Titrasi Potensiometri ( Reaksi Dan Analisa Data )

VIII. REAKSI KIMIA

Arginin + H2SO4

O O

C – OH C – OH

2NH2 – C – H + H2SO4 2NH3+ - C – H + SO4

2-

(CH2)3 (CH2)3

NH NH

C = N+H C = N+H

NH2 NH2

Arginin + NaOH

O O

C – OH C – OH

2NH2 – C – H + NaOH 2NH3+ - C – H + Na- + H2O

(CH2)3 (CH2)3

NH NH

C = N+H C = N+H

NH2 NH2

IX.ANALISA DATA

A. Titrasi dengan NaOH 2 M

1. Menghitung Volume Koreksi dan % Koreksi

V koreksi = (v.titran pada asam amino – V.titran pada air)

% koreksi = V titran pada asamamino−V titran pada air

V titran pada airx100 %

a. Glisin

Volume NaOH pada Glysin = 2,5 ml, pH=11,03

NaOH pada blanko = 1 ml, pH=12,03

Volume koreksi= 2,5 ml- 1 ml= 1,5 ml

% koreksi = 2,5−1

1x 100 %=150 %

b. Asam Glutamat

Volume Asam Glutamat = 2,5 ml, pH=12

Volume Asam Glutamat blanko=1 ml, pH= 12,03

Volume Koreksi= 2,5 ml- 1 ml=1,5 ml

%koreksi=2,5−1

1x 100 %=150 %

c. Alanin

Volume Alanin = 3 ml, pH=12,02

Volume Alanin=1 ml, pH= 12,03

Volume Koreksi= 3 ml- 1 ml=2 ml

%koreksi=3−1

1x 100 %=200 %

d.Arginin

Volume Arginin = 1 ml, pH=12,02

Volume Arginin=1 ml, pH= 12,03

Volume Koreksi= 1 ml- 1 ml= 0 ml

%koreksi=1−1

1x 100 %=0 %

2. Menghitung pH secara teori dan secara praktek.

Glisin

Secara teori

Diket :

m NH2CH2COOH dalam 250 ml pelarut = 0,4 gr x 6,25 = 2,5 gram

Mr NH2CH2COOH = 75 gr/mol

V H2O = 250 ml = 0,25 L

Penyelesaian :

n NH2CH2COOH dalam 250 ml pelarut = m

Mr= 2,5 gr

75 gr /mol0,033 mol

M NH2CH2COOH = nV

=0,033 mol0,25 L

=0.132 M

n NH2CH2COOH = V . M =40 ml .0 .132 M=5,28 mol

Pada Volume 2,5 ml

n NaOH = 2.5 ml (2 M) = 5 mmol +NH3CH2COO- + OH- → NH2CH2COO- + H2O

m 5,28 mmol 5 mmol - -

b 5 mmol 5 mmol 5 mmol 5 mmol

s 0,28 mmol 0 mmol 5 mmol 5 mmol

[ H+ ]=10−9,6 0 , 285

= 5,6 x 10−12

pH = 11,25

Asam Glutamat

Secara teori

Diket :

m NH2CHCOOH(CH2)2COOH dalam 250 ml pelarut = 0,4 gr x 6,25 = 2,5gram

Mr NH2CHCOOHCH2CH2COOH = 147gr/mol

V H2O = 250 ml = 0,25 L

Penyelesaian :

n NH2CHCOOH(CH2)2COOH dlm 250 ml pelarut= m

Mr= 2,5 gr

147 gr /mol=0,017 mol

M NH2CHCOOHCH2CH2COOH = nV

=0,017 mol0,25 L

=0,068 M

n NH2CHCOOHCH2CH2COOH= V . M =40 ml .0,068 M=2,72mol

n NaOH = 2.5 ml (2 M) = 5 mmol +NH3CHCOOHCH2CH2COO- + OH- → NH2CHCOOHCH2CH2COO- + H2O


b 2,72 mmol 2,72 mmol 2,72 mmol 2,72 mmol

s 0 mmol 2,28 mmol 2,72 mmol 2,72 mmol

[OH− ]=2 ,282 ,72

= 0 ,83

pOH = - log 0,83= 0,0809

pH = 13,91

Alanin

Secara teori

Diket :

