Report Kimia 1

ABSTRACT

Basic acid-base titration is commonly used to obtain the molarity of a solution given the molarity of other solution that involves neutralization between acid and base. This experiment is done to standardize the sodium hydroxide solution. Besides that this experiment is conducted 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. In the first experiment the volume of NaOH needed to achieve equivalence point is 16.1 mL for the first titration, 17.9 mL for the second titration, and 12.8 mL for the third titration respectively. As for the second experiment the volume of NaOH needed to achieve its equivalence point is 12.5 mL for the first titration, 11.9 mL for the second titration, and 13.0 mL for the third titration respectively. The average molarity of NaOH needed to react with KHP to achieve equivalence is 0.5 M and 0.62 M for the reaction with vinegar. The result obtained in this experiment is 3.76 % for first titration, 3.57 % for second titration, and 3.90 % for the third titration for the mass percent of CH3COOH in each titration and the average of mass percent of CH3COOH is 3.74 %.

INTRODUCTION

The concentration of a solution is the amount of solute in the specific amount of solvent. Thus, solutions with high concentration contain high amount of solute in a given solvent and vice versa. There are two ways to express concentration which is molarity and percent by mass. Molarity is the amount of moles of solute per litre of solution. Percent solute is equal to the mass of solute per mass of solution times by 100%. Vinegar is a solution with high concentrations of acetic acid, CH3COOH. The concentration of solution can be determined by performing titration. By using the known amount of solution needed to titrate the unknown solution and the stoichiometry for that reaction after completed, the concentration of the unknown solution can be calculated.

Molarity (M )=molesof soluteliter of solution (Equation 2-1)

Percent solute= gramsof solutegramsof solution

x100 % (Equation 2-2)

There are two main experiment was done. First was the standardization of sodium hydroxide solution and second was determination of molarity of acetic acid and mass percent in vinegar. In the first experiment, 6 g solid sodium hydroxide, 0.6 M, and 250 ml of sodium hydroxide solution is prepared. Potassium hydrogen phthalate (KHP) solution is prepared by dissolving 1.5 g of KHP with 30 mL distilled water. The titration is done by using this KHP and sodium hydroxide solution. Based on the result the volume of sodium hydroxide solution

used to neutralized the KHP can be obtained and based from the volume, the molarity of sodium hydroxide solution can be calculated. In the second experiment 10 mL of vinegar is diluted with 100 mL of water in 250 mL beaker. The same step of titration with experiment one is done. The vinegar is titrated with sodium hydroxide solution and based on the pH and sodium hydroxide solution added, the volume of sodium hydroxide solution used to neutralized vinegar can be obtained and next the molarity of acetic acid.

OBJECTIVES

Standardization of sodium hydroxide solution. Determine the molarity of a solution and the percent by mass of acetic acid in vinegar

by titration with a standardized sodium hydroxide solution.

THEORY

The equivalence points is said to be the point where the mole of acid in the solution is equal to the mole of the base in the titration. In this experiment the stoichiometry is reach when 1 mole of sodium hydroxide solution is used to neutralized 1 mole of acetic acid.

NaOH(aq)+CH3CO2H(aq) NaCH3CO2(aq)+H2O(l)

The sudden change in pH of the solution when titrate indicate the titration has reach the equivalence point. pH in a solution is related to its hydrogen ion concentration.

pH=-log10[H3O+]

pH scale indicate the acidity or basicity of a solution. Solutions with pH is higher than 7, the solution are basic solution. When the pH is equal to 7 means that the solution is neutral and when the pH indicator is lower than 7, means that the solution is acidic.

The experiment is initiated by taking the pH of KHP before titrated with standardize sodium hydroxide solution. Hydrogen ion are neutralized for every adding of sodium hydroxide solution. As hydrogen ion concentration decrease, the pH reading increase. When equivalence point is reach, further drop of sodium hydroxide solution into the KHP, sudden increase in pH. The volume used to titrated until equivalence point is reach can be determined based on the graph when the pH is equal to 7. For second experiment the same step is taken to titrated. Instead of using KHP, vinegar is used.

KHC8H4O4(aq)+NaOH(aq) KNaC8H4O4(aq)+H2O(l)

CH3COOH(aq)+NaOH(aq) NaCH3COO(aq)+H2O(l)

METHODOLOGY

APPARATUS

Beaker, 250 ml Balance Retort stand pH meter Magnetic stirrer Volumetric flask Pipette Burette

MATERIALS

NaOH, 0.15 M KHP, 1.5 grams Vinegar Distilled water

PROCEDURE

Experiment 1 : Standardization of sodium hydroxide solution

1. 250 mL of approximately 0.6 M sodium hydroxide solution was prepared from the NaOH solid. The solution was prepared in a beaker. The calculation was check with the laboratory instructor prior to prepare the solution. Calculations were tabulate.

