acid base

ACID-BASE TITRATION

1. Objectives

a. Student can prepare acid standard solution

b. Student can determine the normally of the acid/base using a standard

solution

c. Student can determine the equivalence point using titration curve

2. Introduction

An acid-base titration is the determination of the concentration of an acid or

base by exactly neutralizing the acid or base with an acid or base of known

concentration. This allows for quantitives analysis of the concentration of an

unknown acid or base solution. It makes use of the neutralization reaction that

occurs between acids and bases will react if their formula are known. Acid base

titrations can also be used to find percent purity of chemical.

Alkalimetry is the specialized analytic use of acid base titration to determine

the concentration of a basic (synonymous of alkaline) substance. Acidimetry,

sometimes spelled acidometry is the same concept of specialized analytic acid

base titration, but for an acidic substance.

Before starting the titration of a suitable pH indicator must be chosen. The

equivalence point of the reaction, the point at which equivalence amount of the

reactant have reacted, will have a pH dependent on the relative strengths of the

acid and base used. The pH of the equivalence point can be estimated using the

following rules :

A strong acid will react with a stong base to form a neutral (pH=7) solution

A strong acid will react with a weak base to form an acidic (pH<7) solution

A weak acid will react with a strong base to form a basic (pH>7) solution

When a weak acid reacts with a weak base, the equivalence point solution will

be basic if the base is stronger and acidic if the acid is stronger. If both are equal

strength then the equivalence pH will be neutral. However, weak acids are not

often titrated against weak bases because the colour change shown with the

indicator is often quick, and therefore very difficult for the obsever to see the

change of colour.

The pH curve diagram below represents the titration of a strong acid with a

strong base.

As we add strong base to a strong acid, the pH increase slowly until we near the

equivalence point, where the pH increases dramatically with a small increase in the

volume of base added. This is due to the logarithmic nature of the pH system

pH=−log¿. At the equivalence point, the pH is 7 as expected passing the equivalence

point by adding more base initially increases the pH dramatically and eventually slopes

off. At the equivalence point : V 1 × N1=V 2 × N2 (chemwiki.ucdavis.edu)

1. The initial pH is “approximated for a weak acid solution in water using the

equation” pH=−log √Ka × M a

2. The pH before the equivalence point

pH=pKa+ log¿

3. At the equivalence point

pH + pOH = 14

KaxKb = 10−14

At equivqlence Na x Va = Nb x Vb

pH=14+log √ M a × M b × Kw

( M a+M b ) K a

4. After the equivalence point

pH=14+logM b ×V b−M a× a

(V a+V b )

3. Apparatus and Reagent

Apparatus : Reagent :

1) 100 ml volumetric flask 1) oxalic acid crystal

2) 100 ml erlenmeyer 2) distilled water

3) 50 ml burette 3) 1 M NaOH

4) Clamps and statif 4) 1 M HCl

5) 10 ml volumetric pipet 5) 1 M CH3COOH

6) pH meter/universal indicator 6) PP indicators

4. Procedure

5. Data Observation

a. Preparation of primary standart solution of H2C2O4(COOH)2.2H2O

Mass of oxalic acid : 1.2607 grams

Volume of solution : 100 ml

Normality of oxalic acid : 0.2 N

b. Determine the concentration of NaOH

Volume of oxalic acid : 10 ml

Volume of NaOH (1) : 19.8 ml


Average volume of NaOH : 19.6 ml

c. Determination of the concentration of HCl using secondary standard

solution of NaOH

Volume of HCl : 10 ml




d. Titration Curve

Relationship of the volumeof titrant with pH in acid base titration

Table 1. pH of NaOH added to 10 ml of 0.1 M HCl

No.

Volume of NaOH added to 10 ml of 0.1 M

HCl pH

1 0 1

2 1 1

3 2 2

4 3 2

5 4 2

6 5 10

7 6 10

8 7 10

9 8 10

10 9 7

11 10 10

12 11 10

13 12 10

14 13 10

15 14 10

0 2 4 6 8 10 12 14 160

2

4

6

8

10

12

Graph 1. Relationship betweenpH and volume NaOH that added to HCl solution

Vol NaOH

pH

Table 2. pH of NaOH added to 10 ml of 0.1 M CH3COOH

No.