m NH2CHCH3COOH dalam 250 ml pelarut = 0,4 gr x 6,25 = 2,5gram

Mr NH2CHCH3COOH = 89gr/mol

V H2O = 250 ml = 0,25 L

Penyelesaian :

n NH2CHCH3COOH dalam 250 ml pelarut = m

Mr= 2,5 gr

89 gr /mol=0,028 mol

M NH2CH2COOH = nV

=0,028 mol0,25 L

=0,112 M

n NH2CH2COOH = V . M =40 ml .0,112 M=4,48mol

n NaOH = 3 ml (2 M) = 6 mmol +NH3CHCH3COO- + OH- → NH2CHCH2COO- + H2O




[OH− ]= 1 ,524 , 48

= 0 ,33

pOH = - log 0,33 = 0,48

pH = 13, 52

Arginin

Secara teori

Diket :

m NH2C5N3H11COOH dalam 250 ml pelarut = 0,4 gr x 6,25 = 2,5 gram

Mr NH2C5N3H11COOH = 174 gr/mol

V H2O = 250 ml = 0,25 L

Penyelesaian

n NH2C5N3H11COOH dalam 250 ml pelarut = m

Mr= 2,5 gr

174 gr /mol=0,014 mol

M NH2C5N3H11COOH = nV

=0,014 mol0,25 L

=0,056 M

n NH2C5N3H11COOH = V . M =40 ml . 0,056 M=2,24 mol

n NaOH = 1 ml (2 M) = 2 mmol+NH3C5N3H11COO- + OH- → NH2C5N3H11COO- + H2O

m 2,24 mmol 2,24 mmol - -

b 2,24 mmol 2 mmol 2 mmol 2 mmol

s 0 mmol 0,24 mmol 2 mmol 2 mmol

[OH− ]=0 , 242

= 0 ,12

pOH = - log 0,12 = 0,92

pH= 13,08

3. Menghitung % kesalahan

% Kesalahan = |pHteori−pHpraktekpHteori

|x 100%

Glisin

Pada volume 2,5 ml, pH praktek=11,03, pH teori= 11,25

% kesalahan =

11 , 25−11 ,0311 , 25

x 100 %=1 , 95 %

Asam Glutamat

Pada volume 2,5 ml, pH praktek=12, pH teori= 13,91

% kesalahan =

13 , 91−11 , 0313 , 91

x 100 %=20 ,7 %

Alanin

Pada volume 3 ml, pH praktek=12,02, pH teori= 13,52

% kesalahan =

13 , 52−11 , 0313 , 52

x100 %=18 , 41%

Arginin

Pada volume 1 ml, pH praktek=12,02, pH teori= 13,08

% kesalahan =

13 , 08−11 ,0313 , 08

x 100 %=15 , 67 %

4. Volume NaOH untuk mencapai pH 12

pH = 12

pOH = 14 -12 = 2

[OH-] = 10-2

[NaOH] = 10-2 = 0,01 M

n NaOH mula – mula = x L (2M) = 2x+NH3CH2COO- + OH- → NH2CH2COO- + H2O

m 0,00525 2x - -

b 0,00525 0,00525 0,00525 0,00525

s - x – 0,00525 mol 0,00525 mol 0,00525 mol

[ NaOH ] = 2 x− 0 ,00525x + 0 ,04

0 ,01 = 2 x− 0 ,00525x + 0 , 04

0 ,01 x + 0 ,0004= 2x− 0 ,00525

0 ,0004 + 0 , 00525= 2 x− 0 ,01x

0 ,00565= 1 ,99 x

x≈ 0 ,00284

V NaOH = 0,00284 L = 2,84 ml

Kurva Titrasi Glisin dengan NaOH secara Praktek

1 2 3 4 5 6 7 8 9 100

1

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12

13

14

B . Titrasi dengan H2SO4 2 M

1.Menghitung Volume Koreksi dan % Koreksi

V koreksi = (v.titran pada asam amino – V.titran pada air)

