vinegar titration lab 1

7/23/2019 Vinegar Titration LAB 1

http://slidepdf.com/reader/full/vinegar-titration-lab-1 1/22

ABSTRACT

Vinegar is a liquid consisting mainly of acetic acid (CH 3COOH) and water or basically it is a

dilute solution of acetic acid. Vinegar is a very common household item. Because of its

chemical roerties vinegar can be used in a variety of ways! to clean coffee ma"ers# as a

salad dressing# as a disinfectant# as a reservative# and in coo"ing.

$he concentration of a solution is defined as the amount of solute in a given amount of

solvent. $o determine the molarity of a solution and the ercent by mass of acetic acid in

vinegar# titration method is used with the standardi%ed sodium hydro&ide solution. 'll

measurements such as volumes# lengths# weights have associated errors. tandardi%ation of

sodium hydro&ide solution#*+ m, of aro&imately +.- sodium hydro&ide solution is

reared from /aOH solid. 0etermination of acetic acid concentration in vinegar# + m,

vinegar is transferred to a clean# dry 1*+ m, bea"er using a + m, volumetric iette. 2irstly

before we start the rocedure# we had been wear glove and a goggle as recaution ste to

avoid any inuries from haen. $he standardi%ation of sodium hydro&ide solution# we titrate

the 4H5 with /aOH solution. $he conclusion of this e&eriment# we can determine the

molarity of the 4H5 and the acetic acid by "nowing the equivalent oint. 6ach student in the

lab must be well reared before entering the lab. $he aaratus used in the e&eriment must

be suitable.



Introduction

$itration is a rocess which a small increment of a solution of "nown concentration are added

to a secific volume of a solution of un"nown concentration until the stoichiometry for that

reaction is attained.

$his e&eriment is to determine the molarity of a solution and the ercent by mass of acetic

acid in vinegar by titration with a standardi%ed sodium hydro&ide solution. $he e&eriment is

divided into two arts where the first one is to reare standardi%ing sodium hydro&ide

solution which will be titrated to otassium hydrogen hthalate# 4HC7H8O8(aq) solution and

the second one is to determining the molarity and ercentage by mass of the acetic acid in

vinegar.

$he concentration of acetic acid in vinegar may be e&ressed in two secific terms# as a

molarity or as a mass ercent where9

olarity () :

moles of solute

liter of solution

5ercent solute :grams of solute

grams of solution .

;n this case# we used9

olarity :moles of acetic acid (mol)

volume of vinegar (L)

ass ercent :mass of acetic acid

mass of vinegar & ++

By erforming titration# both molarity and ercent by mass of acetic acid can be determined.

$he urose of titration is to determine the equivalence oint of the reaction. ;n order to

"now where the equivalence is reached# an indicator is used to usually signal the end of

reaction also "nown as the endoint. $he equivalence oint is reach when the added quantity

of one reactant is the e&act amount necessary for stoichiometric reaction with another

reactant.



Aims

$o determine the molarity of a solution and the ercent by mass of acetic acid in vinegar by

titration with the standardi%ed sodium hydro&ide solution.

Theory

Burette are used to erform titrations in which they accurately deliver a measureable volume

of a liquid. ' tyical burette has the smallest calibration unit of +. m,# therefore# volume

disense should be estimated to the nearest +.+ m,.

;n this e&eriment# the equivalence oint is reached when the moles of acid in the solution

equals to the moles of base added in titration. 2or acid<base titration# the "nown chemical

reaction in general is!

'cid = base water = salt

and for the titration of vinegar in this e&eriment secifically reaction will be used is!

CH3CO1H(aq) = /aOH(aq) /aCH3CO1(aq) = H1O(l)

where the /aOH will be the standard reactant solution and acetic acid the calculated

un"nown reactant.

$he sudden change in H of solution indicates the titration has reached the equivalence oint.

H is a measure of hydrogen ion concentration# >H3O=?. H is defined as!

H : < log >H3O=

?

