BIOCHEMISTRY( SFA 2073 )

LAB REPORT

NAME : HUMAIRAH BINTI MOHD FADZIR

MATRIC NO. : 1070292

PROGRAM : BACHELOR OF SCIENCE (HONOUR) FOOD BIOTECHNOLOGY

EXPERIMENT : EXPERIMENT 1

DATE : 7TH JANUARY 2009

LECTURER : DR.NIK NORMA BINTI NIK MAHMOOD

PRACTICAL TITLE :

1.1 DETERMINATION OF ACID CONCENTRATION

OBJECTIVE (s) :

1) To prepare potassium hydrogen phthalate solution.

2) To standardize NaOH solution.

3) To determine the concentration of a weak acid solution by titration with a

standardized solution of sodium hydroxide.

APPARATUS / CHEMICALS :

Burette

Erlenmeyer flask

Volumetric flask ( 250 cm3 )

Pipette ( 25 cm3 and 10 cm3 )

Weighing boat

Potassium hydrogen phthalate

0.1 M hydrogen phthalate

Unknown ( acid )

Indicators ( phenolphthalein )

METHODOLOGY :

A) Standardization of KHC8H4O4 solution

B) Standardization of NaOH solution

C) Determination of the concentration of the unknown acid solution

RESULTS AND OBSERVATIONS :

1) DATA

A) Standardization of KHC8H4O4 solution

Mass of KHC8H4O4 = 3.2 g

B) Standardization of NaOH solution

Exp. Initial volume of

NaOH ( cm3 )

Final volume of

NaOH ( cm3 )

Volume of NaOH

( cm3 )

1 0.00 16.10 16.10

2 16.10 32.10 16.00

3 32.10 48.40 16.30

C) Determination of the concentration of the unknown acid solution

Exp. Initial volume of

NaOH ( cm3 )

Final volume of

NaOH ( cm3 )

Volume of NaOH

( cm3 )

1 0.00 24.20 24.20

2 24.20 48.30 24.10

3 0.00 48.40 24.20

2) CALCULATIONS :

A) Standardization of KHC8H4O4 solution

Moles of = ____________ 3.2 g__________________ _

KHC8H4O4 (39.10)+(1.008)+(12.01)8+(1.008)4+(16.00)4

= 3.2 g

204.2 g

= 0.0157 mol

Molarity = moles of solute_ M = _n_

Liters of solution V

By using the above formula,

Concentration of KHC8H4O4 = __ 0.0157 mol___

250 Ml/1000 mL

= 0.0628 mol/L #

B) Standardization of NaOH solution

1st titration ,

M KHC8H4O4 V KHC8H4O4 = M NaOH V NaOH

(0.0628mol/L)(25mL /1000mL) = M NaOH (16.10mL /1000mL)

1.57 X 10-3 = (0.0161 L) M NaOH

M NaOH = 0.098 mol #

C) Determination of the concentration of the unknown acid solution

Concentration of NaOH = 0.098 mol+0.098 mol+0.096 mol

3

= 0.097 mol

Volume of NaOH = 24.20 mL +24.10 mL +24.20 mL

3

Exp.Volume of

NaOH ( cm3 )

Concentration of NaOH

( mol )

1 16.10 0.098

2 16.00 0.098

3 16.30 0.096

= 24.17 mL

M unknown V unknown = M NaOH V NaOH

M unknown (25mL/1000mL) = (0.097mol)( 24.17mL/1000mL)

M unknown = 0.094 mol #

DISCUSSION :

From the experiment that had been done, solutions of NaOH can be prepared by

either dissolving solid NaOH pellets in water or by diluting a concentrated solution of

NaOH. However, the exact concentration of the solution prepared by these methods

cannot be calculated from the weighed mass or using the dilution equation for two

reasons:

1) Solid sodium hydroxide is hygroscopic ("water-loving"). Pellets of NaOH

exposed to air will increase in mass as they become hydrated so the actual mass of

pure NaOH is not accurately known.

2) Sodium hydroxide in solution reacts with carbonic acid and its concentration

decreases over time. The acid is formed when small amounts of CO2 gas (which

is always present in air) dissolves in solution.

