analytical chem EXP 1 OCE

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UNIVERSITY MALAYSIA SABAH

GROUP 1

20 JANUARY 2010

SK10401

AMALI KIMIA ANALISIS 1

EXPERIMENT 1: DETERMINATION OF ACID CONTENT IN VINEGAR-

VOLUMETRIC TITRATION

NAME MATRIX NUMBER HO SIE YEE BS 09110359

MELISA SUDIN BS 09110445 NABELLA MUSTAPA BS09110309

SHERONA ANNE SIMON BS 09110439



EXPERIMENT 1: DETERMINATION OF ACID CONTENT IN VINEGAR-

VOLUMETRIC TITRATION

INTRODUCTION:

The principal component of vinegar besides water is the weak acid acetic acid,

CH 3COOH

(K a = 1.78 x 10 5). In this experiment, the concentration of acetic acid in a vinegar sample will be

determined by titrating the acetic acid with the strong base sodium hydroxide. The

solution/sample to be titrated is placed inside a conical flask while the titrant in a burette. The

stoichiometry of the neutralization reaction is as follows:

CH 3COOH + NaOH CH 3COONa + H 2O

The stoichiometry of the reaction is one mole of CH 3COOH reacts with one mole of

NaOH. The endpoint of the titration is reached when enough NaOH has been added to react

stoichiometrically with the acid. This is usually determined by using an indicator solution (e.g

phenolftalein) which show different colour under acid (colourless) and basic (pink) conditions.

The concentration of the titrant (NaOH) should be known accurately and this can be done by standardization. Typically, the base solution is standardized with solid potassium hydrogen

phthalate (kPH), whic h is a weak monoprotic acid. It reacts with NaOH as shown below:

C

C

C

C

C

C

C

OK

OH

H

H

H

H O

O

Potassium Ion

C

C

C

C

C

C

C

OK

C

ONa

O

OH

H

H

H

(aq) + H 2O (l)(aq) + NaOH (aq)



The stoichiometry of the reaction is one mole of potassium hydrogen phthalate reacts

with one mole of NaOH. At the end point, the number of moles of acid (KHP) in solution is

equal to the number of moles of base (NaOH) added. Subsequently, the concentration of the

NaOH solution can be calculated.

Result/ calculation:

1) W eight of NaOH required:

Molarity (M) =

n = MV

= (0.1M)

= 0.01 mol

Mass (m) = n × M W

= 0.01 mol × 40 g/ mol

= 0.400 g

2) W eight of KHC 8H4O4 = average weight taken

=

= 0.503g

3) No. of moles of KHC 8H4O4 in solution =

=

= 2.4645 × 10 -3 mol

4) Colour of solution before titration : colourless

5) Colour of solution at endpoint : pale pink



6) Standardization of NaOH solution :

Titration Volume of NaOH (mL)

T1 20.9

T2 20.4

T3 20.5

Average volume (mL) of NaOH =

=

= 20.6 mL

No. of moles NaOH that reacted with KHC 8H4O4:

KHC 8H4O4 (aq) + NaOH (aq) KHC 8H3O4 Na (aq) + H 2O (l)

From the equation,

1 mol of KHC 8H4O4 1 mol of NaOH

2.4645 × 10 -3 mol of KHC 8H4O4

= 2.4645 × 10 -3 mol of NaOH

Concentration of NaOH (moles/ L) =

=

= 0.120 moles/ L



7) Determination of acetic acid

Titration Volume of NaOH (mL)

V1 V2 T1 32.5 16.0

T2 32.2 15.8

T3 31.4 15.6

Sample/ brand: V 1


=

= 32.03 mL

No. of moles NaOH that reacted with CH 3COOH:

No. of moles = molarity × volume

= 0.120 moles/ L ×

= 3.844 × 10 -3 moles

No. of moles of CH 3COOH in solution:

CH 3COOH + NaOH CH 3COONa + H 2O

From the equation,

1 mol of NaOH 1 mol of CH 3COOH

3.844 × 10 -3 moles of NaOH

= 3.844 × 10 -3 moles of CH 3COOH

Concentration (moles/ L) of CH 3COOH in diluted sample:

From 100 mL diluted sample, 20 mL is used for the titration.

M1 = 0.120 moles/ L M 2 = ?

V1 = 32.03 mL V 2 = 20.0 mL

M1V1 = M 2V2



M2 =

=

= 0.1922 moles/ L

Concentration (moles/ L) of CH 3COOH in concentrated sample:

In 100 mL of diluted sample contained 20 mL of concentrated sample.

Therefore, in 20 mL of diluted sample contained =

= 4 mL concentrated sample

M1 = 0.1922 moles/ L M 2 = ?

