VIETNAM NATIONAL UNIVERSITY – HOCHIMINH CITY
INTERNATIONAL UNIVERSITY
CHEMISTRY LABORATORY
EXPERIMENT 2
pH AND BUFFERS
Instructor: M.Sc. Le Nguyen Thien Phuc
GROUP 4
March 2023
REPORT
EXPERIMENT 2: pH AND BUFFERS
Group: 04 Class: Thus - PM Date: ___09/03/2023_________
Group members:
Full name Student ID Declaration of Signature
Contribution (%)
1 Nguyễn Trần Linh Giao BTBCIU21101 16,67
2 Lương Trần Bảo Châu BTFTIU21149 16,67
3 Nguyễn Hoàng Giang ITITIU21192 16,67
4 Nguyễn Tiến Luân ITITIU21014 16,67
5 Nguyễn Quỳnh Nga IELSIU19206 16,67
6 Nguyễn Thị Minh Châu ITITIU21164 16,67
Total score: _______/100
Part 1. Introduction
- Acid:
+ A chemical that separates in water to generate hydronium (H3O+) ions.
+ Acid serves as a proton donor in the Lewis-Bronsted theory.
+ The H+ from an acid combines with water in an aqueous solution to form H3O + (a
hydronium ion).
+ Weak acid only partially dissociates in an aqueous solution and produces very little
or no H3O+, whereas strong acid totally dissociates in water to produce the
hydronium ion.
HA (aq) + H2O ⇌ H3O+ (aq) + A-(aq) Ka
- Base:
+ A chemical that separates in water to generate hydroxide (OH-) ions.
+ A base is a proton acceptor according to the Lewis-Bronsted theory.
+ A base absorbs a proton from the water in an aqueous solution to form OH- (a
hydroxide ion).
+ Weak base only partially dissociates in an aqueous solution and produces little to
very little OH-, whereas strong bases totally dissolves in water to produce
hydroxide ions.
A-(aq) + H2O ⇌ HA (aq) + OH-(aq) Kb
- Formulas:
+ Ka x Kb = Kwater = 1.0 x 10-14 at 25oC
+ Ka = ¿ ¿
+ Kb = [ HA ] ¿ ¿
+ pKa = -log(Ka)
+ pKb = -log(Kb)
- A weak acid and its conjugate weak base, or a weak base and its conjugate weak acid,
are solutions that make up a buffer. Buffer can withstand significant pH fluctuations
caused by the addition of H + or OH-. The Henderson-Hasselbach equation can be used
to calculate pH changes in buffer solutions:
pH = pKa + log ¿
- The moles of solute after being diluted in solution are equal to the moles of solute in
the initial solution:
ni = nf
- Based on the concentration formula, the moles of solute are equal to the solution
volume multiply the concentration of the solution.
Mi x Vi = Mf x Vf
Part 2. Materials and Methods
A. Materials
- 1 Volumetric Flask (The value of Volumetric Flask must be equal to Vf).
- 1 Pipette (The volumetric pipette is highly recommended due to its high accuracy
measurement).
- A container containing the amount of known concentration.
- Solvent (must be the same as the solvent of the initial solution).
B. Methods
1. pH of DEIONIZED WATER
Step 1: Pour about 50 mL of the room temperature deionized water into a 150 mL beaker.
Step 2: Continuously stir the water for 20 seconds. Then, record the pH.
Step 3: Repeat the stirring and measurement process at 20 second intervals, recording each
time and pH value, until there is no appreciable change in the pH.
Step 4: Repeat the experiments at least three times.
2. pH of STRONG ACID
Step 1: Pipet 10 mL of 0.1M HCl into a 250 mL beaker and measure the pH.
Step 2: Pour 90 mL of distilled water, measure the change in pH.
Step 3: Add 10 mL of 0.10 NaOH. Then, record the pH.
Step 4: Add 90 mL of 0.01 NaOH and record the pH.
3. pH of WEAK ACID
Step 1: Record pH, Ka of 20mL of 0.1M CH3COOH.