H2CO3(aq) + NaOH(aq) → H2O(l) + Na+(aq) +HCO3−(aq)

The water used to make the NaOH solution can be boiled to expel the dissolved

CO2 gas but this time consuming procedure is often not possible in a short laboratory

period. A stock solution of NaOH can be made in advance with boiled water but will

reabsorb CO2 over a period of time unless stored in airtight containers. Therefore, to

know the exact concentration of a freshly made NaOH solution or one that has stood in

air for some time, the solution needs to be standardized. That is, the exact concentration

must be determined by titrating a known mass of a primary standard acid with the NaOH

solution.

A primary standard is a substance used to determine the concentration of a

solution. A primary standard should be available in pure form at reasonable cost, have a

high equivalent weight to minimize weighing errors, be stable at room temperature, easy

to dry, and hygrophobic (should not easily absorb water when exposed to air). And most

important, the primary standard should react with the solute of the solution being

standardized in a simple straightforward way where a balanced chemical equation can be

written. The primary standard reagent commonly used to standardize NaOH is potassium

hydrogen phthalate ("KHP", KHC8H4O4). A monoprotic acid with a molecular weight

of 204.22 g/mol, 1 mole of KHP reacts with 1 mole of NaOH.

KHC8H4O4(aq) + NaOH(aq) → H2O(l) + Na+(aq) + K+ (aq) + C8H4O42−(aq)

To remove any loosely bound waters of hydration, KHP is normally heated at

110°C for one hour then cooled in a desiccator before use. The exact mass (and number

of moles of acid) is determined by weighing the dried acid on an analytical balance. The

acid is then dissolved in water in an Erlenmeyer flask and an indicator is added. The

NaOH solution is added from a burette until the endpoint (the volume of base when the

indicator changes color) is reached. If the indicator has been chosen correctly, the

endpoint should be very close to the equivalence point.

An indicator (abbreviated HIn) is a weak organic acid and is used in small enough

portions so it does not interfere with the titration. Dissociation of the indicator in solution

is shown below:

Moles of H+ = Moles of -OH (at the equivalence point)

The indicator chosen for this experiment is phenolphthalein. When excess acid

(H+) is present, as the case in an acidic solution, the protonated form of phenolphthalein

(HIn) predominates. When excess base is present, H+ is removed, and the

phenolphthalein anion (In−) predominates. HIn and In− are different colors in solution.

As stated before, the volume when enough base has been added to react exactly with the

acid is the equivalence point. The point at which an indicator changes color is called the

endpoint of the reaction. If the indicator has been chosen correctly, the indicator endpoint

and the titration equivalence point will coincide. The intensity of the color is

concentration dependent, so if a large amount of phenolphthalein is present the color

could be very strong at the end-point. However, the presence of a large amount of

phenolphthalein is not usually the reason a titration mixture has an intense color. It is

more common for a student to miss the end-point. You should stop adding titrant when a

pale color that persists throughout the solution for several seconds is observed. If the

solution is intensely pink, you should either disregard the data or back-titrate the sample.

For determination of concentration of the unknown weak acid, I found that its

concentration is 0.094 mol that approaches 0.1 mol. This was done by calculating its

concentration using calculation in (A) and (B). Since, the titration of the unknown acid

was done with the same NaOH used in (B). Therefore, the concentration of NaOH and

the unknown acid might have the same concentration. It was proved, because the

concentration of NaOH is 0.098 mol approaches 0.1 mol.

For safety precautions when doing this experiment:

1) Safety goggles and aprons must be worn in lab at all times.

2) Sodium hydroxide is a strong base and can cause severe burns;wash all

contaminated areas thoroughly with cold water.

3) Acids and bases are corrosive chemicals and can cause burns to skin and eyes.

Avoid contact and wash any contaminated area thoroughly with cold water.

4) All liquids, when placed in a burette, form a curved meniscus at their upper

surfaces. In the case of water or water solutions, this meniscus is concave, and the

most accurate burette readings are obtained by observing the position of the

lowest point on the meniscus on the graduated scales.