% koreksi = V titran pada asamamino−V titran pada air

V titran pada airx100 %

a. Glisin

Volume H2SO4 pada Glysin = 2 ml, pH=1,77

H2SO4 pada blanko =0,5 ml, pH=1,82

Volume koreksi= 2 ml- 0,5 ml= 1,5 ml

% koreksi = 2−0,5

0,5x100 %=300 %

Volume H2SO4 pada Glysin = 2,5 ml, pH=1,66

H2SO4 pada blanko = 1 ml, pH=1,62


% koreksi = 2,5−1

1x 100 %=150 %

Volume H2SO4 pada Glysin = 4 ml, pH=1,25

H2SO4 pada blanko = 3,5 ml, pH=1,23


% koreksi = 4−3,5

3,5x 100 %=14,28 %

b. Asam Glutamat

Volume H2SO4 pada Asam Glutamat = 1 ml, pH=1,6


Volume koreksi= 1 ml- 1 ml= 0 ml

% koreksi = 1−1

1x 100 %=0 %

Volume H2SO4 pada Asam Glutamat = 1,5 ml, pH=1,52


Volume koreksi= 1,5 ml- 1,5 ml= 0 ml

% koreksi = 1,5−1,5

1,5x100 %=0 %




% koreksi = 2−2

2x 100 %=0 %




% koreksi = 2,5−2,5

2,5x100 %=0 %




% koreksi = 3−3

3x100 %=0 %




% koreksi = 3,5−3,5

3,5x100 %=0 %

c. Alanin

Volume H2SO4 pada Alanin = 2,5 ml, pH=1,61



% koreksi = 2,5−1

1x 100 %=150 %




% koreksi = 3,5−2

2x 100 %=75 %

Volume H2SO4 pada Alanin = 4 ml, pH=1,36


Volume koreksi =4 ml- 2,5 ml= 1,5 ml

% koreksi = 4−2,5

2,5x 100 %=60 %




% koreksi = 4,5−3

3x100 %=50 %

d. Arginin

Volume H2SO4 pada Arginin = 2 ml, pH=1,67



% koreksi = 2−1

1x 100 %=100 %




% koreksi = 3−2

2x 100 %=50 %

Volume H2SO4 pada Arginin = 3,5 ml, pH=1,33



% koreksi = 3,5−2,5

2,5x100 %=40 %




% koreksi = 4−3

3x 100 %=30 %

Volume H2SO4 pada Arginin = 4,5 ml, pH=1,24



% koreksi = 4,5−3,5

3,5x100 %=28,57 %

2. Menghitung pH secara teori maupun praktek

Glisin

Secara teori

Diket :

m NH2CH2COOH dalam 250 ml pelarut = 0,4 gr x 6,25 = 2,5 gram

Mr NH2CH2COOH = 75 gr/mol

V H2O = 250 ml = 0,25 L

Penyelesaian :

n NH2CH2COOH dalam 250 ml pelarut = m

Mr=2,5 gr

75 grmol

=0,033 mol

M NH2CH2COOH = nV

=0,033 mol0,25 L

=0.132 M

n NH2CH2COOH = V . M =40 ml .0 .132 M=5,28 mol

Pada volume 2 ml

n H2SO4 = 2 ml (2 M) = 4 mmol

2+NH3CH2COO- + H2SO4 → 2 +NH3CH2COOH + SO4

2-


b 5,28 mmol mmol 4 mmol 4 mmol


[ H+ ]=10−9,6 1,28 4

=3,2 x 10−11

pH = 10,49

Pada volume 2,5 ml

n H2SO4 = 2,5 ml (2 M) = 5 mmol


2-




[ H+ ]=10−9,6 0,28 5

=5,6 x 10−12

pH = 11,25

Pada volume 4 ml

n H2SO4 = 4 ml (2 M) = 8 mmol


2-




[ H+ ]=2,72 44

=0 ,061

[H+] = 2 x 0,061 = 0,12

pH = 0,92

Asam Glutamat

Secara teori

Diket :

m NH2CHCOOH(CH2)2COOH dalam 250 ml pelarut = 0,4 gr x 6,25 = 2,5gram

Mr NH2CHCOOHCH2CH2COOH = 147gr/mol

V H2O = 250 ml = 0,25 L

Penyelesaian :