Step 2: Record pH, Ka of 20mL of 0.01M CH3COOH.
Step 3: Record pH, Ka of 20mL of 0.001M CH3COOH.
4. pH of SALT
Step 1: Record pH, Ka of 20mL of 0.1M NaCl.
Step 2: Record pH, Ka of 20mL of 0.01M NaCl.
Step 3: Record pH, Ka of 20mL of 0.001M NaCl.
5. pH of BUFFERS
● Buffer A:
Step 1: Add 10 mL 0.1M CH3COOH and 40 mL 0.1M CH3COONa into 50 mL Buffer A.
Then record pH value in Buffer A 2 times.
Step 2: Divide equally 50mL Buffer A into 25mL Buffer A1 and 25mL Buffer A2. Then
record pH value of Buffer A1, A2.
Step 3: Use Pasteur pipette:
Add 10 drops 0.1M HCl into Buffer A1 and record pH.
Add 10 drops 0.1M NaOH into Buffer A2 and record pH.
Step 4: Add more drops 0.1M HCl into Buffer A1 until the pH CHANGES by one unit
from the start, record VHCl (in drops).
Add more drops 0.1M NaOH into Buffer A2 until the pH CHANGES by one unit from the
start, record VNaOH (in drops).
● Buffer B:
Step 1: Add 40 mL 0.1M CH3COOH and 10 mL 0.1M CH3COONa into 50 mL Buffer B.
Then record pH value in Buffer B 2 times.
Step 2: Divide equally 50mL Buffer B into 25mL Buffer B1 and 25mL Buffer B2. Then
record pH value of Buffer B1, B2.
Step 3: Use Pasteur pipette:
Add 10 drops 0.1M HCl into Buffer B1 and record pH.
Add 10 drops 0.1M NaOH into Buffer B2 and record pH.
Step 4: Add more drops 0.1M HCl into Buffer B1 until the pH changes by one unit from
the start, record VHCl (in drops).
Add more drops 0.1M NaOH into Buffer B2 until the pH changes by one unit from the
start, record VNaOH (in drops).
Part 3. Results and Discussion
1. pH OF DEIONIZED WATER
Time Observed pH Discussion
(second)
1st 2nd
(Group 4) (Group 3)
0 8.2 6.30 The pH value of deionized water
decreases when we stir the water
20 6.55 5.32 sample.
40 6.31 5.53 Stirring action helps bring water
more into contact with the air.
60 6.16 5.62
80 6.12 5.71 When we stir deionized water, it
can release ions; water sample
100 6.12 5.78 contacts the air, CO2 gas starts to
dissolve it and form carbonic acid.
120 6.06 5.73 As a result, the pH value of
deionized water reduces when we
140 6.06 5.81
keep stirring the sample.
160 6.03 5.85
180 6.03 -
200 6.03 -
2. pH OF STRONG ACID
Solution Theoretical pH Measured pH Discussion
1st 2nd
(Group 4) (Group 33)
10 mL of 0.1M 1 2.40 2.0 - HCl is a strong acid,
HCl so it dissociates
completely in water.
- pH value is less
than 7
Add 90 mL of 2 2.47 2.64 After adding more
distilled water distilled water, the
concentration of acid
decreases. This leads
to the concentration
of H+ also decreases.
So pH increases but
less than 7.
Add 10 mL of 7 11.50 10.85 pH value increases
0.1M NaOH and closes to 7.
When adding NaOH,
OH- dissolves and
joins with H+ in the
mixture to make H2O.
Add 90 mL of 11.65 11.54 11.97 pH value increases
0.01M NaOH and is larger than 7.
Adding NaOH, the
solution is more
basic, pH > 7.