CONCLUSION :

Concentration of the unknown acid solution is 0.094 mol that is determined by titrating a

standardized NaOH solution with potassium hydrogen phthalate solution by using the

acid-base titration method.

QUESTION (s) :

1) Why NaOH solution need to be standardized?

NaOH solution need to be standardized because :

1) Pure sodium hydroxide is very hygroscopic ( water loving ) which means

it absorbs water readily from the atmosphere. Pellets of NaOH exposed to

air will increase in mass as they become hydrated so the actual mass of

pure NaOH is not accurately known.

2) Sodium hydroxide also reacts over time with carbon dioxide from the

atmosphere to form sodium carbonate ( Na2CO3 ).Older samples of sodium

hydroxide tend to be contaminated with significant quantities of sodium

carbonate.

2) Describe the basic steps involved in a acid-base titration. Why is this technique of

great practical value?

1) First, a known amount of acid is transferred to an Erlenmeyer flask and

some distilled water is added to make up a solution.

2) After that, an indicator is dropped in the acid solution.

3) Next, the base solution is carefully added to the acid solution from burette

until they reach an equivalence point.

4) At equivalence point the acid solution has been neutralized by the added

base solution and the solution is still colourless. If we add just one more

drop of base solution from the burette, the solution at last immediately

turn pink because the solution is now basic.

REFERENCE (s) :

1) Nelson, J., Chemistry: The Central Science, 3rd edition, Prentice-Hall,1985

2) Chang, R. (2007). Chemistry (9th Edition). McGraw Hill.:USA

3) http//www.Jchemed.chem..wisc.edu.com.my

4) http://www.ig.pwr.wroc.pl/minproc/journal/pdf/2003/97-105.pdf

5) http://www.desline.com/articoli/5347.pdf

PRACTICAL TITLE :

1.2 pH MEASUREMENT AND ITS APPLICATION

OBJECTIVE (s) :

1) To use various methods of measuring the pH of acids, bases and salts.

2) To determine the dissociation constant of weak acid.

APPARATUS / CHEMICALS :

Test tubes

Conical flask 250 ml

Pipette 25 ml

Burette

pH paper

0.1 M and 1.0 M HCl

0.1 M and 1.0 M NH3

0.1 M CH3COONa

0.2 M NaOH

0.1 M NH4NO3

0.1 M and 1.0 M CH3COOH

0.1 M NaCl

Indicators (phenolphthalein , universal indicator , methyl violet , methyl orange , alizarin

yellow )

METHODOLOGY :

A) Determination of pH of acidic and basic solutions

B) Determination of pH of salt solutions

C) Determination of dissociation constant of weak acid, Ka

RESULTS AND OBSERVATIONS :

1) DATA

A) Determination of pH of acidic and basic solutions

indicator

solution

pH paperUniversal

indicator

Alizarin

yellowpH meter

Methyl

violet

Methyl

orange

0.1 M HCl 1

Colorless

↓

Brick red

1.15

Colorless

↓

Blue

1.0 M HCl 0

Colorless

↓

Purplish red

0.49

Colorless

↓

Green

0.1 M

CH3COOH3

Colorless

↓

Purple

Colorless

↓

Dark orange

1.0 M

CH3COOH

2 Colorless

↓

Purplish

Colorless

↓

Light orange

blue

0.1 M NaOH 13

Colorless

↓

Dark

orange

1.0 M NaOH 14

Colorless

↓

Lighter

dark orange

0.1 M NH3

Colorless

↓

Light

orange

1.0 M NH3

Colorless

↓

Dark

orange

B) Determination of pH of salt solutions

indicator pH paper Universal indicator

solution

0.1 M NaCl 7

Colorless

↓

Yellow

0.1 M CH3COONa 8

Colorless

↓

Green

0.1 M NH4NO3 6

Colorless

↓

Dark yellow

C) Determination of dissociation constant of weak acid, Ka

Volume of NaOH (0.2 M) used in titration the X = 12.20 ml

Observation after adding titrated X with Y = pink → colourless

pH of the mixture :

i) pH meter = 4.27

ii) pH paper = 5

2) CALCULATIONS :

