DETERMINATION ACETIC ACID

November 11, 2018 | Author: ameyakem | Category: Ph, Titration, Sodium Hydroxide, Molar Concentration, Mole (Unit)
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DETERMINATION ACETIC ACID...

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UNIVERSITI TEKNOLOGI MARA FAKULTI KEJURUTERAAN KIMIA ENGINEERING CHEMISTRY LABORATORY I (CHE485) NAME STUDENT NO. GROUP EXPERIMENT

: : : :

DATES PERFORM SEMESTER PROGRAMME/ CODE SUBMITT TO

: : : :

MOHAMMAD AMIR HAKIM BIN RUSLI 2014294432 GROUP 4 DETERMINATION DETERMINATION OF THE CONCENTRATION OF ACETIC ACID IN VINEGAR SOLUTION 3 OCTOBER 2014 3 EH22O PN. RAFEQAH BINTI RASLAN

Remarks:  No.

Title

Allocated Marks (%)

1

Abstract/Summary

5

2

Introduction

5

3

Objective/Aims

5

4

Theory

5

5

Apparatus

5

6

Methodology/Procedure

10

7

Results

10

8

Calculations

10

9

Discussion

20

10

Recommendations

5

11

Conclusion

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12

Reference / Appendix

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13

Supervisor’s grading

10

TOTAL MARKS

Marks

100

Checked by:

Rechecked by:

--------------------------Date :

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TABLE OF CONTENT

i. abstract ii. Introduction iii. Aims/ objectives iv. Theory v. Apparatus vi. Procedure vii. Result & calculation viii. Sample error calculation ix. Discussion x. Conclusion xi. Recommendation xii. References xiii. Appendices

ABSTRACT

The objective of this experiment is to determine the molarity of a solution and the percent by mass of acetic acid in vinegar by titration with a standardized sodium hydroxide solution. The first experiment, 250 mL of approximately 0.6M sodium hydroxide was perepared from 3.6 grams NaOH solid in a beaker. Then, measured 1.5 grams of KHP was added into another  beaker and 30 mL of distilled water then was added into the beaker and the solution was stirred until KHP dissolved completely. The solution then titrated with NaOH solution and pH with every 1 mL of NaOH solution added was recorded until we reach equivalence point. The steps were repeated with another two solutions for NaOH standardization. Then, the graph of pH versus NaOH was plotted to get volume of NaOH required to neutralize KHP solution for each titration. Molarity of sodium hydroxide for titrations 1, 2 and 3 was calculated. Then, the average molarity of sodium hydroxidewas calculated. For the second experiment which is determination of molarity of acetic acid and mass percent in vinegar, firstly, 10.00 mL of vinegar was transferred into a clean and dry 250 mL beaker by using a 10 mL volumetric pipette. Then, 80 mL of sufficient water was added to cover the pH electrode tip during the titration. Then, 2 mL of NaOH was added into the vinegar solution and the pH of the solution was recorded. pH reading was recorded for every NaOH solution added into the solution until we reach the equivalence point. The steps was repeated twice more. Graph of pH versus NaOH volume added was plotted to get volume of NaOH required to neutralize the vinegar in each titration. The data was recorded. Molarity of acetic acid in vinegar, average molarity of acetic acid and percent by mass of acetic acid in vinegar for titrations 1, 2 and 3 was calculated. The percent by mass also was calculated. For the first experiment, volume of NaOH solution required to neutralize KHP solution in titration 1 is 11.9mL while the titration 2 need 11.8mL . For the second experiment, volume of NaOH solution required to neutralize vinegar solution in titration 1 is 28.5mL while the other titrations need 29.9mL.The conclusion for this experiment is the objective was achieved. From the experiment, we get the average molarity of sodium hydroxide is 0.620 M, the average molarity of acetic acid is 1.8104 M and the average percent by mass of acetic acid is 10.87 % .

INTRODUCTION Concentration of solution is an amount of solute in a given amount of solvent. In a solution, the solvent is the major component while the solute is the minor component. A concentrated solution contains large quantity of solute in a given amount of solvent and vice versa for dilute solution. There are two specific terms to express concentration, molarity and percent by mass. Molarity is the number of moles of solute per volume of solution (in L)

        ( ) Percent by mass is the mass of solute (in g) per 100 grams of solution

   ( )         ( ) Vinegar is dilute solution of acetic acid. The molecular formula for acetic acid is CH3COOH. The molarity and percent by mass of acetic acid in vinegar can be determine by 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 reaction is attained. Knowing the quantity of the known solution required to complete a 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.

OBJECTIVE 

To determine the molarity of the solution and p ercent by mass of acetic acid in vinegar by titration with a standardized sodium hydroxide solution.