n NH2CHCOOH(CH2)2COOH dlm 250 ml pelarut=m

Mr= 2,5 gr

147 gr /mol=0,017 mol

M NH2CHCOOHCH2CH2COOH = nV

=0,017 mol0,25 L

=0,068 M

n NH2CHCOOHCH2CH2COOH=V . M =40 ml .0,068 M=2,72mol

Pada volume 1 ml

n H2SO4 = 1 ml (2 M) = 2 mmol

2+NH3CHCOOH(CH2)2COO+H2SO4→2+NH3CHCOOHCH2CH2COO + SO42-




[ H+ ]=10−9,6 0,72 2

=3,6 x 10−11

pH = 10,44

Pada volume 1,5 ml

n H2SO4 = 1,5 ml (2 M) = 3 mmol

2NH3CHCOOH(CH2)2COO-+H2SO4→2+NH3CHCOOH(CH2)2COOH+ SO42-




[ H+ ]=0,28 41 ,5

=0 ,0067

[H+] = 2 x 0,0067 = 0,013

pH = 1,88

Pada volume 2 ml

n H2SO4 = 2 ml (2 M) = 4 mmol

2+NH3CHCOOH(CH2)2COO-+H2SO4→2+NH3CHCOOH(CH2)2COOH+SO42-




[ H+ ]=1,28 42

=0 ,0304

[H+] = 2 x 0,0304 = 0,0608

pH = 1,21

Pada volume 2,5 ml

n H2SO4 = 2,5 ml (2 M) = 5 mmol





[ H+ ]=1,28 42

=0 ,0304

[H+] = 2 x 0,0304 = 0,0608

pH = 1,21

Pada volume 3 ml

n H2SO4 = 3 ml (2 M) = 6 mmol

2+NH3CHCOOHCH2CH2COO-+H2SO4→2+NH3CHCOOHCH2CH2COOH+SO42-




[ H+ ]=3,28 43

=0 ,0762

[H+] = 2 x 0,0762 = 0,15

pH = 0,82

Pada volume 3,5 ml

n H2SO4 = 3,5 ml (2 M) = 7 mmol





[ H+ ]=4,28 43 ,5

=0 ,098

[H+] = 2 x 0,098 = 0,19

pH = 0,72

Alanin

Secara teori

Diket :

m NH2CHCH3COOH dalam 250 ml pelarut = 0,4 gr x 6,25 = 2,5

gram

Mr NH2CHCH3COOH = 89gr/mol

V H2O = 250 ml = 0,25 L

Penyelesaian :

n NH2CHCH3COOH dalam 250 ml pelarut =

mMr

= 2,5 gr89 gr /mol

=0,028 mol

M NH2CH2COOH = nV

=0,028 mol0,25 L

=0,112 M

n NH2CH2COOH = V . M =40 ml .0,112 M=4,48mol

Pada volume 2,5 ml

n H2SO4 = 2,5 ml (2 M) = 5 mmol

2 +NH3CHCH3COO- + H2SO4→ 2+NH3CHCH2COOH + SO42-




[ H+ ]=0,52 42 ,5

=0 ,012

[H+] = 2 x 0,012 = 0,024

pH = 1,61

Pada volume 3,5 ml

n H2SO4 = 3,5 ml (2 M) = 7 mmol

2+NH3CHCH3COO- + H2SO4→ 2+NH3CHCH2COOH + SO42-




[ H+ ]=2,52 43 ,5

=0 ,057

[H+] = 2 x 0,057 = 0,114

pH = 0,94

Pada volume 4 ml

n H2SO4 = 4 ml (2 M) = 8 mmol

2 +NH3CHCH3COO- + H2SO4→ 2+NH3CHCH2COOH + SO42-




[ H+ ]=3,52 44

=0 ,08

[H+] = 2 x 0,08 = 0,16

pH = 0,79

Pada volume 4,5 ml

n H2SO4 = 4,5 ml (2 M) = 9 mmol

2+NH3CHCH3COO- + H2SO4→ 2+NH3CHCH2COOH + SO42-




[ H+ ]=4,52 44 , 5

=0 ,10

[H+] = 2 x 0,10 = 0,2

pH = 0,69

Arginin

Secara teori

Diket :

