Biochem Lab
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Transcript of Biochem Lab
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