Lab Report 1.docx

TABLE OF CONTENT

NO TITLE PAGE

1 Abstract 2

2 Introduction 3

3 Aims 4

4 Theory 5

5 Apparatus 7

6 Procedure 8

7 Results and Calculations 9

9 Discussion 18

10 Conclusion 20

11 Recommendations 20

12 Reference 20

13 Appendix 21

1

ABSTRACT

Vinegar is a common use especially in cooking. Vinegar is diluted solution of acetic acid.

The molecular formula of acetic acid is CH3COOH. In this experiment, vinegar has been

titrated with NaOH solution. The percent by mass of acetic acid in vinegar is determined by

titration and several calculations involved to determine the molarity of acetic acid present in

vinegar together its mass. From the titration curve, student is able to determine the

equivalence point of the reaction.

2

INTRODUCTION

A solution is a homogenous mixture composed of only one phase. In such mixture, a solute

is a substance dissolved in another substance, known as solvent. The solvent does the

dissolving. The solution more less takes on the characteristics of the solvent including its

phase and the solvent is commonly major fraction of the mixture. The concentration of a

solute in a solution is a measure of how much of that solute dissolved in the solvent.

Concentration of solution is also can be expressed in term of molarity and percent by mass.

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

Molarity (M )=moles of solute (mol )liter of solution(L)

(Equation 1.1)

Percent by mass is the mass in grams of solute per 100 grams of solutions

Percent solute= grams of solutegramsof solution

x 100 % (Equation 1.2)

Acetic acid is a weak acid with molecular formula CH3COOH. Vinegar is the dilute solution

of acetic acid. The molarity and percent by mass of acetic acid in vinegar solution can be

determined by performing titration. Titration is the process of determining the concentration

of a solution by adding to its standard reagent of known concentration in carefully measured

amounts until a reaction of definite and known proportion is completed, as shown by a

colour changes or electrical measurement, and next calculating the unknown 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.

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OBJECTIVES

The aim in this experiment is to determine the molarity and percent by mass of acetic acid in

vinegar by titration with standardized sodium hydroxide, NaOH solution.

4

THEORY

In titration process, a burette is used to dispense a small, quantifiable increment of solution of

known concentration. A typical burette has the smallest calibration unit of 0.1mL , therefore

the volume dispensed from the burette should be estimated to the nearest 0.05mL.

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 1.3

NaOH(aq) + CH3COOH(aq) →NaCH3CO2(aq) + H2O (Equation 1.3)

The sudden change in the solution pH shows 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=−log10 [H 3O ]+¿¿ ( Eqution 1.4)

pH is the measure of acidity or basicity of an aqueous solution. Solutions with a pH less than

7 are said to be acidic and solutions with a pH greater than 7 are basic or alkaline. Pure water

has a pH very close to 7. In this experiment, pH electrodes will be used. The titration 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 ions will be neutralized. when

the concentration of the hydrogen ion is decreases, the pH of the solution will increase. When

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NaOH is added until the acid is neutralized, most of the H3O+ ions are removed from the

solution. The next volume of NaOH that added in the solution will cause a sudden increase in

pH. The volume of based required to completely neutralized of acid is determined from the

equivalence point of titration curve.

In this experiment, vinegar will be titrated with standardized sodium hydroxide solution. The

sodium hydroxide solution is standardized by a primary standard acid solution that is initially

being prepared. Basically, primary standard acid solution is produces by dissolving a weigh

quantity of pure acid or base in a known volume of solution.

Primary standard acid or bases have a several common characteristics to be classified. Firstly,

they must be available in at least 99.9 purities. Second, they must have a high molecular

molar mass to minimize error in weighing. Third, they must be stable upon heating and must

be soluble in the solvent of interest.

