101023214 Titration Curves Report

8/13/2019 101023214 Titration Curves Report

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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.
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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


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


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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101023214 Titration Curves Report

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