Andre Anusta Barus (Chemistry Education 2010)

8/2/2019 Andre Anusta Barus (Chemistry Education 2010)

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THE ASSIGNMENT ABOUT ACID-BASE TITRATION

CREATED BY :

NAME : ANDRE ANUSTA BARUS

ID NUMBER : 4103332001

CLASS : BILLINGUAL CHEMISTRY EDUCATION

SUBJECT : ANALITICAL CHEMISTRY EDUCATION


2/42

PART 1

OH-= (0.0200) = 0.0200 M

H-=

=

=

PH = -log 5 = 12.30

OH-=

(0.0200)

= 0.01760 M

H-=

=

=

PH = -log 5.68

= 12.24

OH-=

(0.0200)

= 0.01540 M

H-=

=

=

PH = -log 5.68 = 12.18

OH-=

(0.0200)

= 0.01320 M


3/42

H-=

=

=

PH = -log 7.57

= 12.12

OH-=

(0.0200)

= 0.01110 M

H-=

=

=

PH = -log 9.009 = 12.04

OH-=

(0.0200)

= 0.00909 M

H

-

=

=

=

PH = -log 1.111 = 11.95

OH- = (0.0200) = 0.00714 M

H-=

=

=


4/42

PH = -log 1.40056 = 11.85

OH-= (0.0200) = 0.00526 M

H-=

=

=

PH = -log 1.90114 = 11.77

OH-=

(0.0200)

= 0.00345 M

H-=

=

=

PH = -log 2.8985

= 11.53

OH-=

(0.0200)

= 0.00169 M

H- = = = PH = -log 2.8985 = 11.22


5/42

OH-=

(0.0200)

= 0.00084 M

H-=

=

=

PH = -log 1.1904 = 10.92

OH-=

(0.0200)

= 0.00017 M

H- = = = PH = -log 5.8823 = 10.22

OH-=

(0.0200)

= 0.00004 M

H-=

=

=

PH = -log 5 = 9.22


6/42

PART 2

H20 H+

+ OH-

(x) (x)

Kw = x2 x = 1.0 x 10-7M PH = -log 1.0 x10 -7

PART 3

[H+] = (0.100) = 2 x 10

-5M

PH = - log [H+] = - log 2 x 10

-5= 4.78 M

[H+] = (0.100)

= 1.7 x 10

-4M

PH = - log [H+] = - log 1.7 x 10

-4M = 3.78

[H+] = (0.100)

= 8.3 x 10

-4M

PH = - log [H+

] = - log 8.3 x 10-4

M = 3.08

[H+] = (0.100)

= 8.3 x 10

-4M


7/42

PH = - log [H+] = - log 8.3 x 10

-4M = 3.08

[H+] = (0.100) = 1.64x 10

-3M

PH = - log [H+] = - log 1.64 x 10

-3M = 2.79

[H+] = (0.100)

= 3.23x 10

-3M

PH = - log [H+] = - log 3.23 x 10

-3M = 2.49

[H+] = (0.100)

= 4.76x 10

-3M

PH = - log [H+

] = - log 4.76 x 10-3

M = 2.32

[H+] = (0.100)

= 6.25x 10

-3M

PH = - log [H+] = - log 6.25 x 10

-3M = 2.20

[H+] = (0.100)

= 7.69x 10

-3M

PH = - log [H+] = - log 7.69 x 10

-3M = 2.11


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[H+] = (0.100)

= 9.09x 10

-3M

PH = - log [H+] = - log 9.09 x 10

-3M = 2.04

TABLE

The position The addition The concentration of The concentration of pH

in curve volume of HBr OH-doensnt react excess H

+

0.00 0.02000 - 12.30

1.00 0.01760 - 12.24

2.00 0.01540 - 12.18

3.00 0.01320 - 12.12

4.00 0.01110 - 12.04

5.00 0.00909 - 11.95

Part I 6.00 0.00714 - 11,85

7.00 0.00526 - 11.72

8.00 0.00345 - 11.53

9.00 0.00169 - 11.22

9.50 0.00084 - 10.92

9.90 0.00017 - 10.22

9.99 0.00002 - 9.22

Part II 10.00 - - 7.00

10.01 - 0.00002 4.78

10.10 - 0.00017 3.78

10.50 - 0.00083 3.08

11.00 - 0.00164 2.79

Part III 12.00 - 0.00323 2.49

13.00 - 0.00476 2.32

14.00 - 0.00625 2.20

15.00 - 0.00769 2.11


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16.00 - 0.00909 2.04

B.TTRATION OF WEAK ACID WITH STRONG B ASE

PART 1

HA H+

+ A-(Ka = 10

-6.15)

