NITRATION OF METHYL BENZOATE (ELECTROPHILIC AROMATIC SUBSTITUITION)

Objectives

1)    To prepare and calculate the percentage yield of methyl m-nitrobenzoate by electrophilic aromatic substituition.

2)    To get the melting point of the product.

Introduction

            Aromatic substituition is electrophilic, due to high density in benzene ring.Benzene ring is one of components in most important natural products and other useful products. The species reacting with the aromatic ring is usually a positive ion or the end of a dipole.

            Nitration is one of the most important examples of electrophilic substituition.The electrophile in nitration is the nitronium ion which is generated from nitric acid by protonation and loss of water , using sulphuric acid as the dehydrating agent.The reaction is shown below :

HNO₃ + 2H₂SO₄              NO₂⁺ + H₃O⁺ + 2HSO₄^-

In this experiment we will put nitro group on a benzene ring which already has ester group attached to it.The actual electrophile in the reaction is the nitronium ion , which is generated in the reaction mixture using concentrated nitric acid and concentrated sulphuric acid.The equation has been shown below:

In this experiment , a cool solution of the aromatic ester that has been dissolved in sulphuric acid and nitric acid . This highly exothermic reaction is kept under control by cooling then the mixture is poured into ice.The solid product is isolated by filtration is isolated by filtration and recrystallizationfrom methanol which is very soluble. This is the mechanisms of for nitration of methyl benzoate:

Result and observation

Melting point for 1st= 75˚C - 78˚C

                           2nd= 76˚C - 78˚C

From the literature boling point which is melting point for methyl m-nitrobenzoate whic is 78-80˚C, we can conclude that the product that we get is methyl m-nitrobenzoate.

Mass of methyl benzoate=3.0500g

Mass  product=39.2225 g

Mass filter paper = 0.6963 g

Mass final product = 2.6996 g

The chemical equation :

C₆H₅CO₂CH₃ + H₂NO₃             C₆H₄CO₂CH₃NO₂ + H₂O

Number mole of C₆H₅CO₂CH₃

Molar mass of C₆H₅CO₂CH₃ = 136.1487

Mass of C₆H₅CO₂CH₃ = 3.0500g

Number of mole = 3.0500g / 136.1487

                          = 0.022 mol

1 mol of C₆H₅CO₂CH₃ needed to produce 1 mol of C₆H₄CO₂CH₃NO₂

0.022 mol of C₆H₅CO₂CH₃ needed to produce 0.022 mol of C₆H₄CO₂CH₃NO₂

Molar mass of C₆H₄CO₂CH₃NO₂ = 181.1463

The theoretical yield of C₆H₄CO₂CH₃NO₂ = 0.022 mol x 181.1463 g/mol

                                                                 = 3.9852 g

Actual yield = 38.5262 g

Percentage yield of product = (2.6996 g / 3.9852 g) x 100%

                                             = 67.74%

Discussion

            Nitration is an introduction of nitrogen dioxide into a chemical compound acid . In the process the methyl benzoate was nitrated to form a methyl m-nitro benzoate . The reagents were added very slow to avoid a vigrous reactions and the temperature was maintained low to avoid formation of dinitro product .

            In this experiment , electrophilic aromatic substituitions involved the replacement of a proton on an aromatic ring with an electrophile that becomes substituent. The solvent sulphuric acid protonates the methyl benzoate , creating the resonance stabilized arenium ion intermediate . The electron deficient nitronium ion reacts with the protonated intermidiate meta position . The ester group is the meta deactivator and the reaction takes place at the meta position because the ortho and para positions are destabilized by adjacent positives charges on the resonance structure .The major product is the meta product due to carboxyl and nitro groups both being powerful electron withdrawing groups.

            The actual yield methyl – 3- nitrobenzoate crude product is 2.6996 g while the theoretical yield is 3.9852 g  .The percentage yield that we get is 67.74%.The melting point is 75˚C - 78˚C and 76˚C - 78˚C , the value is closed to the literature value which is 78˚C .