THEORY

A titration mixes two solutions which contain reactants for a known chemical reaction under conditions such that: a) the point at which both reactants have been completely consumed by the known reaction can be detected (end point)  b) the amount of one reactant can be calculated from the known concentration of reactant in a standard solution, the volume of standard solution used, and the balanced known chemical equation.

In this experiment, the equivalence point occurs when the moles of acid in the solution equals the mole of base added in the titration. For example, the stoichiometry amount of 1 mole of the strong base, sodium hydroxide (NaOH), is necessary to neut ralize 1 mole of the weak acid, acetic acid (CH3CO2H), as indicated in the Equation 1 :  NaOH(aq) + CH3CO2H(aq) 

NaCH3CO2(aq) + H2O(l)

(Equation 1)

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 = -log10[H3O ]

(Equation 2)

 pH scale is a method of expressing the acidity or basicity of a solution. Solution with pH < 7 are acidic, pH = 7 are neutral, pH > 7 are basic. pH electrode will be use in this experiment. The titration is initiated by inserting the pH electrode into a beaker containing the acid solution (pH within 3-5). As sodium hydroxide, NaOH, is incrementally add ed 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 a 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.

Equivalence oint

In this experiment, titration of the vinegar sample with a standardize 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 :



Must be available in at least 99.9 purity



Must have a high molar mass to minimize error in weighing



Must be stable upon heating



Must be soluble in the solvent of interest

Most acids and bases are available in primary standard form. To standardize it, titration of the solution with a primary standard should be p erformed. In this experiment, NaOH solution will be titrated with potassium hydrogen phthalate. The reaction equation for this is :KHC8H4O4(aq) + NaOH(aq) 

KNaC8H4O4(aq) + H2O(l)

(Equation 3)

Once the sodium hydroxide solution has been standardized it will be titrated with 10.00mL aliquots of vinegar. The reaction equation for vinegar with NaOH is : CH3COOH(aq) + NaOH(aq)

 NaCH3COO(aq) + H2O(l)

(Equation 4)

Knowing the standardize NaOH concentration and using equation 4, we can determine the molarity and percent by mass of acetic acid in the vinegar solution.

Apparatus & Materials

Beaker , 50-ml burette, magnetic rod, pH meter, conical flask, stirrer, electronic balance, spatula, retort stand, NaOH solution, vinegar, droppers , stirrer, stir bar, buret clamp, pipet 10mL, Funnel, analytical balance, potassium hydrogen phthalate (KHP)

PROCEDURE

PART A 1. 250 mL of 0.6 M sodium hydroxide solution was prepared from NaOH solid. The solution was prepared in a beaker; the calculation was checked by the laboratory instructor prior to preparing the solution. The calculation was recorded. 2. The beaker was placed on the balance and was tare. 1.5 grams of KHP was added to the  beaker. The mass of KHP was recorded to nearest 0.001 g. 30mL of distilled water was added to the beaker. The solution was stir until the KHP completely dissolved. 3. The solution was titrated with NaOH and the p H was recorded with 1 mL addition of  NaOH solution. 4. Steps 1 to 3 were repeated and to more solutions was prepared for NaOH standardization. 5. The graph of pH versus NaOH was plotted. The volume of NaOH required to neutralize the KHP solution in each titration was determined from the plots. 6. The molarity of NaOH for titrations 1, 2 and 3 was calculated. 7. The average molarity of the sodium hydroxide solution was calculated. The resulting sodium hydroxide concentration was used in part B of the experiment. PART B 1. 10.00mL of a vinegar was transferred to a clean, dry 250mL beaker using a 10mL volumetric pipette. 75 to 100 mL of water was added to cover the pH electrode tip during the titration. 2. 2 mL of NaOH was added to the vinegar solution and the pH was recorded. 3. The above steps was repeated twice more. 4. The graph of pH versus NaOH volume added was plotted and the volume of NaOH required to neutralize the vinegar in each titration was determined from the plots. The data was recorded. 5. The molarity of acetic acid in vinegar for titration 1, 2 and 3 was calculated. 6. The average molarity of acetic acid for each solution was calculated. 7. The percent by mass of acetic acid in vinegar for titrations 1, 2 and 3 was calculated. 8. The percent by mass of acetic acid in vinegar was calculated.

