point is determined at 13.10 mL of NaOH droplet that resulting the 0.5698 M of NaOH. For the last experiment of Part A, the mass of KHP weighed is 1.5180 g with the same 16 mL droplet of NaOH. The equivalence point is achieved at 12.10 mL of NaOH with 0.6144 M of NaOH. Average molarity for sodium hydroxide is calculated to be 0.6406 M.

For the experiment in Part B which is to determine the concentration of acetic acid in vinegar, 10 mL of vinegar are use to titrate with the 20 mL droplet of sodium hydroxide. During this experiment, the equivalence point is constant for all three different concentration gained from the first experiment that is at 15.10 mL volume of NaOH. This shows that the NaOH solution had been successfully standardized by titration with KHP previously. Titration 1 for the first experiment shows 1.1139 M of CH3COOH with 6.690 percent CH3COOH by mass. Titration 2 shows 0.8604 M CH3COOH with 5.168 percent CH3COOH by mass and Titration 3 shows 0.9277 M CH3COOH with 5.572 percent CH3COOH by mass. Average molarity of acetic acid calculated thus equal to 0.9673 M CH3COOH and the average percent by mass of acetic acid in vinegar are equal to 5.81 %.

CONCLUSIONAs a conclusion, the concentration of acetic acid in vinegar can be determined by applying an acid-base titration. However, the base must first be standardizing with primary acid (KHP) so that the concentration of NaOH neutralized can be determined first. The concentrations of the neutralized solution are determined at the equivalence point where the sudden changes of pH value happen, that is when KHP are titrated with NaOH. To obtain the molarity and percent by mass of acetic acid for experiment in Part B, the equivalence point also need to be recognized plus these concentration of acetic acid are depend on the concentration of NaOH derived from the standardization process in Part A for further calculation.RECOMMENDATION Reduce parallax error during any measurement was taken. Wear Personal Protective Equipment (PPE) when handling the chemical and apparatus in the laboratory. Avoid any bubbles in the burette (by bubbling) before start the titration test because the air contain will interrupt the measurement. Conduct the titration more than once so that the accurate average reading can be taken.

Reject any flask or beaker that contains any contaminant because it will affect the pH value (such as wet beaker will contain water droplet that will mixed with acid thus reduce the acidity)

Does not turn the magnetic stirrer to the highest speed of stirring because the KHP solution will decrease in volume due to rapid movement of stirrer.

Prepare excess volume of NaOH solution due to that the equivalence point cannot be expected (the equivalence point will happen at any volume / droplet of NaOH added into the acid)

REFERENCE Robert H.Perry, Don W.Green, Perrys Chemical Engineers Handbook, McGraw Hill,1998. Engineering Chemistry lab. Dalwani, M., N. E. Benes, et al. (2011) Effect of pH on the performance of

polyamide/polyacrylonitrile based thin film composite membranes. Journal of Membrane Science, 372(1-2): 228-238. Nilsson, M., G. Tragardh, et al. (2008). The influence of pH, salt and temperature on

nanofiltration performance. Journal of Membrane Science, 312(1-2): 97-106. Su, M., D.-X. Wang, et al. (2006) Rejection of ions by NF membranes for binary

electrolyte solutions of NaCl, NaNO3, CaCl2 and Ca(NO3)2. Desalination, 191(1-

3): 303-308.

APPENDICES

15

_1379627817.xlsChart1

3.44

3.6

3.91

4.07

4.24

4.39

4.4

4.54

4.69

4.83

5.02

5.25

5.67

6.32

12.28

12.69

12.7

pH

pH

Sheet1

VolumepH

03.44

13.6

23.91

34.07

44.24

54.39

64.4

74.54

84.69

94.83

105.02

115.25

125.67

136.32

1412.28

1512.69

1612.7

Sheet1

Volume of NaOH (mL)

pH

Graph of pH Value Versus Volume of NaOH

Sheet2

Sheet3

_1379628432.xlsChart1

3.05

3.07

3.25

3.51

3.69

3.86

3.98

4.14

4.24

4.36

4.49

4.62

4.8

5

5.2

5.57

11.17

pH

pH

Sheet1

VolumepH

03.05

13.07

23.25

33.51

43.69

53.86

63.98

74.14

84.24

94.36

104.49

114.62

124.8

135

145.2

155.57

1611.17

1712.11

1812.35

1912.57

2012.66

Sheet1

Volume of NaOH (mL)

pH


Sheet2

Sheet3

_1379628679.xlsChart1

3.1

3.14

3.2

3.5

3.64

3.81

3.95

4.13

4.2

4.35

4.39

4.5

4.76

5.01

5.1

5.6

11.08

pH

pH

Sheet1

VolumepH

03.1

13.14

23.2

33.5

43.64

53.81

63.95

74.13

84.2

94.35

104.39

114.5

124.76

135.01

145.1

155.6

1611.08

1712.09

1812.4

1912.6

2012.74

Sheet1

Volume of NaOH (mL)

pH


Sheet2

Sheet3

_1379628939.xlsChart1

3.12

3.5

3.84

4.05

4.22

4.36

4.48

4.6

4.71

4.8

4.95

5.15

5.28

5.45

5.68

6.05

10.65

pH

pH

Sheet1

VolumepH

03.12

13.5

23.84

34.05

44.22

54.36

64.48

74.6

84.71

94.8

104.95

115.15

125.28

135.45

145.68

156.05

1610.65

1711.65

1811.92

1912.04

2012.13

Sheet1

Volume of NaOH (mL)

pH


Sheet2

Sheet3

_1379628023.xlsChart1

3.5

3.71

3.92

4.1

4.25

4.42

4.55

4.71

4.87

5.05

5.27

5.6

6.22

12.29

12.7

12.71

12.93

pH

pH

Sheet1

VolumepH

03.5

13.71

23.92

34.1

44.25

54.42

64.55

74.71

84.87

95.05

105.27

115.6

126.22

1312.29

1412.7

1512.71

1612.93

Sheet1

pH

Volume of NaOH (mL)

pH


Sheet2

Sheet3

_1379627240.xlsChart1

3.52

3.83

4.19

4.5

4.65

4.81

5.01

5.18

5.53

6.04

6.06

12.26

12.27

12.71

12.92

12.97

13.01

pH

pH

Sheet1

VolumepH

03.52

13.83

24.19

34.5

44.65

54.81

65.01

75.18

85.53

96.04

106.06

1112.26

1212.27

1312.71

1412.92

1512.97

1613.01

Sheet1

pH

Volume of NaOH (mL)

pH


Sheet2

Sheet3

Lab 1-Acetic Acid in Vinegar Faris(1)

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Transcript of Lab 1-Acetic Acid in Vinegar Faris(1)