Potassium hydrogen phthalate (KHP), KHC8H4O4, and oxalic acid, (COOH)2, are common

primary standard acid. Sodium carbonate (Na2CO3) is the commonly use base. Besides that,

HCI, CH3COOH, NaOH and KOH are also several others primary standard form. To

standardize one of these acidic or basic solution, they must be titrated with primary standard

form. In this experiment, NaOH solution will be titrated with KHP. The reaction can be form

in this equation:

KHC8H4O4 (aq) + NaOH (aq) → KNaC8H4O4 (aq) + H2O(l) Equation 1.5

Once the NaOH has been standardized it will be titrated with aliquots of vinegar. The

reaction equation of vinegar with NaOH is :

CH3COOH(aq) + NaOH (aq) ) →Na CH3COO(aq) + H2O(l) Equation 1.6

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APPARATUS

Burette, pH meter, retort stand, conical flask, 250 mL beaker, weighing balance, magnetic stirrer, 10 mL volumetric pipette, measuring cylinder.

Chemicals

Sodium hydroxide, NaOH solid; potassium hydrogen phthalate, KHP; distilled water, vinegar.

Figure 1 Apparatus set up for titration process

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PROCEDURE

A) Standardization of sodium hydroxide solution

1. A 250 ml of 0.6M NaOH is prepared in beaker from NaOH pellets.

2. Next, KHP is weighed for 1.5 grams in beaker which mass of the KHP is nearest to 0.001g.

Then, 30ml of distilled water is added into the beaker. The solution is stirred until it

completely dissolved.

3. The solution of KHP is the titrated with NaOH. The pH changes for every 1ml additions of

NaOH is added.

4. Steps 1 until 3 is repeated and two more solution of KHP are prepared for NaOH

standardization..

5. The graph of pH versus NaOH is plotted. From the graph, the volume of NaOH needed to

neutralized the KHP solution for each titration.

6. The molarity of NaOH for each titration also has been calculated.

7. The average molarity of NaOH solution is calculated too.

B. Molarity of acetic acid and mass percent in vinegar

1. 10ml of vinegar is transferred to a clean , dry 250ml beaker by using a 10ml volumetric

pipette. A sufficient water about 75ml to 100ml is added to cover the pH electrode tip during

the titration.

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

3. The steps are repeated twice more.

4. The graph of pH versus NaOH is plotted. From the graph, the volume of NaOH needed to

neutralized the vinegar solution for each titration.

5. The molarity of acetic acid in vinegar for each titration also has been calculated.

6. The average molarity of acetic acid solution is calculated too.

7. Next, the calculation of percent by mass of acetic acid in vinegar is also be performed for

the three titrations respectively.

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RESULTS

1) Table of KHP titrated to NaOH

Titration 1 Titration 2 Titration 3

Average

Mass of KHP (g) 1.499 1.5011 1.499Volume of NaOH(mL)

pH

0 3.62 3.76 3.64 3.671 3.91 3.9 3.9 3.902 4.08 4.09 4.08 4.083 4.26 4.28 4.28 4.274 4.33 4.37 4.3 4.335 4.52 4.5 4.52 4.516 4.65 4.63 4.66 4.657 4.77 4.77 4.78 4.778 4.9 4.99 4.99 4.969 5.09 5.05 5.11 5.08

10 5.24 5.25 5.29 5.2611 5.63 5.49 5.59 5.5712 6.14 5.97 6.27 6.1313 11.32 11.29 11.55 11.39

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2) Graph of KHP titrated with NaOH

10

0 1 2 3 4 5 6 7 8 9 10 11 12 130

2

4

6

8

10

12

Titration 1: pH KHP Vs Volume NaOH

Volume of NaOH(ml)