C) Determination of dissociation constant of weak acid, Ka

Reaction between CH3COOH and NaOH is :

CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

1 mol CH3COOH ≈ 1 mol NaOH

The number of moles of NaOH in 12.20 mL is :

12.20mL X 0.200mol NaOH X 1 L = 2.44 X 10-3mol

1 L NaOH 1000 mL

The number of moles of CH3COOH in 25.00 mL is :

50.00mL X 0.100mol CH3COOH X 1 L = 5.0X 10-3mol

1 L CH3COOH 1000 mL

CH3COOH(aq) + NaOH(aq)→CH3COONa(aq)+ H2O

Initial (mol) 5.0X 10-3 2.44 X 10-3 0

Changes (mol) -2.44 X 10-3 -2.44 X 10-3 +2.44 X 10-3

Final (mol) 2.56 X 10-3 0 +2.44 X 10-3

Concentration of CH3COOH = 2.56 X 10 -3 X 1000mL

50.00mL 1L

= 5.12 X 10-2 M

Concentration of CH3COONa = 2.44 X 10 -3 X 1000mL

62.20mL 1L

= 3.92 X 10-2

Ka = [H + ] [A - ]

[HA]

= [H + ] [CH 3COO - ]

[CH3COONa]

[H+] = [CH3COONa] Ka

[CH3COO-]

= (5.12 X 10 -2 )(1.8 X 10 -5 )

(3.92 X 10-2)

= 2.35 X 10-5

Therefore,

pH = -log (2.35 X 10-5)

= 4.63 #

DISCUSSION :

In this experiment, we have used two methods in determining the pH of acidic

and basic solutions which is by using pH meter and indicators (pH meter, universal

indicator, methyl violet, methyl orange and alizarin yellow).

The pH at which the color of an indicator changes is called the transition interval.

Chemists use appropriate indicators to signal the end of an acid-base neutralization

reaction. When the reaction is complete, that is, when there is no excess of acid or base

but only the reaction products, that is called the endpoint of the titration. The indicator

must change color at the pH which corresponds to that endpoint.

The indicator changes color because of its own neutralization in the solution.

Different indicators have different transition intervals, so the choice of indicator depends

on matching the transition interval to the expected pH of the solution just as the reaction

reaches the point of complete neutralization. Phenolphthalein changes from colorless to

pink across a range of pH 8.2 to pH 10. Methyl orange changes from red and yellow to

orange at pH 3.7. Those are the two most common indicators, but others are available for

much higher and lower pH values. Methyl violet, for example, changes from yellow to

blue at a transition interval of pH 0.0 to pH 1.6. Alizarin yellow R changes from yellow

to red at a transition interval of pH 10.0 to pH 12.1. Other indicators are available

through most of the pH range, and can be used in the titration of a wide range of weak

acids and bases.

Paper dipped in a mixture of several indicators and then dried is called pH paper,

useful for obtaining the approximate pH of a solution.

A universal indicator indicates the pH of a substance in a more complex way than

most pH indicators. Instead of changing colors if the pH of a substance it is exposed to is

above or below a certain pH point, a universal indicator changes to a color that is

calibrated to indicate the pH level of the substance.

1.Veryacidic-Red

2.Acidic-Orange/Yellow

3.Neutral-Green

4.Basic/base/alkali-Blue

5. Very basic/base/alkali - Purple

Methyl violet 2B (simply called methyl violet) is used in chemistry as a pH

indicator to test pH ranges from 0 to 1.6. At the acid end of its measuring range, it takes

on a yellow color. At the alkaline end, it becomes bluish-violet. Methyl violet can be

supplied as pre-made pH testing paper, or it can be supplied as pure crystals and

dissolved in the sample being checked.

Methyl orange is one of the indicators commonly used in titrations. In an alkaline

solution, methyl orange is yellow while in an acidic solution, methyl orange is orange in

color.

Alizarine Yellow R (Chrome Orange, Mordant Orange 1) is a yellow colored azo

dye made by the diazo coupling reaction. It usually comes as a sodium salt. In its pure

form it is a rust-colored solid. It is a pH indicator.