Calculation:
- 10 mL of 0.1M HCl:
n H +¿ ¿ = 0.1 x 0.01 = 1x10−3 mol
n 1 x 10
−3
[H+] = H +¿ ¿ = = 0.1M => pH = -log[H+] = -log(0.1) = 1
V 0.01
- Add 90 mL of distilled water: (distilled water dilutes the initial solution
concentration of HCl)
M HCl before x V HCl before = M HCl after x V HCl after
⇔ 0.1 x 10 = M HCl after x 100
⇔ M HCl after = 0.01
⇒ 100 mL of 0.01M HCl
n H +¿ ¿ = 0.01 x 0.1 = 1x10−3 mol
n 1 x 10
−3
[H+] = H +¿ ¿ = = 0.01M ⇒ pH = -log[H+] = -log(0.01) = 2
V 0.1
- Add 10 mL of 0.1M NaOH:
100 mL of 0.01M HCl ⇒ n H +¿ ¿ = 0.01 x 0.1 = 1x10−3 mol
10 mL of 0.1M NaOH ⇒ nOH −¿ ¿ = 0.1 x 0.001 = 1x10−3 mol
Since n H +¿ ¿ = nOH −¿ ¿, pH = 7
- Add 90 mL of 0.01M NaOH
When add 10mL of 0.1M NaOH, pH of solution is 7 ⇒ neutral solution
So, when add 90 mL of 0.01M NaOH ⇒ nOH −¿ ¿= 0.01 x 0.09 = 0.9x10−3 mol
−3
⇒ [OH-] = 0.9 x 10 = 4.5x10−3 M ⇒ pOH = -log[OH-] = -log(4.5x10−3 ) = 2.35
0.1+0.01+0.09
⇒ pH = 14 - 2.35 = 11.65
3. pH OF WEAK ACID
Solution Measured pH Ka Discussion
1st 2nd
(Group 4) (Group 3)
0.1M acetic acid 3.38 3.24 2.537 × 10-6 CH3COOH is a weak acid
since it only partially dissolves
and produces a little number
of H+ ions. Compared to other
strong acids, at higher pH
value. It may be observed that
0.01M acetic 3.82 4 1.668 ×10-6 the pH values rise as the acid
acid solution is diluted. So, the pH
of solution A < solution B <
solution C because the
concentration of acid in
solution A > solution B >
solution translate C
0.001M acetic 4.47 4.14 3.423 ×10-6
acid
Calculation:
Ka of 0.1M acetic acid:
CH3COOH ⇄ CH3COO- + H+
Initial: 0.1M 0M 0M
Change: -x (M) +x (M) +x (M)
Equilibrium: 0.1 - x (M) +x (M) +x (M)
We have:
So, the average
Applying the same method, we have:
Ka of 0.01M acetic acid:
So, the average
Ka of 0.001M acetic acid:
So, the average
4. pH OF SALTS
Solution Predicted pH Measured pH Discussion
1st 2nd
(Group 4) (Group 3)
0.1M NaCl 7 7.17 6.4 -Since NaCl is a salt
synthesized from a
strong base and a strong
acid, the solution is
neutral and has a pH of
approximately 7.
0.1M >7 7.7 6.32 -CH3COONa is a salt
CH3COONa created from a weak acid
and a strong base, and
the solution's pH is
greater than 7.
0.1M NH4Cl <7 5.48 5.45 -NH4Cl is a salt formed
by combining a strong
acid and a weak base; its
solution is acidic, and its
pH value is less than 7.
Calculation:
- 0.1M NaCl
NaCl is a salt synthesized from a strong base and a strong acid. Therefore, NaCl is neutral
and has a pH of 7.
- 0.1M CH3COONa
CH3COONa → Na+ + CH3COO-
0,1M 0,1M
CH3COO + H2O ⇌ CH3COOH + OH-
-
Initial: 0.1M
After: 0.1-x x x
pH = 7.7 → [H+] = 1.99 x10-8 → [OH-] = 5.012 x10-7
−7 2
x
2
(5.012 ×10 )
Kb =¿ ¿= = = 2.512 x 10-12
0 ,1−x 0.1−(5.012× 10−7 )
- 0.1M NH4Cl
NH4Cl ⇌ NH4+ + Cl-
0.1M 0,1M
NH4+ ⇌ NH3 + H+
Initial: 0,1
After: 0.1-x x x
pH = 5.48 → [H+] = 3.311 x 10-6
−6 2
x
2
(3.311× 10 )
Ka=¿ ¿= = −6
=1.0963 ×10−10
0 ,1−x 0 ,1−(3.311× 10 )
Comment:
- In this experiment, the pH values of three solutions were significantly different from what
was predicted. It could be because the amount of strong base or strong acid added was
insufficient to achieve the desired pH and the quality of the chemicals was not as expected.