PH

0 1 2 3 4 5 6 7 8 9 10 11 12 130

2

4

6

8

10

12


Volume of NaOH (ml)

pH

0 1 2 3 4 5 6 7 8 9 10 11 12 130

2

4

6

8

10

12

14


Volume NaOH (ml)

pH

12.5 mL NaOH at equivalence point

12.5 mL NaOH at equivalence point

12.5mL NaOH at equivalence point

3) Calculations

Standardization of sodium hydroxide solution

1. Calculations for preparing 250mL of approximately 0.6M NaOH solution

1M of 1L NaOH → 40 g/mol NaOH

0.6 M of 1L NaOH →0.6M1M

∗40 g NaOH=24 g NaOH

0.6M of 250ml NaOH→0.25 L

1L∗24 gNaOH=6 g NaOH

3. Moles and molarity NaOH

i) Titration 1

Moles of KHP used

= 1.499 gKH C8H 4O4×1mol KH C8H 4O4

204.2 g KHC8H 4O 4

=0.007340g KH C8H 4O4

Moles of NaOH required to neutralized the moles of KHP

= 0.007340mol KHP×1mol NaOH1mol KHP

=0.007340mol NaOH

Molarity of NaOH solution

=12.50mLNaOH ×1 L

1000mL=0.01250L NaOH

M = molof NaOHLof Solution

=0.007340mol NaOH0.01250 LNaOH

=0.5872mol NaOHLsolution

=0.5872M

ii) Titration 2

Moles of KHP used

= 1.5011 g KHC8H 4O 4 x1mol KHC8H 4O 4

204.2 g KHC8H 4O 4

=0.007351g KH C8H 4O4


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= 0.007351 gmol KHP x1mol NaOH1mol KHP

=0.007351mol NaOH

Molarity of NaOH

= =12.50mLNaOH ×1 L

1000mL=0.01250L NaOH




=0.5880M

iii) Titration 3

Moles of KHP used

= 1.499 gKH C8H 4O4×1mol KH C8H 4O4

204.2 g KHC8H 4O 4

=0.007340g KH C8H 4O4


= 0.007340 gmol KHP×1mol NaOH1mol KHP

=0.007340mol NaOH

Molarity of NaOH

=12.50mLNaOH ×1 L

1000mL=0.01250L NaOH




=0.5872M

Average molarity of NaOH

=Titration1+Titration2+Titration3

3

= 0.5872+0.5880+0.5872

3

= 0.5875 M

Error = Theoriticaldata−Obtained data

Theoriticaldata×100 %

12

= 0.6−0.5875

0.6 x 100%

= 2.08%

B) Molarity of acetic acid and mass percent in vinegar

1. Table of data

Titration 1 Titration 2 Titration 3 AverageVolume of NaOH(mL) ph

0 2.50 2.55 2.40 2.481 3.06 3.05 3.01 3.042 3.20 3.19 3.25 3.213 3.35 3.34 3.29 3.334 3.59 3.56 3.50 3.555 3.76 3.75 3.69 3.736 3.87 3.80 3.88 3.857 3.99 3.98 3.99 3.998 4.09 4.11 4.10 4.109 4.18 4.20 4.17 4.18

10 4.27 4.29 4.25 4.2711 4.36 4.39 4.38 4.3812 4.45 4.47 4.45 4.4613 4.55 4.56 4.57 4.5614 4.65 4.60 4.64 4.6315 4.76 4.70 4.73 4.7316 4.83 4.87 4.85 4.8517 4.90 4.95 4.93 4.9318 5.04 5.08 5.02 5.0519 5.17 5.19 5.11 5.1620 5.22 5.25 5.23 5.2321 5.43 5.49 5.42 5.4522 5.56 5.58 5.60 5.5823 6.47 6.47 6.44 6.4624 11.04 11.13 11.11 11.0925 11.19 11.21 11.18 11.19

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2. Graph of vinegar titrated with NaOH

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0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 250

