Vinegar Titration LAB 1
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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.
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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.
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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.
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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
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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.
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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
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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
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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 of the titration curve
slope
Volume of NaOH (ml)
Slope o! the gra!ient
Figure 2: titration, 2n! trial
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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 of the titration curve
slope
Volume of NaOH (ml)
Slope of the gra!ient
Figure ": titration, "r! trial
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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 "
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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 of the titration curve
slope
Volume of NaOH (ml)
Slope of the gra!ient
Figure #: titration, 1t trial
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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 of the titration curve
slope
Volume of the NaOH (ml)
Slope of the gra!ient
Figure $: titration, 2n! trial
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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 of the titration curve
slope
Volume of NaOH (ml)
Slope of the gra!ient
Figure %: titration, "r! trial
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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
$oles of #HP used in titration(
n KH C 8 H
4O
4 ) 1(482 g KH C 8 H 4O4 *
1 mol of KHC 8 H 4 O4
204.2g K H C 8 H 4O 4
) 0(007258 +ol KH C 8 H 4O4
$oles of &a'H re,uired to neutralie the +oles of #HP(
n NaOH ) 0(007258 +ol
KH C 8 H 4O4 *1 molNaOH
1 molKHC 8 H 4 O4
) 0(007258 +ol NaOH
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$olarit% of &a'H solution
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
L solution ) 0(5583 $ &a'H
Titration "&
$oles of #HP used in titration(
n KH C 8 H 4 O 4 ) 1(497 g
KH C 8 H 4O4 *
1 mol of KHC 8 H 4 O4
204.2g K H C 8 H 4O 4
) 0(007331 +ol KH C 8 H 4O4
$oles of &a'H re,uired to neutralie the +oles of #HP(
n NaOH ) 0(007331 +ol
KH C 8 H 4O4 *
1 molNaOH
1 molKHC 8 H 4 O4
) 0(007331 +ol NaOH
$olarit% of &a'H solution
13(01 +- &a'H *1 L
1000 mL=¿
0(01301 - &a'H
$ )
mol NaOH
L of solution=0.007331mol NaOH
0.01301 L NaOH )
0.5635mol NaOH
L solution ) 0(5635 $ &a'H
./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
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$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
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Titration 2
$oles of &a'H that reacted
19(01 +- of &a'H *1 L
1000mL
=¿0(01901 - &a'H
0(01901 - &a'H *0.5621 mol NaOH
1 L NaOH solution=0.01068mol NaOH
he +ole ofCH 3 COOH
neutralied !% +oles of &a'H
0.01068mol NaOH *
1 molCH 3COOH
1 molNaOH =0.01068 mol
CH 3 COOH
$olarit% of theCH 3 COOH solution
10 +-CH 3 COOH
*1 L
1000 mL=0.010 LCH 3COOH solution
$ )
molCH 3 COOH
Lof solution )
0.01068molCH 3 COOH
0.010 L solution =1.068 M CH 3 COOH
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
*
1.068 M CH 3 COOH
1 L solut ion *
60.6 g CH 3COOH
1 molCH 3 COOH ) 0(6472 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 =0.6472 g CH
3COOH
10.00 g CH 3COOH * 100) 6(472 CH 3 COOH
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Titration "
$oles of &a'H that reacted
19(00 +- of &a'H *1 L
1000 mL=¿
0(01900 - &a'H
0(01900 - &a'H *0.5621 mol NaOH
1 L NaOH solution=0.01068mol NaOH
he +ole ofCH 3 COOH
neutralied !% +oles of &a'H
0.01068mol NaOH *
1 molCH 3COOH
1 molNaOH =0.01068 mol
CH 3 COOH
$olarit% of theCH 3 COOH solution
10 +-CH 3 COOH
*1 L
1000 mL=0.010 LCH 3COOH solution
$ )
molCH 3 COOH
Lof solution )
0.01068molCH 3 COOH
0.010 L solution =1.068 M CH 3 COOH
he +ass of acetic acid in the solution
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
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
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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$. The molarity for titration ! is !.#!& -% titration & is !.#+ - and titration " is
!.#+. The mass and percentage mass for titration !% titration & and titration " are
#.+!"" g% +.!""3% #.+,4& g% +.,4&3% #.+,4& g% +.,4&3 respectively. The average
percentage mass is +."*23. 5rom the theory% the percentage of vinegar is !.2+"3. 6tis less than the calculation. The pattern of the graph from the theory is the same as
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the titration ! and titration &. The different of the percentage mass of theory and the
calculation is due to an error. 6n this experiment% it happens that parallax error is
occurred from the preparation of KHP solution and the reading of urette.
Conclusion- The conclusion of this experiment% we can determine the molarity of
the KHP and the acetic acid y )nowing the equivalent point. 7e can also calculate
the mass of the acetic acid and percentage of acid if the molarity and equivalence
point are )nown. The average percentage mass of the acetic acid in this experiment
is +."*23. while the theory percentage mass is !.2+"3.
Recommendations- 8ach student in the la must e well prepared efore entering
the la. The apparatus used in the experiment must e suitale. 1efore use the
apparatus% ma)e sure to calirate it first to avoid an error happen. 5ollow all the
procedures correctly and carry out the experiment seriously to avoid unnecessary
things from happen. /lways wear a glove and goggle for safety. 0epeat the titration if
necessary.
References
9. :purloc) '&#!"(% :tandardization of a 1ase% NaOH ;lass Notes. 0etrieved
Octoer !+% &#!" from
http<==homepages.ius.edu=dspurloc=c!&!=wee)!!.html
/cetic /cid ;ontent of >inegar 'n.d.(% 0etrived Octoer !4% &#!" from
http<==firstyear.chem.usyd.edu.au=$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
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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.
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/lways read the fluid level in the uret from the ottom of the meniscus.