RESULT AND CALCULATION

Titration 1 100.69 102.19 1.5

Mass of beaker (g) Mass of beaker + KHP (g) Mass of KHP

Titration 2 100.72 102.22 1.5

a) Volume of NaOH to neutralize KHP Titration

Volume of NaOH to neutralize KHP (ml)

1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

st

n

2 4.74 4.96 5.15 5.35 5.47 5.61 5.75 5.86 6.01 6.16 6.37 6.55 7.01 12.36 12.81 12.95 13.04

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

4.88 5.06 5.24 5.36 5.51 5.62 5.75 5.87 6.02 6.18 6.34 6.60 7.05 12.50 12.88 13.03 13.13

 b) Volume of NaOH to Neutralize vinegar 1

Volume of NaOH to neutralize vinegar (ml)

V(ml) 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36

st

n

2 pH 3.17 3.88 4.25 4.47 4.67 4.84 4.97 5.09 5.21 5.35 5.46 5.61 5.77 5.99 6.31 7.45 12.11 12.43

12.55

V(ml) 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36

pH 3.29 4.02 4.34 4.57 4.75 4.90 5.05 5.18 5.30 5.43 5.55 5.70 5.86 6.06 6.32 7.03 12.03 12.49 12.65

a) Standardization of sodium hydroxide solution Titration 1 Volume of NaOH (ml) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

pH 4.74 4.96 5.15 5.35 5.47 5.61 5.75 5.86 6.01 6.16 6.37 6.55 7.01 12.36 12.81 12.95 13.04

Titration of KHP with NaOH 14.00 12.00 10.00 8.00    H    p

6.00

Titration 1

4.00 2.00 0.00 0

2

4

6

8

10

Volume of NaOH (mL)



When pH is 7, the volume NaOH is 11.9mL

12

14

16

18

Titration 2 Volume of NaOH (ml) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

pH 4.88 5.06 5.24 5.36 5.51 5.62 5.75 5.87 6.02 6.18 6.34 6.60 7.05 12.50 12.88 13.03 13.13

Titration of KHP with NaOH 14.00 12.00 10.00 8.00    H    p

Column1

6.00

Titration 2 4.00 2.00 0.00 0

2

4

6

8

10

Volume of NaOH (mL)



When pH is 7, the volume NaOH is 11.8mL

12

14

16

18

Calculation st

1  titration a) Mole KHP :

         = 7.35x10⁻3 mole =

 b) Mole NaOH required

       =7.35x10⁻3 mole =

c) Molarity =

      =      

=0.6176 M NaOH nd

2  titration a) Mole KHP : =

        

= 0.007345 mole KHP  b) Mole NaOH required =

      

=0.007345 mole NaOH c) Molarity =

   =         

=0.6224 M NaOH

Average molarity

 (   )  ( ) = 0.620 M

 b) Molarity of acetic acid and mass percent in vinegar Titration 1 Volume of NaOH (ml) 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36

 pH 3.17 3.88 4.25 4.47 4.67 4.84 4.97 5.09 5.21 5.35 5.46 5.61 5.77 5.99 6.31 7.45 12.11 12.45 12.55

Titration of Vinegar with NaOH 14.00 12.00 10.00 8.00    H    p

6.00

Titration 1

4.00 2.00 0.00 0

5

10

15

20

25

Volume of NaOH (mL)



When pH is 7, the volume of NaOH is 28.5mL

30

35

40

Titration 2 Volume of NaOH (ml) 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36

 pH 3.29 4.02 4.34 4.57 4.75 4.90 5.05 5.18 5.30 5.43 5.55 5.70 5.86 6.06 6.32 7.03 12.03 12.49 12.65

Titration of Vinegar with NaOH 14.00 12.00 10.00 8.00    H    p

6.00

Titration 2

4.00 2.00 0.00 0

5

10

15

20

25

Volume of NaOH (mL)



When the pH is 7, the NaOH is 29.9mL

30

35

40

Calculation st

1  titration a) Moles of NaOH that reacted :  b)

         = 0.01767 mol NaOH Molarity of the CH₃COOH :



1 mole of NaOH Ξ 1 moles of CH COOH 0.01767 mole of NaOH Ξ 0.01767 moles of CH COOH



 ₃  ₃      



   CH₃COOH

c) Calculate the mass of acetic acid in the solution

 ₃           ₃    ₃

d) Percentage by mass of acetic acid in the solution

₃              ₃    

nd

2  titration a) Moles of NaOH that reacted :  b)

         = 0.018538 mol NaOH Molarity of the CH₃COOH :



1 mole of NaOH Ξ 1 moles of CH COOH 0.018538 mole of NaOH Ξ 0.018538 moles of CH COOH



 ₃   ₃      



   CH₃COOH

c) Calculate the mass of acetic acid in the solution

 ₃    ₃          ₃

d) Percentage by mass of acetic acid in the solution

₃              ₃    

Average molarity of acetic acid

)    (   = 1.8104 M ₃ Average percentage

)   ()  (   = 10.87%

Sample calculation for standardization a base with NaOH

Let, equivalence point =

 

Mass of KHP = Y grams (g) 

moles of KHP (KCH3H4O4) used in the titration = Y g KHP



     

= V mol of KHP 

 Number of sodium hydroxide (NaOH) used

 = Z mol  = V mol KHP



Volume of solution = Y mL



=Y L 

     

  

Molarity of NaOH solution Molarity ( M ) =

=

   (  )    (  )

   

= A Molal (M) Sample calculation for determine the acetic acid concentration in vinegar by titration with standardized base.