H is a measure of acidity or al"alinity of a solution. H electrode will be used in this

e&eriment. 's /aOH is incrementally added to the acid solution# some of hydrogen ions will

be neutrali%ed. 's the hydrogen ion concentration decreases# the H of the solution will

gradually increase. @hen sufficient /aOH is added to comletely neutrali%ed the acid# the

ne&t dro of /aOH added will cause a sudden shar increase in H. Here# the volume of base

required to comletely neutrali%ed the acid is determine at the equivalence oint of titration.



Apparatus

1*+m, bea"er

Burette

easuring cylinder

Aetort stand

tirrer

1*+ m, 6rlenmeyer flas"

@eighing machine

2ilter funnel

Material

0istilled water

4H5 solid

/aOH solid

Vinegar



Methodology/Procedure

tandardi%ation of sodium hydro&ide solution

. 'ro&imately 1*+ m, of +.- sodium hydro&ide solution is reared from /aOH solid

in a bea"er. $he calculation to rearing the solution is chec"ed with the laboratory

instructor. $he calculation is recorded.

1. ' 1*+ m, bea"er is weighted and the mass is recorded to the nearest +.++g. .* grams of

4H5 is added to the bea"er. $he mass of the bea"er and the 4H5 is recorded to the nearest

+.++g. $he mass of 4H5 by difference is calculated and recorded. 3+ m, of distilled water is

added to the bea"er. $he solution is stirred until the 4H5 has dissolved comletely.

3. $he 4H5 solution is neutrali%ing by titrate it with the /aOH solution. $he volume of

/aOH required to neutrali%e the 4H5 is recorded each every m, until the 4H5 is

neutrali%ed. $he H is recorded using H measuring instrument.

8. stes to 3 was reeated twice. $he data was recorded.

0etermination of 'cetic 'cid Concentration ;n Vinegar

. +.++ m, vinegar is transferred to a clean# dry 1*+ m, bea"er using a + m, volumetric

iette. ufficient water was added# * to ++ m,# to cover the H electrode ti during

titration.

1. $he volume of /aOH required to neutrali%e the vinegar is recorded each every m, until

the vinegar is neutrali%ed. $he H value of solution was recorded in each addition of m,

/aOH.

3. te and 1 is reeated to erform a second titration of vinegar with the standardi%e

/aOH.

8. $he grah of H vs volume /aOH added is lotted. 2rom the lot# the volume of /aOH

required to neutrali%e vinegar in each titration is determined and recorded.



Results

tandardi%ation of sodium hydro&ide solution

$itration $itration 1 $itration 3

ass of bea"er (g) +.++ +.++ +.++

ass of bea"er =

4H5 (g)

.*+ .873 .87

ass of 4H5 (g) .*++ .871 .8

Volume of /aOH to

neutrali%e the 4H5

solution (m,)

3.+ 3.++ 3.+

Table 1

Volume /aOH (ml) H <st trial H <1nd trial H <3rd trial

+ 8.+- 8.* 8.*

8. 8.7 8.7

1 8.- 8.8 8.*

3 *. *.3 *.

8 *.1 *.1* *.1-

* *.3 *.3* *.8- *.8 *.8 *.*

*.- *.- *.-*

7 *.- *.3 *.73

*. *.7 *.7

+ -.+7 -.+1 -.8

-.38 -.1 -.3-

1 -.- -.*3 -.3

3 . 7.*3 .78

8 1.*- 1.3 1.8-

* 1.7 1.-3 1.- 1. 1. 1.7

3 1.* 1.

7 3.+ 1. 1.*

3.1 3.+8 3

1+ 3.- 3.+7 3.+8

1 3. 3.1 3.+

11 3.11 3.8 3.

13 3.1* 3. 3.3

18 3.1 3. 3.*

1* 3.17 3.1 3.Table 2



0 5 10 15 20 250

2

4

6

8

10

12

14

Titration curve

Ph

Volume of NaOH (ml)

pH

5 6 7 8 9 10111213141516171819200

2

4

6

8

10

12

Slope of the titration curve

slope

Volume of NaOH (ml)

Slope for the gradient

Figure 1: titration, 1t trial



0 5 10 15 20 25 30

0

2

4

6

8

10

12

14

Titration curve

pH

Volume of NaOH (ml)

pH

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

200

2

4

6

8

10

12


slope

Volume of NaOH (ml)