(a-x) (x) (x)

= Ka x = 1.19 x 10

-4

[H+

] = 1.19 x 10-4

PH = -log 1.19 x 10-4

PH = 3.93


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PART 2

HA + OH- A

-+ H2O

Initial : (1) ( ) (0)

After Titration : (

) (0) (

)

PH = pKa + log

= 6.15 +

= 4.87

HA + OH- A

-+ H2O

Initial : (1) ( ) (0)After Titration : (

) (0) (

)

PH = pKa + log

= 6.15 +

= 5.20

HA + OH- A

-+ H2O


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Initial : (1) ( ) (0)

After Titration : (

) (0) (

)

PH = pKa + log

= 6.15 +

= 5.55

HA + OH- A

-+ H2O

Initial : (1) (

) (0)

After Titration : () (0) (

)

PH = pKa + log

= 6.15 +

= 5.78

HA + OH- A

-+ H2O

Initial : (1) ( ) (0)


)


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PH = pKa + log

= 6.15 +

= 5.97

HA + OH- A

-+ H2O

Initial : (1) (

) (0)


)

PH = pKa + log

= 6.15 +

= 6.15

HA + OH- A

-+ H2O

Initial : (1) ( ) (0)

After Titration : ( ) (0) ( )

PH = pKa + log


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= 6.15 +

= 6.33

HA + OH- A

-+ H2O

Initial : (1) ( ) (0)

After Titration : (

) (0) (

)

PH = pKa + log

= 6.15 +

= 6.52

HA + OH- A

-+ H2O

Initial : (1) ( ) (0)


)

PH = pKa + log AHA

= 6.15 +

90010

1010

= 7.10


14/42

HA + OH- A

-+ H2O

Initial : (1) (95010

) (0)

After Titration : (0510

) (0) (95010

)

PH = pKa + logAHA

= 6.15 +950100510

= 7.43

HA + OH- A

-+ H2O

Initial : (1) (9

90

10 ) (0)

After Titration : (01010

) (0) (99010

)

PH = pKa + logAHA

= 6.15 +

99010

01010

= 8.15


15/42

PART 3

A-+ H2O HA + OH

-

(a-x) (x) (x)

Kb =

a

HA + OH- A

-+ H2O

Initial : (1) (1) (0)

Chemical reaction : (0) (0) (1)

[A] = 0.020 500500100 = 0.0167M

x2

ax = Kb =K

Ka= 1.43 x 10

-8 x = 1.54 10 -5 M

PH = - log [H] = - log Kx

- log10x1014154x10

5 = 9.18

PART 4

[OH] = 0.100 50500 1010

= 1.66 x 10-4

M

PH = -logK

OH = 10.22


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[OH] = 0.100 50500 1010 = 1.66 x 10

-4M

PH = -logK

OH = 10.91

[OH] = 0.100 50500 1010 = 1.66 x 10

-4M

PH = -logK

OH= 11.21

[OH] = 0.100 50500 1010 = 1.66 x 10

-4M

PH = -logK

OH = 11.50

[OH] = 0.100 50500 1010 = 1.66 x 10

-4M

PH = -logK

OH = 11.67

[OH] = 0.100 50500 1010 = 1.66 x 10

-4M

PH = -logK

OH = 11.79


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[OH] = 0.100 50500 1010 = 1.66 x 10