m NH2C5N3H11COOH dalam 250 ml pelarut = 0,4 gr x 6,25 = 2,5 gram

Mr NH2C5N3H11COOH = 174 gr/mol

V H2O = 250 ml = 0,25 L

Penyelesaian

n NH2C5N3H11COOH dalam 250 ml pelarut =

m

Mr= 2,5 gr

174 gr /mol=0,014 mol

M NH2C5N3H11COOH = nV

=0,014 mol0,25 L

=0,056 M

n NH2C5N3H11COOH = V . M =40 ml . 0,056 M=2,24 mol

Pada volume 2 ml

n H2SO4 = 2 ml (2 M) = 4 mmol

2+NH3C5N3H11COO- + H2SO4 → 2+NH3C5N3H11COOH + SO4

2-




[ H+ ]=2,76 42

=0 ,065

[H+] = 2 x 0,065 = 0,13

pH = 0,88

Pada volume 3 ml

n H2SO4 = 3 ml (2 M) = 6 mmol


2-




[ H+ ]=3,76 43

=0 ,087

[H+] = 2 x 0,087 = 0,174

pH = 0,75

Pada volume 3,5 ml

n H2SO4 = 3,5 ml (2 M) = 7 mmol


2-




[ H+ ]=4,76 43 ,5

=0 ,109

[H+] = 2 x 0,109 = 0,218

pH = 0,66

Pada volume 4 ml

n H2SO4 = 4 ml (2 M) = 8 mmol


2-




[ H+ ]=5,76 44

=0 ,13

[H+] = 2 x 0,13 = 0,26

pH = 0,58

Pada volume 4,5 ml

n H2SO4 = 4,5 ml (2 M) = 9 mmol


2-




[ H+ ]=6,76 44 , 5

=0 ,15

[H+] = 2 x 0,15 = 0,3

pH = 0,52

3. Menghitung persentase % kesalahan

Glisin

H2SO4 =2 ml, pH praktek = 1,77 ; pH teori = 10,49

% kesalahan = |10 ,49−1 ,7710 ,49

|x100 %=83 ,12 %

H2SO4 =2,5 ml, pH praktek = 1,66 ; pH teori = 11,25

% kesalahan = |11 , 25−1 ,6611 , 25

|x100 %=85 , 24 %


% kesalahan

= |0 , 92−1 , 25

0 ,92|x 100 %=3 , 86 %

Asam Glutamat


% kesalahan = |10 ,44−1,610 ,44

|x100%=84 ,67 %


% kesalahan = |1 ,88−1 ,52

1 ,88|x100 %=19 , 14 %


% kesalahan = |1 ,21−1 , 42

1 , 21|x100 %=17 ,35 %


% kesalahan = |1 ,21−1 ,35

1 , 21|x100%=11 ,57 %


% kesalahan = |0 ,82−1 ,29

0 ,82|x 100 %=57 , 31 %


% kesalahan = |0 ,72−1 ,25

0 ,72|x100 %=73 ,61%

Alanin


% kesalahan = |1 ,61−1 ,61

1 ,61|x100 %=0 %


% kesalahan = |0 ,94−1 ,47

0 ,94|x100 %=56 , 38 %


% kesalahan = |0 , 79−1 , 36

0 ,79|x 100 %=72 ,15 %


% kesalahan = |0 , 69−1 ,29

0 ,69|x 100 %=86 , 95 %

Arginin


% kesalahan = |0 , 88−1 ,67

0 ,88|x100 %=89 , 77 %


% kesalahan = |0 ,75−1 , 41

0 ,75|x100 %=88%


% kesalahan = |0 ,66−1 ,33

0 ,66|x100 %=101 ,51 %


% kesalahan = |0 ,58−1 ,31

0 ,58|x100 %=125 ,86 %


% kesalahan

= |0 ,52−1 , 24

0 ,52|x 100 %=138 ,46 %

4. Volume H2SO4 untuk mencapai pH 1,2

[H+] = 10-1,2 ; H2SO4] = 10-1,2 /2 = 0,0315 M

n H2SO4 mula – mula = x L (1 M) = x


2-

m 0,00525 x - -

b 0,005 25 0,002625 0,00525 0,002625

s - x – 0,002625 mol 0,00525 mol 0,002625 mol

[ H 2 SO4 ] = x− 0 ,002625x + 0 , 04

0 ,0315 = x− 0 ,002625x + 0 ,04

0 ,0315 = x− 0 ,002625x + 0 ,04

0 ,0315 x + 0 , 00126= x− 0 , 002625

0 ,00126 + 0 , 002625= x− 0 ,0315 x

0 ,003885= 0 ,9685 x

x≈ 0 ,00401

V H2SO4 = 0,00401 L = 4,01 mL

Kurva Titrasi Glisin dengan NaOH secara Praktek

1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

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Titrasi Potensiometri ( Reaksi Dan Analisa Data )

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