Volume of NaOH added

to 10 ml of 0.1 M CH3COOH pH

1 0 3

2 1 4

3 2 4

4 3 4

5 4 5

6 5 5

7 6 4

8 7 3

9 8 9

10 9 6

11 10 10

12 11 10

13 12 10

14 13 10

15 14 10

0 2 4 6 8 10 12 14 160

2

4

6

8

10

12

Graph 2. Relationship between pHand volume NaOH that added to 10 ml CH3COOH

Vol NaOH

pH

6. Calculation


1) Mass of oxalic acid : 1.2607 grams

2) Volume of solution : 100 ml

3) Molarity of oxalic acid :

M oxalic acid=m

Mr . V

M oxalic acid=1.2607

126.07 ×0.1

M oxalic acid=0.1 M

4) Normality of oxalic acid :

Noxalic acid=M × valence

Noxalic acid=0.1× 2

Noxalic acid=0.2 N


1) Volume of oxalic acid : 10 ml

2) Volume of NaOH (1) : 19.8 ml


4) Average volume of NaOH : 19.6 ml

5) Normality of NaOH (N2) :

2 NaOH (aq )+ (COOH )2( aq) → (COONa )2 (aq )+2 H 2 O

V 1 × N1=V 2 × N2

10 ×0.2=19.6 × N2

N2=0.102 N

6) Molarity of NaOH :

M 2=0.102

1

M 2=0.102 M


solution of NaOH

1) Volume of HCl : 10 ml



4) Average volume of NaOH : 12.4 ml

5) Normality of HCl (NHCl) :

V 1 × N1=V 2 × N2

10 × N1=12.4 ×0.102

N2=0.126 N

6) Molarity of HCl :

M 2=0.126

1

M 2=0.126 M

d. Titration Curve

1) Titration between NaOH (strong base) with HCl (strong acid)

1.1) V NaOH=0ml

V HCl=10 ml

M NaOH =0.102 M

M HCl=0.126 M

nNaOH=¿0 mmol

nHCl=1.26 mmol

HCl + NaOH ≠ NaCl + H2O

M HCl=1.2610

M HCl=0.126 M

¿

¿

¿

pH=−log¿¿

pH=−log [0.126 ]

pH=0.9

1.2) V NaOH=1ml

V HCl=10 ml

M NaOH =0.102 M

M HCl=0.126 M

nNaOH=¿0.102 mmol

nHCl=1.26 mmol

HCl + NaOH → NaCl + H2O

Mula-mula 1.26 0.102 - -

Reaksi 0.102 0.102 0.102 0.102

Sisa 1.158 0 0.102 0.102

M HCl=1.158

11M HCl=0.105 M

¿

¿

¿

pH=−log¿¿

pH=−log [0.105 ]

