101023214 titration curves report
TRANSCRIPT
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Titration Curves
Chem 18.1 Experiment
8
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Titration
method of analysis that will allow youto determine the equivalence point of
a reaction and therefore the precise
quantity of reactant in the titration flask
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Titrant
solution in the burette
usually a strong base or acid
Analyte
solution being titrated
often the unknown
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Titration Set-up
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Titration Curves
are the graphs obtained by plotting thepH of the reaction mixture against the
volume of base (or acid) added during
the titration of either an acid by a baseor vice versa.
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If the pH of an acid solution is plotted
against the amount of base added
during a titration, the shape of thegraph is called a titration curve. All
acid titration curves follow the same
basic shapes.
http://0.tqn.com/d/chemistry/1/0/f/g/satitration.JPGhttp://0.tqn.com/d/chemistry/1/0/e/g/watitration.JPGhttp://0.tqn.com/d/chemistry/1/0/f/g/satitration.JPG -
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Equivalence point
The volume of titrant and pH at whichthe amount of acid equals the amount
of base present in the original solution,
or the amount of base equals theamount of acid present in the original
solution.
Not the same as end point
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Results and
Discussion
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Titration of Weak Acid with Strong Base
0
2
4
6
8
10
12
14
0 3 5 9 12 16 19 21 23 25 32 38 40 48 51 53 55 60
pH
mL of titrant
Titration of 25mL of 0.1M HOAc with O.1M NaOH
Equivalence point:
volume: 50.5 mL
8.96-pH
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0
2
4
6
8
10
12
14
0 5 12 19 23 32 40 51 55
pH
mL of titrant
Titration of 25mL of
0.1M HOAc with O.1M
NaOH
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Discuss the titration curves
obtained in Part A
0
24
6
8
10
12
14
0 5 12 19 23 32 40 51 55
pH
mL of titrant
Titration of 25mL of
0.1M HOAc with O.1M
NaOHCurve begins at low
pH
Near the equivalencepoint there is a rapid
increase in pHEquivalence point
higher than 7, indicating
the presence of basic
salt
Curve ends at high
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Titration of Strong Acid with Strong Base
0
2
4
6
8
10
12
14
0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 77 78 80
pH
mL of titrant
Titration of 25mL of 0.1M HCl with 0.1M NaOH
Equivalence point:
volume: 77.5 mL
7.5-pH
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0
2
4
6
8
10
12
14
0 10 20 30 40 50 60 70 77 80
pH
mL of titrant
Titration of 25mL of 0.1M
HCl with 0.1M NaOH
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Solution of strongacid will always
have a lower initialpH
Before reaching eq.pt, pH increases
slowly but as itapproaches thesaid point, solutionexperiencessteeper rise in pH
Theoretically, theeq. pH of a strongacid-strong basetitration is 7
0
2
4
6
8
10
12
14
0 10 20 30 40 50 60 70 77 80
pH
mL of titrant
Titration of 25mL of 0.1M
HCl with 0.1M NaOH
Tit ti f W k B ith St
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Titration of Weak Base with Strong
Acid
0
1
2
3
4
5
6
7
8
9
10
0 3 5 6 7 8 9 10 11
pH
mL of titrant
Titration of 25mL of 0.1M NH4OH with 0.1M HCl
Equivalence point:
volume: 6.5 mL
pH: 5.355
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0
2
4
6
8
10
0 3 5 6 7 8 9 10 11
pH
mL of titrant
Titration of 25mL of 0.1MNH4OH with 0.1M HCl
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Pattern of weak base-
strong acid titration
quite similar with thatof weak acid-strong
base, except that the
pH of the solution
decreases as titration
continues
Equivalence point less
than 7 Cation of weak base
(NH4+) will undergo
hydrolysis
0
1
2
3
4
56
7
8
910
0 3 5 6 7 8 9 10 11
pH
mL of titrant
Titration of 25mL of
0.1M NH4OH with 0.1M
HCl
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Weak Acid and Weak Base
Curve begins at higher
acidic pH and ends at
low basic pH.
Thereis not a great pHchange at the
equivalence point (pH ~
7) making this a verydifficult titration to
perform.
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Guide Questions
Construct the theoretical titration curvesfor the following by plotting pH versusthe volume of the titrant computed afterevery 5mL of the titrant has been added.
A. 25 mL 0.1M HOAc with 0.1 M NaOH
B. 25 mL 0.1M HCl with 0.1 M NaOH
C. 25 Ml 0.1M NH4OH with 0.1 M HCl
Show sample computations.
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25 mL 0.1M HOAc with 0.1 M NaOH
mL ofHOAc mL ofNaOH totalvol
(mL)pH
25 0 25 2.8725 5 30 4.1425 10 35 4.5625 15 40 4.7425 20 45 5.3525 25 50 8.7225 30 55 11.9
525 35 60 12.2
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25 mL 0.1M HCl with 0.1 M NaOH
mL
of
HClmL ofNaOH totalvol
(mL)pH
25 0 25 1.0025 5 30 1.1825 10 35 1.3025 15 40 1.4825 20 45 1.7825 25 50 7.0025 30 55 11.9525 35 60 12.22
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Calculations
B. 25 mL 0.1M HCl with 0.1M NaOH
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25mL 0.1M NH4OH with 0.1M HCl
mL ofNH4O
HmLof
HCltotalvol
(mL)pH
25 0 25 11.1325 5 30 9.8625 10 35 9.4325 15 40 9.0825 20 45 8.6525 25 50 5.2825 30 55 2.0425 35 60 1.78
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Calculations
C. 25 mL 0.1M NH4OH with 0.1M HCl
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