            As we know sulphuric acid are extreamly corrosive and can cause severe burns while nitric acid is one of the strong oxidizing agent so we need to wear gloves while doing an experiment. Methanol is toxic and we need to use in well – ventiled space only for example fume board . All of the chemical material should be discarded properly in the container provided.

Conclusion

The methyl m-nitrobenzoate was prepared. The theoretical yield is 3.9852 g  while tha actual yield is 2.6996 g so we get the percentage yield is 67.74%. The melting point of our product is 75˚C - 78˚C and 76˚C - 78˚C. From the given physical constant we know that the literature melting point of methyl m-nitrobenzoate is 78 - 80˚C , so we can conclude that the product we get is methyl m-nitrobenzoate.

ACETYLSALICYLIC ACID

OBJECTIVE

The purpose of this experiment is to was to prepared acetylsalicylic acid

INTRODUCTION

  +                   +

(salysilic acid)       (acetic anhydride)                    (acetysalicylic acid)       (acetic acid)

            From the chemical equation the acetysalicylic acid can be prepared by the reaction between salysilic acid and acetic anhydride.In this rection , the hydroxyl group from benzene ring was reacted with acetic acid to form ester functional group , so this reaction is referred as esterification reaction.Concentrated sulphuric acid is the catalyst in this experiment which make the reaction is complete.Crystallization process was required in this experiment because there is some impurities such as unreacted salysilic acid and acetic anhydrate presence with the acetysalicylic acid.So the process is needed to makesure there is ni impurities in the last product.

            Thus the separation of acetysalisilic acid from other materials was accomplished and the product also washed with distilled water after the crystal was formed because the water can decrease the solubity of acetysalicylic acid and dissolves the left impurities.

            Recrystallization process was formed to more purify the product.Methanol was used as solvent in order to prevent the decomposition of salicylic acid because water can make aspirin partially decompose while heating the solution.If the aspirin decompose , we cannot get the correct result for this experiment.The impurities which is salicylic acid will decompose from incomplete reaction with acetic anhydrate, so after recrystallization, there is no impurities in the last product.

RESULT AND OBSERVATION

Mass of salicyclic acid = 2.0011 g

Mass of filter paper = 0.5595 g

Mass of watch glass = 59.3232 g

Total mass = 64.9764 g – 59.3232 g – 0.5595 g

                   =5.0937 g

Mass after recrytallization = 4.4081 g

Mass of crude product =5.0937 g - 4.4081 g

                                     = 0.6856 g

C₇H₆O₃ + C₄H₆O₃               C₉H₈O₄ + CH₃COOH

Molar mass of salicylic acid = 138.121 g/mol

Mass of salicylic acid = 2.0011 g

Number of mole of salicylic acid = 2.0011 g / 138 .121

                                                    = 0.001449mol

1 mol of salicyclic acid react with 1 mol of acetic anhydride to produce 1 mol of acetylsalicylic acid

0.001449 mol of salicyclic acid react with 0.001449 mol of acetic anhydride to produce 0.001449  mol of acetylsalicylic acid

Theoretical yield of acetylsalicyclic acid = 0.001449 mol x 180.157

                                                               = 2.610 g

Percentage yield of acetylsalicyclic acid = (0.6856 g / 2.610 g) x 100%

                                                                = 26.29%

Result for ferric chloride test

Test tube A = Colourless solution change to purple solution in colour

Test tube b = colourless solution change to violet solution in colour

Test tube c = There is no change in colour which is colourless solution.

DISCUSSION

            This experiment was performed to prepared acetylsalicyclic acid . As we know acetylsalicyclic acid is a drug that is widely used as an antipyretic agent that is to reduce fever ,as an analgesic agent which is to reduce pain and as an anti-inflammatory.Chemically acetylsalicyclic acid is an ester.Esters are the products of the reaction of acids with alcohols , as shown as from the balance chemical equation below :

+            +

(salysilic acid)       (acetic anhydride)                    (acetysalicylic acid)       (acetic acid)

            We  use an acetic anhydride instead of an acetic acid because the anhydride react with water to form acetic acid. There is the formula to calculate the theoretical yield of the acetysalicyclic acid :

            From the result the theoretical yield for this experiment is 2.610 g while the actual yield is 0.6856 g .the percentage yield we get is 26.29%.There is large different percentage yield of this experiment to 100%.There maybe error occur when the temperature required for on the step of the experiment that is 50˚C , we do not maintained the temperature so the reaction do not completely react.