2. A beaker was placed on the balance and tare it. 1.5 g of KHP was add to the beaker. The mass of KHP was recorded to the nearest 0.001 g. 30 mL of distilled water was added to the beaker. The solution was stir until the KHP has dissolve.

3. This solution was titrated with sodium hydroxide solution and the pH was recorded for every 1 mL addition of sodium hydroxide solution.

4. Step 1-3 was repeated and two more solutions for sodium hydroxide standardization was prepared.

5. Graph of pH versus sodium hydroxide was plotted and the volume of sodium hydroxide required to neutralize the KHP solution in each titration was calculated.

6. The molarity of sodium hydroxide 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 experiment.

Experiment 2 : Molarity of acetic acid and mass percent in vinegar

1. 10 mL of vinegar was transfer to a clean, 250 mL beaker using a 10 mL volumetric pipette. 100 mL of water was added to cover the pH electrode tip during the titration.

2. 1 mL of sodium hydroxide solution was added to the vinegar solution and the pH was recorded.

3. The above steps was repeat twice more.4. The graph of pH vs volume sodium hydroxide solution added was plotted and the

volume of sodium hydroxide solution used to neutralized vinegar in each titration was determined.

5. The molarity of acetic acid in vinegar for titration 1,2, and 3 was calculated.6. The average molarity of acetic acid in each titration 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 acids in vinegar was calculate.

RESULT

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 230

2

4

6

8

10

12

14

Graph pH versus volume of NaOH

titration 1

titration 2

titration 3

volume of NaOH

pH

Figure 1 : Titration of KHP solutions with NaOH

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 180123456789

10111213

Graph pH versus volume of NaOH

titration 1titration 2titration 3

volume of NaOH

pH

Figure 2 : Titration of Vinegar solutions with NaOH

CALCULATIONS

Experiment 1 :

1)Calculations for preparing 250 mL of approximately 0.6 M sodium hydroxide solution.

25 mL NaOH × 1 L

1000mL = 0.25 L NaOH

Molarity, M = molesof solutevolume of solution

0.6 M NaOH = molesof NaOH0.25 LNaOH

Moles of NaOH = 0.15 mol

Mass NaOH = moles of NaOH × molar mass NaOh

=0.15 mol NaOH × 40 g/mol

=6 g solid NaOH

Titration 1 Titration 2 Titration 3Mass of KHP (g) 1.5058 1.5046 1.5075Volume of NaOH to neutralize the KHP solution (mL)

16.1 17.9 12.8

3)Calculation of molarity of sodium hydroxide for each titration

Titration 1 :

Mol KHP = mass

molar mass

= 1.5058 g

204.2g /mol

= 0.00734 mol

1 mol KHP = 1 mol NaOH

Molarity NaOH = molvolume

= 0.00734mol

0.0161 L

= 0.46 M

Titration 2 :

Mol KHP = mass

molar mass

= 1.5046 g

204.2g /mol

= 0.00737 mol



= 0.00737mol

0.0179 L

= 0.461M

Titration 3 :

Mol KHP = mass

molar mass

= 1.5075 g

204.2g /mol

= 0.00738 mol



= 0.00738mol

0.0128 L

= 0.58 M

4)Calculation of the average molarity of sodium hydroxide for each titration.

Average molarity of NaOH = 0.461+0.46+0.58

3

=0.5 M

Experiment 2 :

Titration 1 Titration 2 Titration 3Volume of NaOH to neutralize the vinegar solution (mL)

12.5 11.9 13.0

2)Calculation of molarity of acetic acid in each titration

Titration 1:

Volume of NaOH = 12.5 mL

Mol of NaOH = 0.0125 L × 0.5 mol/L

= 0.00625 mol

From the equation :

CH3COOH(aq) + NaOH(aq) NaCH3COOH(aq) + H2O(l)

1 mol NaOH = 1 mol CH3COOH

Thus 0.00625 mol NaOH = 0.00625 mol CH3COOH

Molarity of CH3COOH = molof CH 3COOHVol of CH 3COOH

= 0.00625mol

0.01 L

= 0.625 M

Titration 2:


Mol of NaOH = 0.0119 L × 0.5 M

= 0.00595 mol

From the equation :





= 0.00595mol

0.01 L

= 0.595 M

Titration 3:


Mol of NaOH = 0.013 L × 0.5 M

= 0.0065 mol

From the equation :





= 0.0065mol

0.01 L

= 0.65 M

3)Calculation of average molarity of acetic acid

Average molarity of acetic acid = 0.65+0.595+0.625

3

= 0.62 M

4)Calculation of percent by mass of acetic acid in vinegar

Titration 1 :