Slope o! the gra!ient

Figure 2: titration, 2n! trial



0 5 10 15 20 25 30

0

2

4

6

8

10

12

14

Titration curve

pH

Volume of NaOH (ml)

pH

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

200

2

4

6

8

10

12


slope

Volume of NaOH (ml)

Slope of the gra!ient

Figure ": titration, "r! trial



0etermination of acetic acid in vinegar

$itration $itration 1 $itration 3

Volume of /aOH required

to neutrali%ed vinegar

(m,)

7.++ .+ .++

Volume of /aOH H <st trial H <1nd trial H <3rd trial

+ 3.1 3.7 3.17

3.-* 3.- 3.-*

1 8.+1 8.+* 8.+7

3 8.13 8.18 8.1

8 8.37 8.3 8.88

* 8.*7 8.*1 8.*

- 8.-7 8.-* 8.

8. 8. 8.7

7 8.77 8.7- 8.

8. 8.- *.+

+ *.+7 *.+* *.

*.- *.3 *.

1 *.17 *.11 *.17

3 *.8 *.38 *.3

8 *.*3 *.8* *.*

* *.- *.* *.-3

- *.7* *. *.

-.7 *.77 *.

7 -. -.- -.1-

.** -.1 .-1

1+ .- .* .

1 1.* 1.+1 1.3

11 1.1 1.1 1.1-

13 1.3- 1.31 1.37

18 1.83 1.83 1.88

1* 1.87 1.8 1.*

1- 1.*3 1.*- 1.**

1 1.* 1.- 1.*

17 1.- 1.-8 1.-3Table "



0 5 10 15 20 25 30

0

2

4

6

8

10

12

14

Titration curve

pH

Volume of NaOH (ml)

pH

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

0

2

4

6

8

10

12


slope

Volume of NaOH (ml)


Figure #: titration, 1t trial



0 5 10 15 20 25 30

0

2

4

6

8

10

12

14

Titration curve

pH

Volume of NaOH (ml)

pH

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

0

2

4

6

8

10

12


slope

Volume of the NaOH (ml)


Figure $: titration, 2n! trial



0 5 10 15 20 25 30

0

2

4

6

8

10

12

14

Titration curve

pH

Volume of NaOH (ml)

pH

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

0

2

4

6

8

10

12


slope

Volume of NaOH (ml)


Figure %: titration, "r! trial



Calculation

Calculation for standardiing a !ased "ith #HP

$olarit% of &a'H solution in each titration

Titration 1&

$oles of #HP used in titration(

n KH C 8 H 4 O 4 ) 1(500 g

KH C 8 H 4O4 *

1 mol of KHC 8 H 4 O4

204.2g K H C 8 H 4O 4

) 0(007346 +ol KH C 8 H 4O4

$oles of &a'H re,uired to neutralie the +oles of #HP(

n NaOH ) 0(007346 +ol

KH C 8 H 4O4 *

1 molNaOH

1 molKHC 8 H 4 O4

) 0(007346 +ol NaOH

$olarit% of &a'H solution

13(01 +- &a'H *

1 L

1000 mL=¿ 0(01301 - &a'H

$ )mol NaOH

L of solution=

0.007346 mol NaOH

0.01301 L NaOH )0.5646 mol NaOH

L solution ) 0(5646 $ &a'H

Titration 2


n KH C 8 H

4O

4 ) 1(482 g KH C 8 H 4O4 *


204.2g K H C 8 H 4O 4

) 0(007258 +ol KH C 8 H 4O4


n NaOH ) 0(007258 +ol

KH C 8 H 4O4 *1 molNaOH

1 molKHC 8 H 4 O4

) 0(007258 +ol NaOH




13(00 +- &a'H *1 L

1000 mL=¿

0(01300 - &a'H

$ )mol NaOH

L of solution=

0.007258 mol NaOH

0.01300 L NaOH )0.5583mol NaOH


Titration "&


n KH C 8 H 4 O 4 ) 1(497 g

KH C 8 H 4O4 *


204.2g K H C 8 H 4O 4

) 0(007331 +ol KH C 8 H 4O4


n NaOH ) 0(007331 +ol

KH C 8 H 4O4 *

1 molNaOH

1 molKHC 8 H 4 O4

) 0(007331 +ol NaOH


13(01 +- &a'H *1 L

1000 mL=¿

0(01301 - &a'H

$ )