            There also may occur error when we cut the filter paper , the pen ink may contaminated our product so we need to make sure that we cut all the filter paper which contained the ink.While we are doing experiment we should take care of some precaution.As we know the acetic anhydride can causes severe burns and harmful to inhaled.We should wear gloves when handling the acetic anhydride solution and handle it in the fume board carefully.

The next experiment which is test of ferric chloride purity. From the result that we obtained from the experiment :

Test tube A = Colourless solution change to purple solution in colour

Test tube b = colourless solution change to violet solution in colour

Test tube c = There is no change in colour which is colourless solution.

            Ferric chloride do not react with aspirin . It will react with salicylic  acid which is used to synthesize aspirin . In this light , adding an aqueous ferric chloride solution into a sample of aspirin is a better way to see if there any unreacted salicylic acid which is not react .A purple colour indicate the unreacted salicylic acid , so a purple colour should not exhibit any color change. For test tube A and B salicylic acid was added  so the ferric chloride was reacted with the salicylic acid added while in the test tube C salisylic acid doesn’t added to the test.So there is no impurities exists in the product of the experiment.

CONCLUSION

            Acetylsalicylic acid was prepared . The actual yield we obtained from the experiment is 0.6856 g while the theoretical yield we calculated is 2.610 g.The percentage yield is 26.29%.From the ferric chloride test , we can conclude that there is no impurities in the product of the experiment.

THE PREPARATION OF 1-BROMOBUTANE FROM 1- BUTANOL

Objective ;

The purpose of this experiment is to prepare 1- butanol

Introduction ;

            The most generally uses classes of synthetic organic reactions is nucleophilic substituition. This is a second order nucleophilic substituition, S_N2. As we know that the reaction required a nucleophile , an electrophile and a leaving group in order to apply the experiment in nucleophile second order sustituition .

CH₃CH₂CH₂CH₂-OH₂⁺ + Br^-               CH₃CH₂CH₂CH₂-Br + H₂O

The  mechanism S_N2 shown as below:

            Bromide is an exellent nucleophile and the electrophile is a 1˚ alkyl group, but hydroxide is poor leaving group due to its negative charge and its basicity. We a few choices to make the OH^- become better leaving group. First, we react the alcohol with p-toluenesulfonyl chloride which will convert OH^-  into sulfonic acid ester to react with sodium bromide to produce 1-bromobutane. Second , we react with the alcohol with phosphorus tribromide(PBr₃), which convert the OH^- to “P(OH)X₂” leaving group and also which produces free bromide ions which react with electrophile , replacing the new leaving group .Lastly, we are using a strong acid to protonate the OH^- group in the presence of the bromide ion , which changes the leaving group in the presence of the bromide ion , which changes the leaving the group from hydroxide to water , and allows the bromide to react in the same mixture . In this experiment we are using the third method to prepare the 1-bromobutane.We will separate and purify the product using simple distillation.

            We will carry out a reaction with the specific purpose of making new compound which is a synthetic reaction .It is our responsibility to really understand the experimentcarefully. In this experiment , sodium bromide and 1-Butanol are dissolved in water . Sulphuric acid is added cautiously which generates hydrobromic acid , which turn reacts with the alcohol upon heating to make 1- Bromobutane.