pH=0.98

1.3) V NaOH=2ml

V HCl=10 ml

M NaOH=0.204 M

M HCl=0.126 M

nNaOH=¿0.102 mmol

nHCl=1.26 mmol


Mula-mula 1.26 0.204 - -

Reaksi 0.204 0.204 0.204 0.204

Sisa 1.056 0 0.204 0.204

M HCl=1.056

12

M HCl=0.088 M

¿

¿

¿

pH=−log¿¿

pH=−log [ 8.8× 10−2 ]pH=2−0.94

pH=1.06

1.4) V NaOH=3ml

V HCl=10 ml

M NaOH =0.306 M

M HCl=0.126 M

nNaOH=¿0.102 mmol

nHCl=1.26 mmol


Mula-mula 1.26 0.306 - -

Reaksi 0.306 0.306 0.306 0.306

Sisa 0.954 0 0.306 0.306

M HCl=0.954

13

M HCl=0.073 M

¿

¿

¿

pH=−log¿¿

pH=−log [7.3 ×10−2 ]pH=2−0.86

pH=1.14

1.5) V NaOH=4 ml

V HCl=10 ml

M NaOH =0.102 M

M HCl=0.126 M

nNaOH=0.408 mmol

nHCl=1.26 mmol


Mula-mula 1.26 0.408 - -

Reaksi 0.408 0.408 0.408 0.408

Sisa 0.852 0 0.408 0.408

M HCl=0.852

14

M HCl=0.060 M

¿

¿

¿

pH=−log6× 10−2

pH=2−0.78

pH=1.22

1.6) V NaOH=5ml

V HCl=10 ml

M NaOH =0.306 M

M HCl=0.126 M

nNaOH=¿0.51 mmol

nHCl=1.26 mmol


Mula-mula 1.26 0.501 - -

Reaksi 0.501 0.501 0.501 0.501

Sisa 0.75 0 0.501 0.501

M HCl=0.7515

M HCl=0.05 M

¿

¿

¿

pH=−log¿¿

pH=−log [5 ×10−2 ]pH=2−0.7

pH=1.3

1.7) V NaOH=6ml

V HCl=10 ml

M NaOH =0.306 M

M HCl=0.126 M

nNaOH=¿0.612 mmol

nHCl=1.26 mmol


Mula-mula 1.26 0.612 - -

Reaksi 0.612 0.612 0.612 0.612

Sisa 0.648 0 0.612 0.612

M HCl=0.648

16

M HCl=0.040 M

¿

¿

¿

pH=−log¿¿

pH=−log [ 4 ×10−2 ]pH=2−0.6

pH=1.4

1.8) V NaOH=7ml

V HCl=10 ml

M NaOH =0.306 M

M HCl=0.126 M

nNaOH=¿0.714 mmol

nHCl=1.26 mmol


Mula-mula 1.26 0.714 - -

Reaksi 0.714 0.714 0.714 0.714

Sisa 0.546 0 0.714 0.714

M HCl=0.546

17

M HCl=0.032 M

¿

¿

¿

pH=−log¿¿

pH=−log [3.2 ×10−2 ]pH=2−0.5

pH=1.5

1.9) V NaOH=8ml

V HCl=10 ml

M NaOH =0.306 M

M HCl=0.126 M

nNaOH=¿0.816 mmol

nHCl=1.26 mmol


Mula-mula 1.26 0.816 - -

Reaksi 0.816 0.816 0.816 0.816

Sisa 0.444 0 0.816 0.816

M HCl=0.444

18

M HCl=0.024 M

¿

¿

¿

pH=−log¿¿

pH=−log [2.4 ×10−2 ]pH=2−0.38

pH=1.62

1.10) V NaOH=9ml V HCl=10 ml

M NaOH =0.102 M M HCl=0.126 M

nNaOH=¿0.918 mmol nHCl=1.26 mmol


Mula-mula 1.26 0.918 - -

Reaksi 0.918 0.918 0.918 0.918

Sisa 0.342 0 0.918 0.9

M NaOH =0.34219

M NaOH=0.018 M

¿

¿

¿

pH=2−log 1.8

pH=2−0.25

pH=1.75

1.11) V NaOH=10 ml V HCl=10 ml

M NaOH =0.102 M M HCl=0.126 M



Mula-mula 1.26 1.02 - -

Reaksi 1.02 1.02 1.02 1.02

Sisa 0.24 0 1.02 1.02

M NaOH =0.2420

M NaOH =0.012 M

¿

¿

¿

pH=−log1.2 ×10−2

pH=2−0.079

pH=1.92

1.12) V NaOH=11ml V HCl=10 ml

M NaOH =0.102 M M HCl=0.126 M



Mula-mula 1.26 1.122 - -