There could also be a technical issue with the pH meter or contamination after usage.
- Comparing the results of two experimental groups: The observed experimental
measurement results for pH in two groups are not significantly different.
5. pH OF BUFFERS
Buffer Volume Volume [Acid] [Base] Calculated Measured pH
(mL) 0.1M (mL) 0.1M pH
CH3COOH CH3COONa 1st 2nd
(Group 4) (Group 3)
A 10.0 40.0 0.02M 0.08M 5.36 5.87 5.3
B 40.0 10.0 0.08 M 0.02M 4.14 4.89 4.53
Calculation pH:
Mixture A :
nCH3COOH = 0.1M x 0.01L = 0.001 (mol)
nCH3COONa = 0.1M x 0.04L = 0.004 (mol)
CH3COOH →CH3COO- + H+
0.001 x x (mol)
CH3COONa →CH3COO- + Na+
0.004 0.004 0.004 (mol)
n 0.001
CM [Acid] = = = 0.02 M
V 0.01+0.04
n 0.004
CM [Base] = = = 0.08 M
V 0.01+0.04
0.08 x
We have Ka [CH3COOH] = 1.76x10-5 = ¿ ¿=
0.02
→ x = 4.4x10-6
pH = -log[x] = -log[4.4x10-6 ] = 5.36
For mixture B, apply the same method of calculation.
Discussion:
Due to the preparation being precise, there is no significant difference between the
calculated pH and the recorded pH.
- The result when measuring compare with the result calculated are approximately
equal.
- Weak acid CH3COOH dissociates partly in the solution:
CH3COOH ↔ CH3COO- + H+
While CH3COONa dissociates completely in the solution:
CH3COONa ↔ CH3COO- + Na+
- The solution remains an acidic one.
● Part I: Addition of 10 drops 0.1 M HCl
Buffer pH from pH after Total volume HCl Discussion
the start, adding 10 (drops) to change pH
pHo drops HCl by one unit (pHo-1)
A 5.85 5.81 400 drops When HCl is added to
the solution, which
increases H+, it reacts
with CH3COOH-.
More CH3COOH is
produced, and the
+
excess H is removed.
B 4.54 4.49 36 drops The pH value slightly
decreases when added.
Discussion:
○ When adding HCl, which means an increase in the concentration of H +, leads to
the decreasing of the pH value.
○ Compared with the theory volume, it is slightly different from the reality volume.
The laboratory differs in condition.
* Part II: Addition of 10 drops 0.1 M NaOH
Buffer pH from pH after Total volume Discussion
the start, adding 10 NaOH (drops) to
pHo drops change pH by one
NaOH unit (pHo+1)
A 5.84 6.05 33 drops When adding NaOH to the
solution, which means
OH- adding, it reacts with
CH3COOH.
CH3COONA and its ion,
CH3COO-, are formed.
B 82drops
4.53 4.63
The pH value of the
solution slightly increased
after the addition.
Discussion:
- Because the concentration of H+ decreases when adding NaOH to the buffer, results in
the increase of the pH value.
- Compared with the theory volume, it is slightly different from the reality volume. The
laboratory differs in condition.
Part 4. Conclusions
- Due to the pH, a property (acid, base, neutral) of a solution can be forecasted which is
very important and meaningful in practice. Prediction of reaction will be more accurate if
correctly value the property.
- Strong acids and bases can totally dissolve in water, but weak acids and bases can only
do so partially. As a result, a balance between the dissociated and undissociated forms is
established.
- Either a weak acid and its conjugate base or a weak base and its conjugate acid will form
a buffer that maintains the range of pH.
END.