RESULT AND OBSERVATION

Melting point 98 ˚c

Mass of 100-ml round bottom flask=41.6120g

Mass of anhydrous CaCl₂ = 3.3874-2.3858

                                          = 1.0016 g

Mass of 1 – bromobutane = 43.5112 g – 41.6120 g

                           = 1.8992 g

CH₃CH₂CH₂CH₂-OH₂⁺ + Br^-               CH₃CH₂CH₂CH₂-Br + H₂O

Molar mass of 1-butanol = 136.904

Mass sodium bromide = 17.0405 g

No of mol 1-butanol = 17.0405 / 136.904

                               = 0.1245 mol

1 mol of 1-butanol need to react with 1 mol Br^- to produce 1 mol 1-bromobutane

0.1245 mol of 1-butanol need to react with 0.1245 mol Br^- to produce o.1245 mol 1-bromobutane

Theoretical yield = 0.1245 x 136.904

                          = 17.0405 g

Actual yield = 1.8992 g

Percentage yield = (1.8992 g / 17.0405 g) x 100%

                            = 11.15%

DICUSSION

           

            S_N2 mechanisms was shown as above. S_N2 reaction always occur in with inversion configuration at the substrate carbon.The nucleophile approaches the substracte carbon from the back side with respect to the leaving group.In this experiment we are using water as an solvent which called as protic solvent because it has a hydrogen atom which it attached to a strongly to electroneagative element.Hydrogen bonding encumbers a nucleophile and hinders its reactivity in a substituion reaction.

The boiling point of this product is 95˚C-98˚C , so the boiling point is nearly to 1- Bromobutane which has 101˚C as it boiling point. As we know alcohol do not undergo nucleophilic substituition reactions because hydroxide are ion is strongly basic and poor leaving group.The leaving group is is a group of atom which depart  with the electron pair used to bond them with the substrate.However , alcohols readily undergo S_N2 because the sulphuric acid was added which protonate the hydroxyl group in the presence of the bromide ion which chenges the leaving group from hydroxide to water , and allows bromide ions to react with it in the same mixture .

            The sulphuric acid serves as two purposes which is to increases the amount of protonated alcohols present in the reaction mixture and to help tie up the water molecules generated in the reaction shifting the equilibrium in favor of the alkyl bromide .An alternative and more covenient method involves the in situ generation oh hydrobromic acid by the addition of concentrated sulphuric acid to an aqueous soluition sodium bromide.

            From the calculation , the percatage yielid for this experiment is 11.15% which is very large different from 100%. It is because while we are doing an experiment the set up of the distillation apparatus is not correct so small amount of  gas from the 100 ml round bottom flask was release to surrounding so we do not get an accurate an actual mass in this experiment. We also need to wear gloves while doing an experiment because 1- Butanol and 1-Bromobutane both are flammable fluid which ca irritate our skin.Sulphuric acid also very concentrated acid which can effect our skin.Makesure we wash our hand and gloves after handling the substances .

CONCLUSION

            The experiment was involving second order nucleophilic substituition, S_N2.The 1-Bromobutane was prepared from 1-Butanol.The theoretical yield for this experiment is 17.0405 g while the actual yield is 1.8992 g. The percentage yield is 11.15% . the melting point is 95˚C-98˚C nearly to the boiling point of 1-Bromobutane which is 101˚C , so the product in this experiment is 1-Bromobutane

ANALYSIS OF AN UNKNWON SAMPLE(analytical chemistry)

Abstract :

One of the purpose of this experiment is to prepare standard solution of sodium hydroxide.First we determined the volume of NaOH  solution that we should use.We used 10 mL of NaOH solutionWe add 300 mL of distilled water into plastic bottle followed by 10 mL of NaOH solution.Then we mix the solution and continuously adding 100mL distilled water into the plastic bottle continuously.The standard solution was used for the next experiment.

Second purpose of this experiment isto standardise the base against Potassium Hydrogen.We used titration method to determined the molarity of sodium hydroxide.First we weigh the KHP and transfer the sample into the 250 mL conical flask.Then 35 of distilled water was added and swirl the flask to make sure them dissolve in the solution.Using the burette, tittrate sodium hydroxide solution into the conical flask.The onitial and final volume was recorded.From the experiment we have determined the molarity of sodium hydroxide is  0.0002 M.

To analyse the unknown in vinegar sample is one of the objective of this experiment.10 mL of vinegar sample was pipetted into clean 250 mL conical flask and the add 25 mL of distilled water.The sodium hydroxide was placed into burette and the initial reading was recorded.Phenolpthalein was added into the conical flask and tittrate the sodium hydroxide into the conical flak until it achieve the end point.The percentage of acetic acid in vinegar was determined.It is 3.02 % .