Reaksi 1.122 1.122 1.122 1.122

Sisa 0.138 0 1.122 1.122

M NaOH=0.13821

M NaOH =6.57×10−3 M

¿

¿

¿

pH=−log6.57× 10−3

pH=3−0.82

pH=2.18


M NaOH =0.102 M M HCl=0.126 M



Mula-mula 1.26 1.224 - -

Reaksi 1.224 1.224 1.224 1.224

Sisa 0.036 0 1.224 1.224

M NaOH=0.03622

M NaOH=1.63× 10−3 M

¿

¿

¿

pH=−log1.63×10−3

pH=3−0.21

pH=2.79


M NaOH =0.102 M M HCl=0.126 M

nNaOH=¿ mmol nHCl=1.26 mmol


Mula-mula 1.26 1.326 - -

Reaksi 1.26 1.26 1.26 1.26

Sisa 0 0.066 1.26 1.26

M NaOH=0.06623

M NaOH=2.87×10−3 M

¿

¿

¿

pOH=−log 2.87 ×10−3

pOH=3−0.46

pOH=2.54

pH=14−pOH

pH=14−2.54

pH=11.46


M NaOH =0.102 M M HCl=0.126 M



Mula-mula 1.26 1.428 - -

Reaksi 1.26 1.26 1.26 1.26

Sisa 0 0.168 1.26 1.26

M NaOH=0.16824

M NaOH =7×10−3 M

¿

¿

¿

pOH=−log 7 ×10−3

pOH=3−0.84

pOH=2.16

pH=14−pOH

pH=14−2.16

pH=¿11.84

Titration of NaOH-CH3COOH

1.1) V NaOH=0ml V CH3 COOH=10 ml

M NaOH =0.102 M M CH3 COOH=0.1M

nNaOH=¿0 mmol nCH 3COOH=1 mmol

NaOH + CH3COOH ≠ CH3COONa + H2O

¿

¿

¿

1.2) V NaOH=1ml V CH3 COOH=10ml

M NaOH =0.102 M M CH3 COOH=0.1 M

nNaOH=¿0.102 mmol nCH 3COOH=1mmol

NaOH + CH3COOH → CH3COONa + H2O

Mula-mula 0.102 1 - -

Reaksi 0.102 0.102 0.102 0.102

Sisa 0.102 0.898 0.102 0.10

¿

¿

¿

pH=−log H+¿ ¿

pH=−log15.85 ×10−5

pH=5−1.2

pH=3.8



nNaOH=¿0.204 mmol nCH 3COOH=1 mmol


Mula-mula 0.204 1 - -

Reaksi 0.204 0.204 0.204 0.204

Sisa 0.204 0.796 0.204 0.204

¿

¿

¿

pH=−log¿¿

pH=−log7.02×10−5

pH=5−0.84

pH=4.16





Mula-mula 0.306 1 - -

Reaksi 0.306 0.306 0.306 0.306

Sisa 0.306 0.694 0.306 0.306

¿

¿

¿

pH=−log¿¿

pH=−log 4.08 ×10−5

pH=5−0.61

pH=4.39

1.5) V NaOH=4 ml V CH3 COOH=10 ml




Mula-mula 0.408 1 - -

Reaksi 0.408 0.408 0.408 0.408

Sisa 0.408 0.592 0.408 0.408

¿

¿

¿

pH=−log¿¿

pH=−log2.61×10−5

pH=5−0.42

pH=4.58





Mula-mula 0.51 1 - -

Reaksi 0.51 0.51 0.51 0.51

Sisa 0 0.49 0.51 0.51

¿

¿

¿

pH=−log¿¿

pH=−log1.73×10−5

pH=5−0.23

pH=4.77





Mula-mula 0.612 1 - -

Reaksi 0.612 0.612 0.612 0.612

Sisa 0 0.388 0.612 0.612

¿

¿

¿

pH=−log¿¿

pH=−log1.14× 10−5

pH=5−0.05

pH=4.95





Mula-mula 0.714 1 - -

Reaksi 0.714 0.714 0.714 0.714

Sisa 0 0.286 0.714 0.714

¿

¿

¿

pH=−log¿¿

pH=−log7.21×10−6

pH=6−0.86

pH=5.14





Mula-mula 0.816 1 - -

Reaksi 0.816 0.816 0.816 0.816

Sisa 0 0.184 0.816 0.816

¿

¿

¿

pH=−log¿¿

pH=−log 4.06 ×10−6

pH=6−0.61

pH=5.39





Mula-mula 0.918 1 - -

Reaksi 0.918 0.918 0.918 0.918

Sisa 0 0.082 0.918 0.918