V1 = 100 mL V 2 = 20 mL

M1V1 = M 2V2

M2 =

=

= 0.961 moles/ L

Concentration (%) of CH 3COOH in concentrated sample:

Volume of concentrated sample = 4 mLMass of CH 3COOH = no. of moles × molecular weight

= 3.844 × 10 -3 moles × 60.0518 g/ mol

= 0.2308 g

W eight/ volume percent (w/ v) =

=

= 5.77 %

Sample/ brand: V 2


=

= 15.8 mL

No. of moles NaOH that reacted with CH 3COOH:



No. of moles = molarity × volume

= 0.120 moles/ L ×

= 1.896× 10 -3 moles

No. of moles of CH 3COOH in solution:CH 3COOH + NaOH CH 3COONa + H 2O

From the equation,

1 mol of NaOH 1 mol of CH 3COOH

1.896× 10 -3 moles of NaOH

= 1.896 × 10 -3 moles of CH 3COOH

Concentration (moles/ L) of CH 3COOH in diluted sample:From 100 mL diluted sample, 20 mL is used for the titration.

M1 = 0.120 moles/ L M 2 = ?

V1 = 15.8 mL V 2 = 20.0 mL

M1V1 = M 2V2

M2 =

=

= 0.0948 moles/ L

Concentration (moles/ L) of CH 3COOH in concentrated sample:

In 100 mL of diluted sample contained 20 mL of concentrated sample.

Therefore, in 20 mL of diluted sample contained =

= 4 mL concentrated sample

M1 = 0.0948 moles/ L M 2 = ?

V1 = 20 mL V 2 = 4 mL

M1V1 = M 2V2

M2 =



=

= 0.474 moles/ L

Concentration (%) of CH 3COOH in concentrated sample:

Volume of concentrated sample = 4 mLMass of CH 3COOH = no. of moles × molecular weight

= 1.896 × 10 -3 moles × 60.0518 g/ mol

= 0.1139 g

W eight/ volume percent (w/ v) =

=

= 2.85 %

DISCUSSION:

In this experiment, we need to determine the acid content in vinegar by volumetric

titration. For the first part of this experiment, the burette was filled with NaOH solution

and the initial reading was taken. After weighted and dissolved 0.50g of KHC 8H4O4 in

water then 3 drops of phenolphthalein was added. The titration was started. H + ions from

KHC 8H4O4 neutralized OH - ions from the NaOH.

H+ (aq) + OH - (aq) H 2O (l)

Phenolphthalein was added and it is used to detect the present of ion OH - in the solution.

Once the OH - ions was detected by the phenolphthalein the colour of solution turn into

pink colour solution. The solution will turn into darker colour if the OH - ions are

detected. So, the solution is more alkali. Due to the different in charge, OH - is more

attractive to H + ions. W hen all the H + ions were fully being neutralized, that means no

more other ions that OH - can find. Here the phenolphthalein will detect the present of

OH - ions and we can observe that the solution will change to pale pink colour. At this

time, we should slow down the titration and observed the colour of solution until it is

remain unchanged. After the colour of solution remains unchanged, the titration was



stopped. The reading from the burette was recorded. This is to ensure that we know the

amount of NaOH used that being used to neutralized KHC 8H4O4 .

CH 3COOH (aq) + NaOH (aq) CH 3COOH Na (aq) + H 2O (l)

From the equation above we know that 1 mol of CH 3COOH needed 1 mol of NaOH to

neutralize the solution. Thus we can find the standard morality of NaOH.

For the second part of the experiment, the same concept was applied in it. The

KHC 8H4O4 was replaced with unknown vinegar; V 1 and V 2 and same procedures were

applied.

CH 3COOH (aq) + NaOH (aq) CH 3COOH Na (aq) + H 2O (l)

From the equation, 1 mol of CH 3COOH needed 1 mol of NaOH to neutralize the

solution. Both part of experiment were repeated three times to obtain the average value of

NaOH value that had being used.

During the experiment, a few errors occurred while the experiment had been done that

cause our group to get an inaccurate results. Some of the errors include errors while

weighing the NaOH and KHP. The mass is not exactly to as the procedure. Other than

that, we also overshoot a few times during the titration. The concentration (%) of ascetic

acid in V 1 as at the bottle is 5% and for V 2 is 2.5% but after we did our experiment, the

concentration (%) of ascetic acid in V 1 that we get is 5.155 % and for V 2 is 2.568 %. This

is due to the errors that we have done during the experiment.

Conclusion:

From the experiment, we can determine the concentration of acid content in the vinegar

sample. Using the titration method, with two sample of vinegar V 1 and V 2, we know that

V1 has higher concentration percent of acetic acid content in it than in V 2, that is 5.771%

for V 1 and 2.85% for V 2.



References:

Gilbert, Kriss, Foster, Daries (2009), Chemistry , second edition, New York, London :

W .W .Norton &Company

Harvey, D. 2000. Modern Analytical Chemistry . Boston : McGraw Hill

Holt, Rineheart and W inston (2009), Modern Chemistry , Austin: A Harcourt Education

Company

Silberberg (2007), Principles of General Chemistry , New York : The Mc Graw-Hill Companies





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analytical chem EXP 1 OCE