Mass of CH3COOH =0.01 L × 0.625 M CH3COOH × 60.06 g/mol CH3COOH =0.375 g CH3COOH

Percentage by mass ¿massof solute(g)massof solution g¿

¿ x 100%

¿0.372gCH 3COOH10.00gCH 3COOH x 100%

= 3.76 %

Titration 2 :

Mass of CH3COOH = 0.01 L × 0.595 M CH3COOH × 60.06 g/mol CH3COOH =0.357 g CH3COOH

Percentage by mass ¿ massof solute(g)massof solution g¿

¿ x 100%

¿0.357gCH 3COOH10.00gCH 3COOH x 100%

= 3.57 %

Titration 3 :

Mass of CH3COOH = 0.01 L × 0.65 M CH3COOH × 60.06 g/mol CH3COOH =0.39 g CH3COOH

Percentage by mass ¿massof solute(g)massof solution g¿

¿ x 100%

¿0.39gCH 3COOH

10.00gCH 3COOH x 100%

= 3.9 %

5)Calculation of average percent by mass of acetic acid in vinegar

average percent by mass = 3.9 %+3.57 %+3.76 %

3

= 3.74 %

DISCUSSION

This experiment is divided into two experiments. The objective for the first experiment was to standardize sodium hydroxide solution. The main objective for the second experiment was to determine the molarity of acetic acid in the vinegar solution and the percent by mass of acetic acid in vinegar by titration with a standardized sodium hydroxide (NaOH) solution. For the first experiment, 250 mL, 0.6 M sodium hydroxide solution is prepared in the beaker. To know the exact mass of the solid NaOH used, calculation first must be done and as a result from the calculations based on the information 250 mL, 0.6 M sodium hydroxide solution to be prepared, the result shown that the mass solid NaOH used is 6 g. Next the solution to be titrate is prepared. 1.5 g of solid potassium hydrogen phthalate (KHP) is prepared in a beaker and is diluted with 30 mL distilled water. Then the solution is titrated. The pH is taken for every 1 mL of NaOH is titrated and the result is tabulated and transfer into a graph form. From those result the molarity of sodium hydroxide solution for each titration can be calculated and the result from those three titration, the average can be calculated. The result for the molarity of NaOH for the three titration is 0.46 M, 0.461M, and 0.58 M respectively. The average that is calculated from the molarity is 0.5 M. In experiment two, the same way is used to obtained the result. The pH is taken for every 1 mL of NaOH is titrated. From the calculation the value of molarity of acetic acid contain in the vinegar is 0.62 M, 0.65 M, and 0.595 M respectively. The value of average molarity based on the molarity from the three titration is 0.625 M. After calculate the molarity and the average molarirty, the percentage by mass must also be calculate. The value of percent by mass for experiment one is 3.76 %, 3.57 % for titration 2,and 3.9 %. The average of percent by mass of acetic acid is 3.74 %.

A titration is a process which used burette to dispance a small increment of solution with known concentration. Titration usually used to calculate a molarity of a solution of an acid and base by using the stoichiometry of the reaction. When one of the molarity of the solution is known. The calculation must be made when the reaction has reached its equivalence point. Equivalence point occurs when the mole of the acid is equal to the mol of the base in the titration. In this experiment, equivalence point reach when one mole of NaOH is equal to one mole of acetic acid. This point will be known when there are a sudden change in the pH. pH is related to the hydrogen ion concentration in the solution. pH is define as the negative of the logarithm hydrogen concentration. pH is divided into 3 scales that indicates the type of the solution. First, when the pH is smaller than 7 means that the solution is acidic, ph is equal to 7 indicates that the solution is neutral and pH is larger than 7 means that solution is basic. To detect the pH in this experiment, pH indicator can be used. In the second experiment same procedure is used which is titration but instead of KHP, vinegar solution is used.

The result tabulated from the experiment is transfer into graph. The pH of the solution as the y-axis and the volume of NaOH used to titrate the KHP solution for the first experiment and NaOH used to titrate the vinegar solution for the second experiment as the x-axis. This acid-based graph indicates the pH of the solution at the certain time when certain volume of NaOH is titrate. The volume of NaOH used to neutralized the KHP solution or the