-4M

PH = -logK

OH = 11.88

[OH] = 0.100 50500 1010 = 1.66 x 10

-4M

PH = -logK

OH = 11.95TABLE

The position The addition volume of NaOH (ml) pH

Part I 0.00 3.93

Part II 0.50 4.87

1.00 5.20

2.00 5.55

3.00 5.78

4.00 5.97

5.00 6.15

6.00 6.33

7.00 6.52

8.00 6.75

9.00 7.10

9.50 7.43

9.90 8.15

Part III 10.00 9.18

Part IV 10.10 10.22


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10.50 10.91

11.00 11.21

12.00 11.50

13.00 11.67

14.00 11.79

15.00 11.88

16.00 11.95

C.TITRATION OF STRONG ACID WITH WEAK BASE

PART 1

B + H2O BH+

+ OH-

(0.8364-x) (x) (x)

Kb =x2

ax =x2

008364x

1.69 x 109

=

x2

008364x

x2 01410x109x = 1.18 x x2


19/42

[OH] = 1.18 x 105POH = -log 1.18 x 105

= 4.93

PH = 14-4.93

= 9.07

PART 2

Ka =K

K=

1014169x109 = 0.59 x 10

5

Amount of pyridine that titrated =050

1960=0.025


1960 0.97PH = pKa + log

H

PH = - logK

K+ log

0970

025

= -log 059x105 + log 38.80= 5.23 + 1.58


20/42

= 6.81

Ka =K

K=

1014169x109 = 0.59 x 10

5


1960 =0.05


1960 0.94PH = pKa + log

H

PH = - logK

K+ log

094005

= -log 059x105 + log 18.9= 5.23 + 1.27

= 6.50

Ka =K

K=

1014169x109 = 0.59 x 10

5


1960 =0.10

Amount of pyridine that titrated = 19602001960 0.89PH = pKa + log

H


21/42

PH = - logK

K+ log

089010

= -log 0

59

x

10

5 + log 8.9= 5.23 + 0.95

= 6.18

Ka =K

K=

1014169x109 = 0.59 x 10

5

Amount of pyridine that titrated =

4

00

1960 =0.20Amount of pyridine that titrated =

19600501960 0.79

PH = pKa + log

H

PH = - logK

K

+ log07902

= -log 059x105 + log 38.80= 5.23 + 0.596

= 5.826

Ka =K

K=10

14

169x109 = 0.59 x 105


1960 =0.25


22/42


1960 0.98PH = pKa + log

H

PH = - logK

K+ log

098025

= -log 059x105 + log 3.92= 5.23 + 0.593

= 5.823

PART 3

BH+

B + H+

(a-x) (x) (x)

[A-] = 0.08364

= 0.08364 x 0.56 = 0.046 MKa =

x2

ax =x2

0046x

x = 0.52 x 10-3

[H+] =0.52 x 10

-3

PH = - log 0.52 x 10-3


23/42

= 3.27

PART 4

[H+] = (0.1067)

=1.18 x 10

-3M

PH = -log 1.18 x 10-3

M

= 2.93

[H+] = (0.1067)

=2.10 x 10

-3

M

PH = -log 2.10 x 10-3

M

= 2.67

[H+] = (0.1067)

=3.24 x 10-3 M

PH = -log 3.24 x 10-3

M

= 2.5


24/42

[H+] = (0.1067)

=4.50 x 10

-3M

PH = -log 4.50 x 10-3

M

= 2.346

[H+] = (0.1067)

=6.66 x10

-3

M

PH = -log6.66 x10-3

M

= 2.17

[H+] = (0.1067)

=0.01067M

PH = -log0.01067M

= 1.9

TABLE

The position The addition volume of NaOH (ml) pH

Part I 0.00 9.07

Part II 0.50 6.80

1.00 6.50

2.00 6.18


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4.00 5.826

5.00 5.823

Part III 19.60 3.27

Part IV 20.10 2.93

20.50 2.67

21.00 2.5

22.00 2.346

23.00 2.17

25.00 1.9

4.TITRATION OF ACID OR BASE WITH DIFFERENT

STRENGTH

PART 1

B + H2O BH+

+ OH-

(0.10-x) (x) (x)