Objective :

The purpose of this experiment is to prepare the sodium hydroxide solution.Next to standardise the base against Potassium Hydrogen Phthalate(KHP) and to analyse the unknown the vineger sample.

Introduction :

            In this experiment we will learn about primary standard, standardisation and standard solution. Primary standard is a reagent which are very pure , representative of the number of moles the substance contains and easily weighed , for example sodium chloride.Next , in definition standardisation is the condition in which a standard has been succesfully established.While standard solution is solution of accurately known concentration prepared from a primary standard that is weighed accurately and made up to a fixed volume.

            Next process is standardization,sodium hydroxide is not primary standard because it is hydgroscopic.The definition of hygroscopic is material which attract so much water that they will form solutions.Thus accurate solution ca be determined by standardising the solution against a very pure potassium hydrogen phthalate(KPH). Sodium hydroxide is deliquescent so it is not possible to prepare a standard solution of sodium hydroxide by weighing NaOH. Its concentration is then determined by titrating it against a solution of the primary standard, KHP. Standardization of sodium hydroxide is important to determine the acetic content acid of a vinegar sample.

Potassium hydrogen phthalate, KHC₈H₄O₄ is a non-hygroscopic, crystalline, solid that behaves as a monoprotic acid. It is water soluble and available in high purity. Because of its high purity, we can determine the number of moles of KHP directly from its mass and it is referred to as a primary standard. We use the primary standard to determine the concentration of a sodium hydroxide solution.

The acetic acid content of a vinegar may be determined by titrating a vinegar sample with a solution of sodium hydroxide of known molar  concentration . Phenolphthalein will be used as an indicator because it will be colorless before the completion of the reaction but pink after the completion. Phenolphthalein, an organic dye, is colorless in an acid solution and pink in a basic solution. We need to be prepared to search carefully for a point in the titration at which one drop of the NaOH solution will cause the solution being titrated to turn from colorless to a barely discernible pink color. This point is called the end point and indicates the reaction is complete.

           

Procedure :

A)Preparation of the sodium hydroxide solution.

            Firstly ,  300 ml of distilled water was measured using measuring cylinder and placed into the plastic bottle.Placed 10 ml of NaOH stock solution into the plastic bottle.The measuring cylinder was rinsed using distilled water.Then the solution was added into the plastic bottle.The cap was screwed and vigorously swirling the plastic bottle repeatedly.Next , 100ml of distilled water was added into the plastic bottle , so that the contents was mixed throughly each time.Another 100 ml was added into the same bottle and mix.Lastly 100 ml of distilled water also added into the same plastic bottle.The bottle was shaked more than 20 times after the last addition distilled water.

B)Standardisation of the base against Potassium Hydrogen Phthalate.

            1 g sample of dry primary – standard grade potassium hydrogen phthalate(KPH) was weighed using analytical balances.KHP had been dried ealier in an oven at 110˚c for 2 hours and stored in a desiccator prior to use.1 g sample above was used as appearance as a guide accurately to weigh two more sample by difference.The sample was transferred into weighing boat into 250 mL conical flask .The weighing boat was rinsed and solution was poured into the conical flask.35mL of water was measured using measuring cylinder and was placed into the conical flask and the flask was swirled until  the solid was dissolved in the liquid.The side wall of conical flask was rinsed using the distilled water.

            50 mL burette was rinsed using distilled water and was filled with NaOH  stock solution the we had been prepared . Add three drops phenolpthalein into the conical flask.The conical flask was place under the burette and lower the tip well  into it.Start the titration.The stopcock was control gentlely so  the NaOH slowly flow into the conical flask.The solution was gently swirled and  the first permanent pink colour appear red close the stopcock and the reading was taken.Repeat all the step three times and the result was recorded.