¿

¿

¿

pH=−log¿¿

pH=−log1.61×10−6

pH=6−0.21

pH=5.79





Mula-mula 1.02 1 - -

Reaksi 1 1 1 1

Sisa 0.02 0 1 1

M NaOH = nV

M NaOH =0.0220

M NaOH =0.001 M

¿

¿

¿

pOH=−log ¿¿

pOH=−log 10−3

pOH=3

pH=14−pOH

pH=14−3

pH=11





Mula-mula 1.122 1 - -

Reaksi 1 1 1 1

Sisa 0.122 0 1 1

M NaOH = nV

M NaOH =0.12221

M NaOH =0.006 M

¿

¿

¿

pOH=−log ¿¿

pOH=−log 6 ×10−3

pOH=3−log6

pOH=3−0.78

pOH=2.22

pH=14−pOH

pH=14−2.22

pH=11.78





Mula-mula 1.224 1 - -

Reaksi 1 1 1 1

Sisa 0.224 0 1 1

M NaOH = nV

M NaOH =0.22422

M NaOH =0.01M

¿

¿

¿

pOH=−log ¿¿

pOH=−log 10−2

pOH=2

pH=14−pOH

pH=14−2

pH=12





Mula-mula 1.326 1 - -

Reaksi 1 1 1 1

Sisa 0.326 0 1 1

M NaOH = nV

M NaOH=0.32623

M NaOH=0.014 M

¿

¿

¿

pOH=−log ¿¿

pOH=−log 1.4 ×10−2

pOH=2−log1.4

pOH=2−0.15

pOH=1.85

pH=14−pOH

pH=14−1.85

pH=12.15





Mula-mula 1.428 1 - -

Reaksi 1 1 1 1

Sisa 0.428 0 1 1

M NaOH = nV

M NaOH=0.42824

M NaOH=0.018 M

¿

¿

¿

pOH=−log ¿¿

pOH=−log 1.8 ×10−2

pOH=2−log1.8

pOH=2−0.25

pOH=1.75

pH=14−pOH

pH=14−1.75

pH=12.25

Table 3. titration of NaOH-HCl

No.


HCl pH

1 0 0.9

2 1 0.98

3 2 1.06

4 3 1.14

5 4 1.22

6 5 1.3

7 6 1.4

8 7 1.5

9 8 1.62

10 9 1.75

11 10 1.92

12 11 2.18

13 12 2.79

14 13 11.46

15 14 11.84

0 2 4 6 8 10 12 14 160

2

4

6

8

10

12

14

Graph 3. Titration NaOH-HCl

Vol NaOH

pH

Table 4. titration of NaOH-CH3COOH

No.

Volume of NaOH added to 10 ml of 0.1

M CH3COOH pH

1 0 0

2 1 3.8

3 2 4.16

4 3 4.33

5 4 4.58

6 5 4.77

7 6 4.95

8 7 5.14

9 8 5.39

10 9 5.79

11 10 11

12 1111.7

8

13 12 12

14 1312.1

5

15 1412.2

5

0 2 4 6 8 10 12 14 160

2

4

6

8

10

12

14

Graph 4. Titration NaOH-CH3COOH

Vol NaOH

pH

7. Discussion

a. Preparation of primary standart solution of

H2C2O4(COOH)2.2H2O

in this experiment, we used oxalic acid to make primary standard

solution, that was a solution which was made with carefulness high. This

solution was used for standardize a solution of NaOH. This solution was

made with diluted crystal oxalic acid 1,2607 grams in 100 ml of distilled

water. From calculation with formula M= mV × Mr

we knew that the

molarity of this primary standard solution was 0.1 M. and from formula

N=M × valence we knew that the normality was 0.2 N because oxalic acid

had 2 valence. When we diluted H2C2O4 in distilled water, its be exactly on

the line because if a little bit more than levels of oxalic acid to be changed.

