Chm 420 lab report 6
organic chemistry (Universiti Teknologi MARA)
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FACULTY OF APPLIED SCIENCE
LABORATORY REPORT
(CHM420)
NAME: IRFAN NURHADI BIN AZAHAR
TITTLE: ACID AND BASES
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METHOD
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DATA
A. pH using indicator
Table 6.1: pH using indicator
Colour Colour
Colour Ph of colour
Indicator change in change in
change change
sample 1 sample 2
Blue
Thymol blue Red-yellow 1.2-2.8 Pink
Bromophenol Yellow Purple
Yellow-blue 3-4.7
blue
Methyl orange Red-yellow 3.2-4.4 Red Yellow
B. Determination of Ka for a weak acid
Determination of Ka for a weak acid: 4.159 x 10 -10
pH of half neutralized solution of unknown weak acid: 5.04
C. Strong acid-base titration
Table 6.2: Strong acid-base titration
HCl
a added 0 10.0 15 20 23 25 26 27 30 31 33 35
(ml)
Measured 12.1 7.2 3.0 2.6 2.3 2.2 2.1 2.1
b 12.0 12.14 11.99 11.63
pH 2 5 0 4 1 4 5 0
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ABSTRACT
The main purpose of this experiment is to study the properties of acidic/basic
substances using indicators and a pH meter.
Part A of these experiment was used various indicators to determine the ph of
two unknown sample. Thymol blue, bromophenol blue, methyl orange. Those
indicators is use to help determine the ph of the solution. Next experiment is
part B where it conduct the stimulation of strong base and weak acid reaction.
pH meter was used to determine pH of mixed solutions. Part C experiment is
the stimulation of reaction when strong acid and strong base were mixed up
together. As we can see the result for experiment A, Thymol blue
(thymolsulphonephthalein) is a brownish-green or reddish-brown crystalline
powder that is used as a pH indicator. It is insoluble in water but soluble in
alcohol and dilutes alkali solutions. It transitions from red to yellow at pH 1.2
to 2.8 and from yellow to pink for sample 1 and yellow to blue for samole 2.
Bromophenol blue is a dye used as a pH indicator, changing from yellow to
blue over the pH range 3.0 to 4.7 it make transition from blue to yellow for
sample 1 and blue to purple for sample 2. Methyl orange is another form of
dye used as an indicator, changing from red to yellow over the pH range 3.2-
4.4 it make transition from yellow to red for sample 1 and remain yellow for
sample 2. These indicators serve as signal, . So both indicate that the pH for
sample 1 very low . From this , we can conclude that the pH range for sample
1 are lower than 4.4 . The pH sample 2 are higher which greater than 4.7 .
From observation, we can see that thymole blue that we used might be
contaminated with other substances. In experiment part B, it display of strong
base and weak acid reaction stimulated to determine the ph of solution by
using NaOH and unknown weak acid and ph meter was used to determin the
ph from the reaction. By doing the calculation we had determined that the
value of [H+] is 9 .12 x 10-6 6 and Ka is 4.159 x 10-10 . The percent ionization is
0.00456 % . In part c experiment, it conducted the reaction between strong
acid and strong base. There was a sudden change in ph solution when acid is
titrated with base. The pH decrease caused the colour of the solution change
with additional phenolphthalein. The change in pH quite slow in early titration
almost all of tge decreases of PH takes place in immediate vicinity of the end
point.The first part of the graph is when we have the excess amount of
NaOH , the second part is when the amount of NaOH and HCL is the same
and the last part is when the amount of acid is excess . The experiment is a
success. For part A, the pH value for sample 1 is 3, pH value for sample 2 is
8. Part B resulting [H+] is 9.12× 10-6, Ka is 4.159× 10-10 and percent ionisation
is 0.00456%.
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QUESTION
1. Estimate the pH of Sample 1 and Sample 2
2. Which indicators bracketed the pH colour change of sample 1?
3. Which indicators bracketed the pH colour change of sample 2?
4. From the observed pH of the unknown weak acid, calculate
a) [H+] in the solution and Ka
b) Percent ionization
5. Construct a titration curve by plotting measured pH versus volume HCl
(mL) added.
a) What is the pH range for the colour change of phenolphthalein as
shown in the plotted graph?
b) What is the pH of the equivalence point in this titration?
c) Explain why phenolphthalein was used in this experiment?
ANSWER
1. Estimation pH for sample 1 is 3.0 and sample 2 is 8.0
2. In my opinion , methyl orange and bromophenole blue. It is because both
of the indicators turned the colour solution immediately. So the solution is
acidic.
3. For sample 2 , It is methyl orange because when it was dropped into
sample 2 , the solution turned yellow which indicate that the solution is in
basic .
4. A)[H+] in the solution and Ka
pH= -(log [H+]) 5.04
= -(log [H+])
[H+]= shift log(-5.04)
= 9.12 x 10-6.
NaOH dissociated in water to form,
NaOH. Na+ + OH-
Based on the equation,
pH = 5.04
[Na+]= 9.12 x 10-6
[A-]= 9.12 x 10-6
[NaOH] = 0.2
Ka = ([H+][A-]/[HA])
Ka = ([Na+][OH-]/[NaOH])
Ka=(9.12 x10-6 )(9.12 x10-6)/0.2
= 4.159 x10-10
B) Percent ionization
Percent ionization= [H+]/[NaOH] x 100%
= (9.12 x 10-6 / 0.2 ) x 100%
= 0.00456 %
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5. (a) Between 2.31 to 11.63 .
(b) 7
(C) It is because when the pH is between 8.30-10 ,phenolphthalein
change the colour solution, in basic solution it is pink colour and colourless
in acidic solution. For the strong base and strong acid titration, this pH
transition will take place in a fraction of actual neutralization due to the
high strength of base.
GRAPH
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