Alizarine Yellow R (pH indicator)

below pH 10.1 above pH 12.0

10.1 ↔ 12.0

In (A), when pH paper was dipped in 0.1 M HCl, the pH paper change its color

and gives the pH of 1 when compared to the color chart. In 1.0 M HCl, the pH was 0

when compared to the color chart. It means that, the higher the concentration of solutions,

the solutions becoming more acidic, therefore it gives more low pH in 0.1 M HCl, the

universal indicator’s color change to brick red while in 1.0 M HCl, the color change to

purplish red. Based on the pH color chart of universal indicator, both solutions were

acidic however, 1.0 M HCl was more acidic than 0.1 M HCl since the color was more

darker. This was proved when the pH of both solutions were tested using pH meter and

we got that the pH of 0.1 M HCl was 1.15 while the pH of 1.0 M HCl was 0.49. In 0.1 M

HCl, the color of methyl violet change to blue and in 1.0 M HCl the color change to

green. Since, 0.1 M HCl less acidic than 1.0 M HCl therefore the color more to alkaline

end.

In 0.1 M CH3COOH, the pH was 3 and in 1.0 M CH3COOH, the pH was 2 using

pH papers. It means that 1.0 M CH3COOH was more acidic than 0.1 M CH3COOH.

When methyl violet was used, 0.1 M CH3COOH change its color to purple while in 1.0

M CH3COOH its color changes to purplish blue. It means that 1.0 M CH3COOH was

more acidic than 0.1 M CH3COOH because the purple color of 0.1 M CH3COOH was

darker indicates that it was close to the alkaline end. In methyl orange, 0.1 M CH3COOH

change its color to dark orange while 1.0 M CH3COOH became light orange. Since, 1.0

M CH3COOH was more acidic than 0.1 M CH3COOH so, the color must be darker than

0.1 M CH3COOH. However, this was not happened and it might because of some error

that occured during the experiment like the amount of solutions and the drops of indicator

used.

In 0.1 M NaOH the pH was 13 and in 1.0 M NaOH the pH was 14 by using pH

paper. It means that 1.0 M NaOH more basic than 0.1 M NaOH because the

concentration of 1.0 M NaOH was higher. In alizarin yellow, 0.1 M NaOH changed to

dark orange while 1.0 M NaOH changed to lighter dark orange. Since, the color of

alizarin yellow in 1.0 M NaOH was lighter than in 0.1 M NaOH so it did not proved that

1.0 M NaOH was more basic than 0.1 M NaOH. May be the same error like in

CH3COOH solutions occurred that led to this result.

In 0.1 M NH3, the color of alizarin yellow changed to light orange while in 1.0 M

NH3 changed to dark orange. It means that 1.0 M NH3 was more basic than 0.1 M NH3

because the color of alizarin yellow was darker and closed to the alkaline end.

In (B), 0.1 M NaCl the pH was 7, in 0.1 M CH3COONa the pH was 8 and in 0.1

M NH4NO3 the pH was 6. When universal indicator was used, 0.1 M NaCl changed to

yellow, 0.1 M CH3COONa changed to green and 0.1 M NH4NO3 changed to dark yellow.

Based on the color chart, it proved that 0.1 M NH4NO3 was acidic. However, it did not

proved the pH that we got using the pH paper where 0.1 M NaCl was neutral and 0.1 M

CH3COONa was basic because yellow color of 0.1 M NaCl indicates that it is acidic and

green color of 0.1 M CH3COONa indicates it is neutral. These occurred might because of

error in solutions used.

In (C), a mixture of 0.1 M CH3COOH and 0.2 M NaOH produced CH3COONa

salt and water. When this mixture was tested with pH meter, the reading was 4.27.

However, in calculation relating the number of moles to find the pH gives a result of

4.63. The error occurs maybe because of careless mistake during the titration or maybe

because of purity of the solution was doubted.

For safety precautions when doing this experiment:

1) Safety goggles and aprons must be worn in lab at all times.