After that a solution must be shaken at least 12 times because if not a

solution will not homogeny.


The reaction : 2NaOH + (COOH)2 → (COONa)2 + 2H2O

On this experiment we did quantitative analysis to standardize

secondary standard solution. The primary standard solution where on

experiment this time secondary standard solution to be used was

NaOH(sodium hydroxide) and a primary standard solution was

H2C2O4.2H2O (oxalic acid). Based on the results of the experiment can be

known that there has been reaction of acid-base between oxalic acid (as a

weak acid) and NaOH (as a strong base). In the manufacture of standard

solutions of NaOH and indicator used Phenolphtalein (indicators PP). an

indicator used PP because PP colorless with the pH between 8.3-10.0 will

facilitate us in knowing that in the process has now reached an equivalent

point. A change that occur at this titration process was turned into a red

color that constant of the color of the origin of clear. Charge to a color this

happens because it had been reached the equivalent point.

The volume of NaOH required to titrate was 19.6 ml calculated from

the average two times experiment. And in the determination of

concentration NaOH obtained molarity and normality of NaOH 0.102 M

and 0.102 N. it was obtained from formula V 1 × N1=V 2 × N2.


solution of NaOH

The reaction : NaOH + HCl → NaCl + H2O

The next experiment was the determination of the concentration of

a strong acid (HCl) using secondary standard solution (NaOH) or also

called alkalimetri. In this experiment, the indicator PP discolored become

pink color at volume NaOH that titrated about 12.4 ml. that was calculated

from the average two times calculation NaOH previously. Calculation of

concentration of HCl on the analysis of dta obtained concentration of HCl

about 0.126 M. and the normality was about 0.126 N. it was not in

accordance with the actual HCl 0.1 M.

d. Titration Curve

As the discussion on the previous, titration was a way to find out

the concentration of a solution to path of react it with a solution of another,

which was usually in the form of an acid or bases. Titration commonly

performed by adding titrant already known the concentration through

burette on titrant with a given volume sought the concentration. In reaction

between an acid and base titration very useful for measuring variation pH

at various point through chemical reactions. The results was a titration. A

curve titration was a graph as a function of pH with amount of titran that

added.

Equivalent point of titration was the point where titran was added

right react with the whole substance titrated without any titran left. In

other words, at the equivalent point the number of moles titran equivalent

to the number of moles titrant according to stoichiometryc

1) Titration of HCl and NaOH

On this titration, volume HCl that be used was 10 ml with

concentration 0.102 M. Volume NaOH that added was 1 ml, 2 ml, 3

ml,……..,14 ml. until reach equivalent point where moles of acid

equivalent to moles of base. We measured the pH of solution any

addition of NaOH.

From the graph we knew that the equivalent point occurred when

volume NaOH… ml. initially the pH rose little by little. Since they

occur equivalent point of that causes of salt solutions NaCl were

neutral. The addition of 14 ml NaOH changed drastic become 10 (pH).

Salt NaCl formed from strong acid and strong base that was an

electrolyte will not hydrolyzed, because the solution get a neutral

(pH=7). At the theoretical calculation, we obtained volume of NaOH

…. Ml at the equivalent point. The different of result may be because

there were volume of HCl that decreased when we measured the pH.

Then we used indicator universal, not use pHmeter so the occurately

was less. And the top level of indicator universal was just 10 (pH). We

needed more than pH 10.

2) Titration of NaOH and CH3COOH

On this titration, volume CH3COOH that be used was 10 ml with

concentration 0.1 M. volume of NaOH that added was 1 ml, 2 ml, 3 ml,

…., 14 ml. until reached equivalent point where moles of acid equivalent

to moles of base. And measured the pH of solution any addition of NaOH.