C) Analysis of the unknwon  vinegar sample

            10.0mL vinegar  was pipetted into dry pre-weigh 50mL beaker and the beaker was reweigh back.More vinegar was added into the beaker.10.0mL of vinegar sample was  pipetted from the beaker into the conical flask.Three more same solution was also prepares and the side of each conical flask was washed using 25 mL of distilled water.The pipette was filled with the NaOH stock solution. Three drops of phenolpthalein was added into each conical flask and the solution was titrated until it achieve the end point.The titration was repeated for other three solution in the other conical flask.

Data / Result :

EXPERIMENT 2

ANALYSIS OF AN UNKNOWN VINEGAR SAMPLE

A)Preparation of the Sodium Hydroxide Solution

Volume of NaOH taken from the 50% stock solution = 10 mL

B)Standardisation of the base against Potassium Hydrogen Phthalate

	Rough	1	2	3
Weight of KHP	1.0221 g	1.0090 g	1.0388 g	1.0146 g
Final reading of NaOH	24.9 mL	24.4 mL	25.9 mL	25.3 mL
Initial reading of NaOH	50 mL	50 mL	50.0 mL	50 mL
Volume of Naoh(mL) used	25.1 mL	25.6 mL	24.1 mL	24.7 mL
Ratio volume of NaOH/Weigh of KHP	24.56	25.37	23.20	24.34

calculation

There is the equation involving the experiment.

The chemical equation is :

KHC₈H₄O_{4 (aq)}  + NaOH_(aq) à KNaC₈H₄O_{4 (aq)}  +  H₂O_(l)

The net ionic equation is:

HC₈H₄O₄^-1_(aq)  + OH^-_(aq) à C₈H₄O₄^-2 _(aq)  +  H₂O_(l)

Molar mass KHP = 204.23 g/mol

There is formula on calculating of KHP

There is formula on calculating NaOH

C)Analysis of the unknown vinegar sample

 

1)ROUGH

Calculating molarity of KHP

=1.0221 g( 1 mol KHP / 204.23 g) = 0.005 mol KHP

=(0.005 mol KHP / 35 ml H₂O) =0.000143M(mol/L) KHP

Calculating molarity NaOH

=0.005 mol KHP ( 1 mol NaOH / 1 mol KHP) = 0.005 mol NaOH

=(0.005 mol NaOH / 25.1 mL) = 0.0002 M(mol/L) NaOH

2)TITRATION 1

Calculating molarity of KHP

=1.0090 g( 1 mol KHP / 204.23 g) =0.005 mol KHP

=(0.005 mol KHP / 35 ml H₂O) =0.000143M(mol/L) KHP

Calculating molarity NaOH

=0.005 mol KHP ( 1 mol NaOH / 1 mol KHP) = 0.005 mol NaOH

=0.005 mol NaOH /25.6 mL) = 0.0002 M(mol/L) NaOH

3)TITRATION 2

Calculating molarity of KHP

=1.0388 g( 1 mol KHP / 204.23 g) =0.005 mol KHP

=(0.005 mol KHP / 35 ml H₂O) =0.000143M(mol/L) KHP

Calculating molarity NaOH

=0.005 mol KHP ( 1 mol NaOH / 1 mol KHP) = 0.005 mol NaOH

=0.005 mol NaOH /24.1 mL) = 0.0002 M(mol/L) NaOH

4)TITRATION 3

Calculating molarity of KHP

=1.0146 g( 1 mol KHP / 204.23 g) =0.005 mol KHP

=(0.005 mol KHP / 35 ml H₂O) =0.000143M(mol/L) KHP

Calculating molarity NaOH

=0.005 mol KHP ( 1 mol NaOH / 1 mol KHP) = 0.005 mol NaOH

=0.005 mol NaOH /24.7 mL) = 0.0002 M(mol/L) NaOH

c)Analysis of the Unknwon Vinegar Sample

Density of vinegar sample = 0.99249 g/mL

	Rough	1	2
Volume of unknown vinegar(mL)	10 mL	10 mL	10 mL
Final reading of standard NaOH	15.5 mL	14.9 mL	15.1 mL
Initial reading of standard NaOH	50 mL	50 mL	50mL
Volume of standard NaOH used	34.5 mL	35.1 mL	34.9 mL

The chemical equation

CH₃COOH + NaOH             CH₃COO^- Na⁺ + H₂O

Calculation :