vinegar solution must be obtained for further calculation. The result for that problem can be solved using graph. As in this experiment, the volume of NaOH used to titrate the KHP or the vinegar solution when it reach its equivalence point can easier to be determined. As the result, in the first experiment the volume of NaOH needed to achieve equivalence point is 16.1 mL for the first titration, 17.9 mL for the second titration, and 12.8 mL for the third titration respectively. As for the second experiment the volume of NaOH needed to achieve its equivalence point is 12.5 mL for the first titration, 11.9 mL for the second titration, and 13.0 mL for the third titration respectively. This volume is all taken when the pH is 9. The result is used to get the average volume of NaOH used to neutralized the KHP solution or the vinegar solution. This average is further used to calculate the mol of NaOH and divided with its correspondent volume to get its molarity. As the result the average molarity of NaOH needed to react with KHP to achieve equivalence is 0.5 M and 0.62 M for the reaction with vinegar. The percent by mass of acetic acid in the titration for experiment two can be calculated based from the molarity of NaOH calculated before. One mol of NaOH is equal to one mol of CH3COOH. Thus molarity of NaOH is equal to molarity of CH3COOH. Used the molarity of CH3COOH and times with volume of CH3COOH used which is 0.01 L and further times with molar mass of CH3COOH to get the mass of CH3COOH. The percent by mass of acetic acid in vinegar can be obtained by divided the mass obtained before with 10.0 g CH3COOH and times with 100 %. The result obtained in this experiment is 3.76 % for first titration, 3.57 % for second titration, and 3.90 % for the third titration for the mass percent of CH3COOH in each titration. The average of mass percent of CH3COOH is 3.74 %.

During the experiment there are several precaution steps that can be taken to avoid any damage to the apparatus or either wise the result. First, the calculation to get the mass of solid NaOH must be perfectly calculate to get the exact amount of 250 mL, 0.6 M NaOH. To get the mass of solid NaOH that is used to get the solution as in the procedure, first the molarity must be times with volume of the solution that is want to be prepared. Make sure the volume in litre, if the volume in mL make sure to convert the unit first. If not the calculation will be wrong and the result will be affect. Next, when the mass of solid NaOH is known, make sure to use the balance to weigh the solid NaOH. Turn the balance into 0.00 g after the beaker is placed on the weigh. Make sure the mass of solid NaOH used is not included with the mass of the beaker. When weighing make sure the mass of solid NaOH is measured to the nearest 0.001 g. If the mass of solid NaOH is excess or redundant it might disturbed the result also the graph. Hence, the exact volume of KHP used to naturalized the NaOH cannot be obtained. Dissolved the solid NaOH completely to make sure maximum reaction. During titration ensure the exact amount is titrated. Read the scale correctly. Make sure the eyes of the observer is perpendicular to the scale to avoid any parallax error. Repeat the experiment to get the average reading so that can reduce the percentage error. Clean the apparatus before used with distilled water and rinse the apparatus with the solution so that the apparatus is clean from any impurities. Cover the ph electrode tip to make sure the pH indicator can read the pH in the solution completely.

CONCLUSION

As a conclusion the experiment is not perfectly done. This is because the result obtained is not consistent. This can be seen on the graph. The line of pH versus volume of NaOH used to neutralized KHP or vinegar for the first is not parallel with line of titration two and titration three. But the method to calculate the molarity, mass percent, and the stoichiometry equation can be used. If the experiment is done perfectly and the correct calculation step is used the result obtained can be as followed. In the first experiment the volume of NaOH needed to achieve equivalence point is 16.1 mL for the first titration, 17.9 mL for the second titration, and 12.8 mL for the third titration respectively. As for the second experiment the volume of NaOH needed to achieve its equivalence point is 12.5 mL for the first titration, 11.9 mL for the second titration, and 13.0 mL for the third titration respectively. The average molarity of NaOH needed to react with KHP to achieve equivalence is 0.5 M and 0.62 M for the reaction with vinegar. The result obtained in this experiment is 3.76 % for first titration, 3.57 % for second titration, and 3.90 % for the third titration for the mass percent of CH3COOH in each titration and the average of mass percent of CH3COOH is 3.74 %.

RECOMENDATION

In any experiment there are several steps can be taken to avoid any error or to get the perfect result. First do not play in the laboratory while doing the experiment. This is to avoid and careless mistake while doing the experiment such as weighing the materials or while pouring any solutions in apparatus. This is to avoid any corrosion solution from contacting to skin. Other than that to avoid any careless mistake while taking the reading. Clean the apparatus before and after use. This is to make sure there are no other impurities in the solution when the experiment is carried out. Use indicator solution so that any changes or reaction in the solution can be seen.

REFERENCE

(2011), Acid-base titration, retrieved from http://www.dartmouth.edu/~chemlab/ techniques/titration.html on 7 October 2011.

http://www.dartmouth.edu/~chemlab/%20techniques/

http://www.dartmouth.edu/~chemlab/%20techniques/

APPENDICES

Weighing pH indicator

Titration process

Report Kimia 1

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