mol NaOH

L of solution=0.007331mol NaOH

0.01301 L NaOH )

0.5635mol NaOH


./erage +olarit% )0.5646+0.5583+0.5635

3 =¿

0(5621 M NaOH

Calculation for deter+ining the acetic acid concentration in /inegar

Titration 1



$oles of &a'H that reacted

18(00 +- of &a'H *1 L

1000 mL=¿

0(01800 - &a'H

0(01800 - &a'H *0.5621 mol NaOH

1 L NaOH solution=0.01012mol NaOH

he +ole ofCH 3 COOH

neutralied !% +oles of &a'H

0.01012mol NaOH *

1 molCH 3COOH

1 molNaOH =0.01012 mol

CH 3 COOH

$olarit% of the CH 3 COOH solution

10 +-CH 3 COOH

*1 L

1000 mL=0.010 LCH 3COOH solution

$ )

molCH 3 COOH

Lof solution )

0.01012molCH 3

COOH

0.010 L solution =1.012 M CH

3COOH

he +ass of acetic acid in the solution

10 +-CH 3 COOH

*1 L

1000mL ) 0(010 -CH 3 COOH

0(010 -CH 3 COOH

*

1.012 M CH 3COOH

1 L solution *

60.6 g CH 3COOH

1 molCH 3 COOH )0(6133 g

CH 3 COOH

he +ass of the acetic acid solution

10+-CH 3 COOH

solution *

1g CH 3

COOH solution

1mLCH 3COOH solution ) 10(0 CH 3 COOH

solution

he percent !% +ass of the acetic acid in the solution

Percent +ass CH 3

COOH =0.6133 g CH

3COOH

10.00g CH 3COOH x 100 ) 6(133

CH 3 COOH



Titration 2


19(01 +- of &a'H *1 L

1000mL

=¿0(01901 - &a'H

0(01901 - &a'H *0.5621 mol NaOH


he +ole ofCH 3 COOH


0.01068mol NaOH *

1 molCH 3COOH


CH 3 COOH

$olarit% of theCH 3 COOH solution

10 +-CH 3 COOH

*1 L


$ )

molCH 3 COOH

Lof solution )

0.01068molCH 3 COOH

0.010 L solution =1.068 M CH 3 COOH


10 +-CH 3 COOH

*1 L

1000mL ) 0(010 -CH 3 COOH

0(010 -CH 3 COOH

*

1.068 M CH 3 COOH

1 L solut ion *

60.6 g CH 3COOH

1 molCH 3 COOH ) 0(6472 g

CH 3 COOH


10+-CH 3 COOH

solution *

1g CH 3

COOH solution


solution


Percent +ass CH 3 COOH =0.6472 g CH

3COOH

10.00 g CH 3COOH * 100) 6(472 CH 3 COOH



Titration "


19(00 +- of &a'H *1 L

1000 mL=¿

0(01900 - &a'H

0(01900 - &a'H *0.5621 mol NaOH


he +ole ofCH 3 COOH


0.01068mol NaOH *

1 molCH 3COOH


CH 3 COOH

$olarit% of theCH 3 COOH solution

10 +-CH 3 COOH

*1 L


$ )

molCH 3 COOH

Lof solution )

0.01068molCH 3 COOH

0.010 L solution =1.068 M CH 3 COOH


10 +-CH 3 COOH

*1 L

1000mL ) 0(010 -CH 3 COOH

0(010 -CH

3COOH

*

1.068 M CH 3 COOH

1 L solution *

60.6 g CH 3COOH

1 molCH 3 COOH ) 0(6472 g

CH 3 COOH


10+-CH 3 COOH

solution *

1g CH 3

COOH solution


solution




Percent +assCH

3COOH =

0.6472 g CH 3

COOH

10.00 g CH 3COOH * 100) 6(472 CH 3 COOH

./erage +olarit% of acetic acidCH 3 COOH =

1.012+1.068+1.068

3 ) 1(0493 $

./erage percent !% +ass of acetic acid in /inegarCH 3 COOH

)

6.133+6.472+6.472

3 ) 6(359

Discussion- Titrate the KHP with NaOH solution for the standardization of sodium

hydroxide solution. The titration is stopped until the constant read of pH value is

showed. The equivalence point is the point in the titration in which enough standard

solution has een added to react exactly with the sustance eing determined.