2) Sodium hydroxide is a strong base and can cause severe burns; wash all

contaminated areas thoroughly with cold water.

3) Acids and bases are corrosive chemicals and can cause burns to skin and eyes.

Avoid contact and wash any contaminated area thoroughly with cold water.

4) All liquids, when placed in a burette, form a curved meniscus at their upper

surfaces. In the case of water or water solutions, this meniscus is concave, and the

most accurate burette readings are obtained by observing the position of the

lowest point on the meniscus on the graduated scales.

CONCLUSION :

The more concentrated the solution, the more acidic or basic the solution. The pH of the

mixture of 0.1 M CH3COOH and 0.2 M NaOH was 4.63 by calculation.

QUESTION (s) :

1) Calculate the percentage of ionization of acetic acid at various concentrations (0.1

M and 1.0 M). How does the percentage of ionization change with its

concentration ?

a) 0.1 M CH3COOH

CH3COOH → H+ + CH3COO-

Ka = [H + ] [A - ]

[HA]

= [H + ] [CH 3COO - ]

[CH3COOH]

1.8 X 10-5 = x 2

0.1-x

Since, 0.1-x ≈ 0.1 so,

1.8 X 10-5 = x 2

0.1

x = 1.34 X 10-3

Percent ionization = [H + ] X 100%

[HA]0

= 1.34 X 10 -3 X 100%

initial 0.1 0 0

changes -x x x

final 0.1-x x x

0.1

= 1.34%#

b) 1.0 M CH3COOH

CH3COOH → H+ + CH3COO-

Ka = [H + ] [A - ]

[HA]

= [H + ] [CH 3COO - ]

[CH3COOH]

1.8 X 10-5 = x 2

1.0-x

Since, 1.0-x ≈ 1.0 so,

1.8 X 10-5 = x 2

1.0

x = 4.24 X 10-3

Percent ionization = [H + ] X 100%

[HA]0

= 4.24 X 10 -3 X 100%

initial 1.0 0 0

changes -x x x

final 1.0-x x x

1.0

= 0.424%#

2) From the pH values obtained from both concentrations of acetic acid, calculate Ka

for acetic acid. Compare these values with those obtained from Part C.

a) 0.1 M CH3COOH

pH = -log [H+]

3 = -log [H+]

[H+] = 0.001

CH3COOH → H+ + CH3COO-

Ka = [H + ] [A - ]

[HA]

= [H + ] [CH 3COO - ]

[CH3COOH]

= (0.001) (0.001)

0.099

initial 0.1 0 0

changes -0.001 0.001 0.001

final 0.099 0.001 0.001

= 1.01 X 10 -5 #

b) 1.0 M CH3COOH

pH = -log [H+]

2 = -log [H+]

[H+] = 0.01

CH3COOH → H+ + CH3COO-

Ka = [H + ] [A - ]

[HA]

= [H + ] [CH 3COO - ]

[CH3COOH]

= (0.01) (0.01)

0.99

= 1.01 X 10 -4 #

From part C,

initial 1.0 0 0

changes -0.01 0.01 0.01

final 0.99 0.01 0.01

[H+] = 2.35 X 10-5

CH3COOH → H+ + CH3COO-

Ka = [H + ] [A - ]

[HA]

= [H + ] [CH 3COO - ]

[CH3COOH]

= ( 2.35 X 10 -5 ) ( 2.35 X 10 -5 )

0.1

= 5.523 X 10 -9 #

REFERENCES :

1) Chang, R. (2007). Chemistry (9th Edition). McGraw Hill.:USA

2) http://encyclopedia.codeboy.net/wikipedia/m/me/methyl_violet.html

3) http://www.bookrags.com/PH_indicator .

initial 0.1 0 0

changes -2.35 X 10-5 2.35 X 10-5 2.35 X 10-5

final 0.1 2.35 X 10-5 2.35 X 10-5

4) http://www.chemguide.co.uk/physical/acidbaseeqia/indicators.html

5) http://www.teachnet.ie/tburke/2005/ph.html

6) http://www.iscid.org/encyclopedia/Universal_Indicator