Kb=HOH

By substitution we gotx

010x = 311x103[OH

-] = 3.11 x 10

-3M


26/42

POH = - log 3.11 x 10-3

= 2.507

PH = 14 2.507 = 11.49

Or directly by using formula

[H+] =

K

PH = 11.49

PART 2

PH = pKa2 + log

H

= 10.0 + log

9

1

= 11.00

PH = pKa2 + log

H

= 10.0 + log8515

= 10.75


27/42

PH = pKa2 + log

H

= 10.0 + log8

2

= 10.60

PH = pKa2 + log

H

= 10.0 + log7

3

= 10.37

PH = pKa2 + log

H

= 10.0 + log6

4

= 10.18

PH = pKa2 + log

H

= 10.0 + log5

5

= 10.00

PH = pKa2 + log

H

= 10.0 + log4

6


28/42

= 9.82

PH = pKa2 + log

H

= 10.0 + log3

7

= 9.63

PH = pKa2 + log

H

= 10.0 + log 28

= 9.40

PH = pKa2 + log

H

= 10.0 + log

1

9

= 9.05

PH = pKa2 + log

H

= 10.0 + log05

19

5

= 8.72

PH = pKa2 + log

H


29/42

= 10.0 + log0199

= 8.00

PART 3

[H

+

] =

Where Ka1 and Ka2 are disosiation constant of acid

BH22+

.The concentration of BH+

(C) is determined by the

initial concentration of B solution :

C = (0.1 M)

= 0.050 M

Wet got..

[H] = ()()()()

()

= 3.16 x 10-8

M

pH = - log 3.16 x 10-8

= 7.50


30/42

In this case, pH =

PART 4

pH = pKa + log

= 5.0 + log

= 6.995

= 7.00

pH = pKa + log

= 5.0 + log

= 6.278

= 6.28


31/42

pH = pKa + log

= 5.0 + log = 5.95

pH = pKa + log

= 5.0 + log

= 5.6

pH = pKa + log

= 5.0 + log

= 5.37

pH = pKa + log

= 5.0 + log


32/42

= 5.18

pH = pKa + log

= 5.0 + log

= 5.0

pH = pKa + log

= 5.0 + log

= 4.82

pH = pKa + log

= 5.0 + log

= 4.63


33/42

pH = pKa + log

= 5.0 + log = 4.40

pH = pKa + log

= 5.0 + log

= 4.05

pH = pKa + log

= 5.0 + log

= 3.72


34/42

PART 5

C = (0.10 M )

= 0.0333 M

The pH of solution is determined by the disosiation

reaction of BH22+

BH2+2

BH+

+ H+

(C-x) (x) (x)

Ka1 =

= 1.0 x 10-5

x = 5.72 x 10-4

[H+] = 5.72 x 10

-4M

pH = - log 5.72 x 10-4

= 3.24


35/42

PART 6

[H+] = (0.10 M)

= 3.22 x 10-4

pH = - log 3.22 x 10-4

= 2.49

[H+

] = (0.10 M)

= 6.25 x 10-3

pH = - log 6.25 x 10-3

= 2.20

[H+] = (0.10 M)

= 9.09 x 10-3pH = - log 9.09 x 10

-3

= 2.04

[H+] = (0.10 M) = 1.18 x 10-2pH = - log 1.18 x 10

-2

= 1.92


36/42

[H+] = (0.10 M)

= 1.43 x 10-2 M

pH = - log 1.43 x 102-

= 1.85

TABLE

Part The addition volume of HCl pH

Part I 0 11.49

1 11.00

1.5 10.75

2 10.60

3 10.37

4 10.18

Part II 5 10.00

6 9.82

7 9.63

8 9.40

9 9.05

9.5 8.72

9.9 8.00

Part III 10 7.50

10.1 7.0

10.5 6.28

11 5.95

12 5.6

13 5.37


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14 5.18

Part IV 15 5.00

16 4.82

17 4.63

18 4.40

19 4.05

19.5 3.72

Part V 20 3.24

21 2.49

22 2.20

Part VI 23 2.04

24 1.92

25 1.85


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Andre Anusta Barus (Chemistry Education 2010)

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