From the graph, we knew that initially the pH rose little by little from pH

initial 4 (weak acid). The equivalent point occurred when volume of

NaOH was … ml with pH… the addition of slightly alkaline, then the pH

will be up slightly, so including buffer solution. Equivalent point was

obtained at pH>7. It was caused salt formed undergo partial hydrolysis

that was alkaline. At the theoretical calculation, we obtained volume of

NaOH … ml and pH… at the equivalent point (from theoretical graph).

The difference of the result may be because there were volume of

CH3COOH that decreased when we measured the pH. Then we used

indicator universal, not used pHmeter so the occurately was less. And the

top level of universal indicator was just 10. We needed more than pH 10.

Partial hydrolysis in H2O : CH3COONa → CH3COO- + Na+

CH3COO- + H2O → CH3COOH + OH-

8. Conclussion and Suggestion

Conclusion :

1) To get acid standard solution with certain molarity can be obtained

with dilute crystal oxalic acid.

Standard solution are prepared by the solution weight accurately

the substance that has high purity and water with a certain number

of solvent in measuring flask. The standard solutions prepared in

this way are known as primary standard solution, whereas standard

solution that the molarity set by primary standard solution called

secondary standard solution. Titration has formula

V 1 × N1=V 2 × N2

2) a. oxalic acid : 0.1 M ; 0.2 N

b. NaOH : 0.102 M ; 0.102 N

c. HCl ; 0.126 M ; 0.126 N

3) Titration of HCl-NaOH

Equivalent point = pH 7 at volume … ml.

Titration of CH3COOH

Equivalent point = pH 7 at volume … ml.

Suggestion :

Before doing experiment should be understand about the matteri about the

experiment.

Be careful to use the material of the experiment.

Burret should be used in positions that do not leak more accurate data

obtained

9. Reference

Anonym. 2014. Acid-Base Titration. Online at http ://en.wikipedia.org.

Helmenstine, Anne Marie.2010.Laws Of Thermochemistry.Online at

http://chemistry.about.com

Petrucci.2007.Thermochemistry.Online at http://www.pearsoned.ca

10. Question

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http://chemistry.about.com/


Mass of oxalic acid : 1.2607 grams

Volume of solution : 100 ml

Normality of oxalic acid : 0.2 N


Volume of oxalic acid : 10 ml





solution of NaOH

Volume of HCl : 10 ml




d. Titration Curve

Relationship of the volumeof titrant with pH in acid base titration

Table 1. pH of NaOH added to 10 ml of 0.1 M HCl

No.


HCl pH

1 0 1

2 1 1

3 2 2

4 3 2

5 4 2

6 5 10

7 6 10

8 7 10

9 8 10

10 9 7

11 10 10

12 11 10

13 12 10

14 13 10

15 14 10

0 2 4 6 8 10 12 14 160

2

4

6

8

10

12

Graph 1. Relationship betweenpH and volume NaOH that added to HCl solution

Vol NaOH

pH

e.

Table 3. titration of NaOH-HCl

No.


HCl pH

1 0 0.9

2 1 0.98

3 2 1.06

4 3 1.14

5 4 1.22

6 5 1.3

7 6 1.4

8 7 1.5

9 8 1.62

10 9 1.75

11 10 1.92

12 11 2.18

13 12 2.79

14 13 11.46

15 14 11.84

0 2 4 6 8 10 12 14 160

2

4

6

8

10

12

14

Graph 3. Titration NaOH-HCl

Vol NaOH

pH

f. Table 4. titration of NaOH-CH3COOH

0 2 4 6 8 10 12 14 160

2

4

6

8

10

12

14

Graph 4. Titration NaOH-CH3COOH

Vol NaOH

pH

No.


CH3COOH pH

1 0 0

2 1 3.8

3 2 4.16

4 3 4.33

5 4 4.58

6 5 4.77

7 6 4.95

8 7 5.14

9 8 5.39

10 9 5.79

11 10 11

12 1111.7

8

13 12 12

14 1312.1

5

15 1412.2

5

acid base

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