Titration ! gives us the equivalent point of !".#! m$% equivalent point of !".## m$ for

titration & and !".#! m$ for titration ". The reactants are mixed in exact molar

proportions represented y the alanced equation. The standard solution is a

solution for which the concentration 'molarity( is accurately )nown. The molarity for

titration ! is #.*+,+ -% titration & is #.**" -% for titration " is #.*+"* -. The average

molarity for titration !% & and " is #.*+&! -. The average molarity is then used in

P/0T 1 calculation to get the molarity of acetic acid in vinegar. The equivalent point

for acetic acid titration ! is !.## m$% titration & is !2.#! m$% and titration " is !2.##

m$. The molarity for titration ! is !.#!& -% titration & is !.#+ - and titration " is

!.#+. The mass and percentage mass for titration !% titration & and titration " are

#.+!"" g% +.!""3% #.+,4& g% +.,4&3% #.+,4& g% +.,4&3 respectively. The average

percentage mass is +."*23. 5rom the theory% the percentage of vinegar is !.2+"3. 6tis less than the calculation. The pattern of the graph from the theory is the same as



the titration ! and titration &. The different of the percentage mass of theory and the

calculation is due to an error. 6n this experiment% it happens that parallax error is

occurred from the preparation of KHP solution and the reading of urette.

Conclusion- The conclusion of this experiment% we can determine the molarity of

the KHP and the acetic acid y )nowing the equivalent point. 7e can also calculate

the mass of the acetic acid and percentage of acid if the molarity and equivalence

point are )nown. The average percentage mass of the acetic acid in this experiment

is +."*23. while the theory percentage mass is !.2+"3.

Recommendations- 8ach student in the la must e well prepared efore entering

the la. The apparatus used in the experiment must e suitale. 1efore use the

apparatus% ma)e sure to calirate it first to avoid an error happen. 5ollow all the

procedures correctly and carry out the experiment seriously to avoid unnecessary

things from happen. /lways wear a glove and goggle for safety. 0epeat the titration if

necessary.

References

9. :purloc) '&#!"(% :tandardization of a 1ase% NaOH ;lass Notes. 0etrieved

Octoer !+% &#!" from

http<==homepages.ius.edu=dspurloc=c!&!=wee)!!.html

/cetic /cid ;ontent of >inegar 'n.d.(% 0etrived Octoer !4% &#!" from

http<==firstyear.chem.usyd.edu.au=$a-anual=8!#.pdf

1oy :tanton% $in ?hu% ;harles H. /twood '&##( % 8xperiments in @eneral

;hemistry 5eaturing -easure Net @uided 6nquiry% :elfA9irected% and

;apstone :econd 8dition

@. ;aroni '&##,(% 8xperiment with /cid and 1ases from

http://homepages.ius.edu/dspurloc/c121/week11.html

http://firstyear.chem.usyd.edu.au/LabManual/E10.pdf

http://firstyear.chem.usyd.edu.au/LabManual/E10.pdf

http://homepages.ius.edu/dspurloc/c121/week11.html



http<==www.funsci.com=fun"Ben=acids=acids.htmC!!

Titration of >inegar 'n.d.(% 0etrived Octoer !4% &#!" from

http<==www.smc.edu=proDects=&=chemistryB!#Bexperiments=ch!#Btitration.pdf

9r. 7alter :charf% 9r. ;harles -alerich 'n.d.(% 9etermination of /cetic /cid ;ontent

of >inegar

APP!DI"

/pparatus to titrate the vinegar.

http://www.funsci.com/fun3_en/acids/acids.htm#11

http://www.smc.edu/projects/28/chemistry_10_experiments/ch10_titration.pdf

http://www.funsci.com/fun3_en/acids/acids.htm#11

http://www.smc.edu/projects/28/chemistry_10_experiments/ch10_titration.pdf



/lways read the fluid level in the uret from the ottom of the meniscus.

vinegar titration lab 1

Documents