A 40.00 mL aliquot of vinegar requires U mL of A M standardized NaOH solution to reach the equivalence point of the titration. Calculate the molarity and the percent by mass of acetic acid in the vinegar solution. Assume the density of the v inegar solution is 1.00 g/mL.

Equivalence point: U mL 

 Number of moles of NaOH that reacted = y mL

  



= y L NaOH



=y L NaOH

      

= Z mol NaOH 

Z mol NaOH react with T mol of CH3COOH



Volume of solution = 10.0 mL



  

= 0.01 L 

Molarity of acetic acid (CH3COOH)

   (  )    (  )     =   

Molarity ( M ) =



= C Molal (M) Mass of acetic acid

 ( )        ( ) T mol of CH3COOH =     

 Number of moles =

mass = H g 

Mass of solution = 1 mL



     

= 10 g 

Mass percent of acetic acid

   ( )    ( )     =     

Percent solute =

= D % of CH3COOH

 

DISCUSSION

This experiment was constructed to determine the molarity of a solution and the percent by mass of acetic acid in vinegar by titration with a standardized sodium hydroxide solution. The experiment perform by a titration of the sodium h ydroxide into the KHP and vinegar. From the titration, the volume of the sodium hydroxide titrate was recorded to plot the graph. From the graph, we can determine the accurate volume to neutralize the KHP and vinegar solution. From the volume, we can determine the molarity and percent by mass of acetic acid in vinegar by do the several calculation. In this experiment, we get the molarity and the percent by mass of acetic acid in vinegar. Even though we get the molarity and percent by mass of acetic acid, it might be not the accurate one. This is because of the several errors that might be happened during the experiment. One of the error that might be happen during the experiment is parallax error. The reading of the sodium hydroxide solution in the burette may be not accurate. This is because it might have air bubble at the end of the burette. This will give the effect to the volume of sodium hydroxide solution used to neutralize the KHP and the vinegar. Besides that, it might be an error happen during the measurement of pH value. The pH reading taken may not the actual reading because the solution might be not mixed well. This happen  because we take the pH reading after the titration without wait it until it really mixed. It will give effect on the volume of sodium hydroxide solution required to neutralize the KHP a nd the vinegar. Furthermore, it might be an error happen that cause by the surrounding. For example when we want to measure the mass of KHP in the balance. We might not get the accurate 1.5 grams of KHP. This is because of the balance is too sensitive. It can detect the small change in mass. The air molecule might influence the mass of KHP because we not close the surrounding of the  beaker while taking the mass of the KHP. An error also might be happen during we measure the volume of vinegar for titration. We use the volumetric pipette to measure the volume of vinegar. The pipette that we use is not function well. We just take the approximate reading of the pipette.All this error will influence the result that we get.

RECOMMANDATIONS

In a way to get more accurate results, several steps can be done to give us more accurate results. The steps that can be do during the experiment to get more accurate results is make sure the eye is parallel to the reading of burette during w e do the titration. Also make sure there is no air  bubble at the end of the burette. The air bubble must be removed to get the accurate volume of sodium hydroxide. Furthermore, during we measure the mass of KHP, make sure we close the  balance. Therefore, the air molecule will not effected the reading of balance. Another way to get the more accurate result, make sure we mixed the solution well before we take the pH reading of the solution. If the solution not mixed well, it will give the different reading of pH. Moreover, we have to make sure that the apparatus that we use during the experiment must be in good in a way to get more accurate reading.

CONCLUSION

The objective of the experiment was achieved. The molarity and percent by mass of acetic acid in vinegar was determine from the calculation of the results. From the experiment, we get the average molarity of sodium hydroxide is 0.620 M, the average molarity of acetic acid is 1.8104 M and the average percent by mass of acetic acid is 10.87 % .

REFERENCES



ChemCollective (2014) Determine the concentration of Acetic A cid in Vinegar . Retrieved at 10 October 2014 : http://chemcollective.org/activities/autograded/131



TitrationInfo(n.d) Determination of acetic acid in vinegar by titration Retrieved at 10 October 2014 :http://www.titrations.info/acid-base-titration-acetic-acidin-vinegar

 

Eileen Y.L and Cham (2009) Determination of Ethanoic Acid in Vinegar Retrieved at 10 October 2014: http://www.slideshare.net/wkkok1957/determination-ofconcentration-of-ethanoic-acid-in-vinegar-by-titration



Lab Manual (UiTM Shah Alam, 2014) Determination of Acetic Acid in Vineger.



APPENDIXES

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