2

4

6

8

10

12

Titration 1: pH CH3COOH Vs. Volume of NaOH

Volume of NaOH (ml)

pH C

H3CO

OH

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 250

2

4

6

8

10

12


Volume of NaOH (ml)

pH C

H3CO

OH

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 250

2

4

6

8

10

12


Volume of NaOH (ml)

pH C

H3CO

OH

23mL NaOH at equivalence point

23.5mL NaOH at equivalence point

23.5mL NaOh at equivalence point

3. Calculations

i) Titration 1

Moles of NaOH reacted

23mLof NaOH ×1L

1000mL=0.023L NaOH

0.023 LNaOH ×0.5875M1 LNaOH

=0.0135mol NaOH

Molarity of the CH3COOH solution

10mLof C H 3COOH×1L

1000mL=0.010 LC H 3COOH

M = molC H 3COOH

Lof solution=

0.0135molC H 3COOH

0.01L solution=1.35MCH 3COOH

Mass of acetic acid in the solution


1000mL=0.010 LC H 3COOH

0.010 LC H 3COOH ×0.0135molC H 3COOH

1Lsolution×

60.06 gC H 3COOH

1molC H 3COOH=0.00811 gC H 3COOH

Theoretical mass of acetic acid in vinegar

10mLof C H 3COOH×1 gC H 3COOH

1mLC H 3COOH = 10.00 g C H 3COOH

Percent by mass of acetic acid in the solution

15

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 250

2

4

6

8

10

12


Volume of NaOH (ml)

pH C

H3CO

OH

percent mass C H 3COOH=gC H 3COOH

gC H 3COOH solution×100%

percent mass C H 3COOH=0.00811 gC H 3COOH

10.00gC H 3COOH solution×100%

= 0.0811 % C H 3COOH

ii) Titration 2


23.5mLof NaOH ×1 L

1000mL=0.0235L NaOH

0.0235 LNaOH ×0.5875M1 LNaOH

=0.0138mol NaOH



1000mL=0.010 LC H 3COOH

M = molC H 3COOH

Lof solution=

0.0138molC H 3COOH




1000mL=0.010 LC H 3COOH


1Lsolution×

60.06 gC H 3COOH



16






percent mass C H 3COOH=0.00828gC H 3COOH


= 0.0828 % C H 3COOH

iii) Titration 3


23mLof NaOH ×1L

1000mL=0.0235L NaOH

0.0235 LNaOH ×0.5875M1 LNaOH

=0.0138mol NaOH



1000mL=0.010 LC H 3COOH

M = molC H 3COOH

Lof solution=

0.0138molC H 3COOH




1000mL=0.010 LC H 3COOH

17


1Lsolution×

60.06 gC H 3COOH








percent mass C H 3COOH=0.00828gC H 3COOH


= 0.0828 % C H 3COOH

Average mass of C H 3COOH

=Titration1+Titration2+Titration3

3

= 0.081+0.0828+0.0828

3

= 0.0822 g C H 3COOH

DISCUSSION

According to Marie (2013), a chemical reaction is set up between a known volume of a

solution of unknown concentration and a known volume of a solution with a known

concentration in titration. Titration technique is applied in this experiment to achieve the

objective, which is to determine the concentration of acetic acid in vinegar. So, in this

experiment the titration occurs between NaOH solution, KHP and acetic acid in vinegar. In

titration, the titrant is titrating to the analyte using the burette. Thus, the NaOH solution acts

as titrant in this experiment while KHP and vinegar act as analyte. Through this experiment

the pH of the solution was detected by pH meter. Then, when the result was obtained, the

titration curve of acid-base was constructing. Then, the equivalent point is obtained through

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the titration curve. According to Clark (2002), this equivalent point is then was used to

perform the calculation to obtain the molarity and percent by mass.

In this experiment, the exact concentration of 0.6 M sodium hydroxide prepared is

determined. The sodium hydroxide solution was prepared using the NaOH solid. From this

experiment, the equivalence point is used to achieve the objectives of this experiment. In

standardization of sodium hydroxide solution, the equivalent point for Titration 1 is 12.5 mL

while 12.5 mL for Titration 2 and same goes to Titration 3.. So from there, the molarity of

sodium hydroxide is determined which is, 0.5875M NaOH compared to the molarity given

which is 0.6 M NaOH and the percentage error is 2.08% CH3COOH. Thus, we can conclude

there are some error occurs regarding this experiment due to the slightly difference in

percentage error.