Moles NaOH = 0.005 mol

Molar mass acetic acid=60.02

From chemical equation

1 mole of CH₃COOH need 1 mole of NaOH to produce 1 mole CH₃COO^- Na⁺ and H₂O

0.005 mole of CH₃COOH need 0.005 mole of NaOH to produce 0.005 moleCH₃COO^- Na⁺ and H₂O

Mass of  acetic acid = 0.005 x (60.02)

                                   = 0.3001 g

Density of vinegar

W₁=32.1275 g

W₂=42.0524 g

W₁-w₂ =9.9249 g

=9.9249 / 10.0

=0.99249 g/mL

Percentage of acetic acid in the solution is = ( 0.3001 g / 9.9249 g) x 100

                                                                           = 3.02 %

Discussion

            From the first experiment the preparation sodium hydroxide solution.As we know sodium hydroxide cannot be used as a primary standard as it really absorbs water and carbon dioxide from the air.So in order to determine the concentration of the sodium hydroxide solution through titration we use KHP because it is non hygroscopic which is not react with any of the components. KHP is an organic acid that is solid

thus, easily weighed. In a titration the volume of a solution added to a reaction is measured using a buret. A buret is a long tube with a stopcock at one end that can be used to control the flow. Burets are typically calibrated in milliliters.

            The KHP and NaOH are reacted together until one of the two is completely reacted. That point is called the endpoint that is neutral. If additional base is added, the solution will then become basicidic depending on which was added in excess.We cannot visually determine end point without using the indicator. To determine the endpoint, an indicator is added to the reaction. An indicator is a chemical that changes colors at a particular pH. When just a tiny excess of the acid or base is added beyond the completion of the reaction, the indicator changes color. The amount added from the buret at this point is called the endpoint.In this experiment we are using phenolpthalein indicator , which the colour change to pink when the solution is basidic.The first pink colour appear is the end point.There is the chemical equation for the experiment :

KHC₈H₄O_{4 (aq)}  + NaOH_(aq) à KNaC₈H₄O_{4 (aq)}  +  H₂O_(l)

From above chemical equation we know that ;

1 mole of KHC₈H₄O₄ need 1 mole of NaOH to produce 1 mole KNaC₈H₄O₄ and H₂O

0.005 mole of KHC₈H₄O₄ need 0.005mole of NaOH to produce 0.005 mole KNaC₈H₄O₄ and H₂O

Using the number of mole we can get the molarity of NaOH used in this experiment that is 0.0002 M(mol/L) NaOH.

Next the analysis of the vinegar sample. Vinegar is a solution of acetic acid in water. Acetic acid, CH3COOH, is a weak monoprotic acid.there is the chemical equation : CH₃COOH + NaOH                 CH₃COO^- Na⁺ + H₂O

From the above equation we can determine that number of mole of acetic acid by using stoichiometry . We had been calculate the number of mole of sodium hydroxide from the part b experiment.

1 mole of CH₃COOH need 1 mole of NaOH to produce 1 mole CH₃COO^- Na⁺ and H₂O

0.005 mole of CH₃COOH need 0.005 mole of NaOH to produce 0.005 moleCH₃COO^- Na⁺ and H₂O

So by using the number of mole we can get the mass of acetic acid , that is  0.3001g.

There is some precaution in handling experiment.First wear lab coat in all times while working in the laboratory.Both acids and bases can be corrosive to human tissue.When it concentrated it can break down the human tissue.As we know sodium hydroxide are very corrosive and cannot contact directly to our hand, so we can use spatula to handle the substance.The lab coat can protect our body  when handling the chemical. If the acid or bases get into our eyes , flush it out immedietly with lots of water until it the chemical was removed.It ts the same , when other place part of our body was spilled with chemical , flush the area with water and the sink or use the safety shower in the lab.

Conclusion :

            The standard solution of sodium hydroxide has been prepared.From the second experiment the molarity of sodium hydroxide has been determine that is 0.0002 M(mol/L) NaOH.The percentage of acetic acid in the vinegar solution is 3.02%.