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%.