Then, the molarity of acetic acid and percent by mass in vinegar was determined. In this

experiment, the equivalence point used to perform the calculation for Titration 1 is 23 mL

while 23.5 mL for Titration 2 and Titration 3. Thus, the average molarity of acetic acid

among them was obtained which is 1.37M CH3COOH. Next, the percent by mass in vinegar

is 0.0811 % CH3COOH for Titration 1, 0.0828% for Titration 2 and 3 respectively.

Precaution must be taken in every experiment to get an accurate and better result. So, for

the experiment of determination of the concentration of acetic acid in vinegar, there are some

precautions taken to avoid an error and accident during experiment. Firstly, safety gloves and

goggles were wearing when handling the chemicals to avoid any reaction of chemical with

hands or eyes. Next, the reading of the measurement of volume of solution was read

perpendicular with our eyes to avoid any parallax error during taken the reading. It means the

eyes must get at the same level of meniscus. It was applied when the reading of burette and

measuring cylinder taken. In addition, the bubbles in the burette tip must be removed to keep

away from an error in result. It can be done by tapping the side of the burette while the

stopcock is open. However, the contamination and impurities must be avoided for a better

result. This is can be passing up with the proper treat when handling glassware like beaker,

flask and burette. All the glassware must be washed properly with distilled water to keep

away from contamination.

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Moreover, in order to prepare the sodium hydroxide solution from sodium hydroxide solid

and distilled water, the mixture must be stir completely. In addition, the mixture of acid and

base in this experiment also must be stir completely. This precaution must be done to ensure

complete mixing between two solutions. Due to this, it also can affect the pH of the solution.

Besides that, the precaution in using the pH meter must be considered in this experiment

since pH meter is used to detect the pH of the solution. Firstly, the pH electrode must be

cleaned properly after use the pH meter. Next, always keep the pH electrode immersed in the

“storage” solution when not in used. This all must be done due to the accurate or better result

of experiment. If not, the reading is slightly affected by this error.

CONCLUSION

In conclusion, the molarity of a solution is determined, which are 0.3799 M NaOH and 1.081

M CH3COOH. Moreover, the percent by mass of acetic acid in vinegar by titration with the

standardized sodium hydroxide solution is 6.493 %.

RECOMMENDATION

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Future recommendation is needed in every experiment for a future research regarding to this

experiment. Firstly, different molarity of NaOH solution can be used and different percent by

mass of acetic acid in vinegar can be obtained by titration with the standardized sodium

hydroxide solution. Next, the acid and base used in this experiment can be changed with

other types of acid and base as the titrant and analyte in titration process. Moreover, the

indicator can be used to obtain the pH of the solution replacing the pH meter. Indicator is a

special chemical that change colour if there is a change in the pH caused by adding an acid or

alkali. So, the result obtained will be compared between the uses of pH meter n indicator.

Besides that, other types of titration curve can be determined for future recommendation.

Different types of acid-base titration indicate different types of titration curve. For example,

strong acid-strong base, strong acid-weak base, weak acid-strong base, and weak acid-weak

base. Lastly, addition of number of titration can be done. Titration 3 and Titration 4 can be

added for a better result.

REFERENCE

1. Clark, J. (2002). pH (Titration) Curves. Retrieved from www.chemguide.co.uk

2. Marie, H. (2013). Titration Basics. Retrieved from chemistry.about.com

3. Standardization of sodium hydroxide solution. Retrieved from chemtech.org

4. Titration of the Weak Acid Potassium Hydrogen Phthalate (KHP). Retrieved from

chemlab.truman.edu

5. Chang, R. (2010). Acid-Base Titrations. Chemistry 10th Edition (pp 723-732)

21

http://chemlab.truman.edu/

http://chemtech.org/

http://www.chemguide.co.uk/

APPENDIX

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