A Report - Laboratory experiment 2 : pH and Buffers | Bài báo cáo học phần Chemistry Laboratory | Trường Đại học Quốc tế, Đại học Quốc gia Thành phố Hồ Chí Minh

According to the Arrhenius theory, an acid is a substance that dissociates in water to form hydronium ion (H3O+), and a base is a substance that dissociates in water to form hydroxide (OH-) ions. For the Lewis-Brønsted theory, an acid is a proton donor, and a base is a proton acceptor. In an aqueous solution, the H+ from an acid is associated with water to form H3O+ (a hydronium ion), while a base accepts a proton from water to form OH- (a hydroxide ion). Strong acid/strong base is completely dissociated in water to produce hydronium ion/hydroxide ion, respectively. Weak acid/base dissociates only partially in an aqueous solution and forms little or very little H3O+ /OH-. Tài liệu giúp bạn tham khảo, ôn tập và đạt kết quả cao. Mời bạn đón xem.

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Profile 0136_Trần Thảo Vy
Internaonal University, Vietnam Naonal University - HCMC 1
GENERAL CHEMISTRY LABORATORY
REPORT
2
EXPERIMENT 2: pH AND BUFFERS
Group: 04 Submission date: 10/14/2023 Score: _________
No.
Full name
Student ID
Contribution (%)
1
Le Thi Thu Ha
IELSIU21194
100%
2
Le Thanh Thai
IELSIU20412
100%
3
Le Huu Phat
IELSIU20386
100%
4
Le Phuong Thao
IELSIU20418
100%
5
Dinh Nam Dan
BEBEIU21199
100%
Internaonal University, Vietnam Naonal University - HCMC 2
GENERAL CHEMISTRY LABORATORY
I. Introduction
Theories
According to the Arrhenius theory, an acid is a substance that dissociates in water to form
hydronium ion (H
3
O
+
), and a base is a substance that dissociates in water to form hydroxide
(OH
-
) ions. For the Lewis-Brłnsted theory, an acid is a proton donor, and a base is a proton
acceptor. In an aqueous solution, the H
+
from an acid is associated with water to form H
3
O+ (a
hydronium ion), while a base accepts a proton from water to form OH
-
(a hydroxide ion).
Strong acid/strong base is completely dissociated in water to produce hydronium ion/hydroxide
ion, respectively. Weak acid/base dissociates only partially in an aqueous solution and forms
little or very little H
3
O
+
/OH
-
.
Acid: HA (aq) + H
2
O ⇌ H
3
O+(aq) + A
-
(aq) Ka Base: A
-
(aq) + H
2
O
⇌ HA (aq) + OH
–
(aq) Kb
We have: K
a
x K
b
= K
water
= 1.0 x 10-14 at 25
o
C
The pH scale is a compact way to specify the acidity of a solution: pH = - log[H
3
O
+
]
Therefore:
• Acidic solution: pH < 7 or [H
3
O
+
] > [OH
-
]
• Basic solution: pH > 7 or [H
3
O
+
] < [OH
-
]
• Neutral solution: pH = 7 or [H
3
O
+
] = [OH
-
]
Strong acids and strong bases are completely dissociated in water to produce hydrogen ions or
hydroxide ions, respectively. Weak acids dissociate only partially and form little or very little
H
+
. A buffer is a solution of a weak acid or weak base and its conjugate weak base or weak
acid, respectively. Buffers have the function that resists a large change in pH on the addition of
H
+
or OH
-
. This is because the weak base, A
-
, will react with added H
+
and the weak acid, HA,
will react with added OH
-
. Changes in pH of buffer solutions can be determined using the
Henderson-Hasselbach equation:
Internaonal University, Vietnam Naonal University - HCMC 3
GENERAL CHEMISTRY LABORATORY
A pH meter can be used to measure the pH of prepared solutions. Different classes of chemicals
behave differently when dissolved in water. By doing this experiment, you will gain a better
understanding of strong acids and strong bases, weak acids and weak bases, salts and buffers.
Dilution is the process of reducing the concentration of a solution by adding solvent into that
solution. In fact, the moles of solute after being diluted in solution are equal to the moles of
solute in the initial solution.
Furthermore, based on the concentration formula, we can know that the moles of solute = the
solution volume x the concentration of the solution. Therefore:
The protocol to make a standard solution-solution dilution:
a)
Calculation: First, determine the volume of initial solvent needed for dilution by
substituting the given values into the formula M
i
x V
i
=M
f
x V
f
. Finally, V
i
can be
obtained.
b)
Equipment:
+ 1 Volumetric Flask (The value of Volumetric Flask must be equal to V
f
.
+ 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).
Internaonal University, Vietnam Naonal University - HCMC 4
GENERAL CHEMISTRY LABORATORY
c) Dilution process
(1) First take the needed volume of initial solution (V
i
) by pipette
(2) Transfer the needed volume into volumetric flask
(3) Add solvent into the volumetric flask until the solution reaches the marked level of
flask (meniscus)
(4) Close the cap and shake the flask gently
Experimental 1. pH OF DEIONIZED WATER
• Step 1: Pour 50mL deionized water.
• Step 2: Stirring rod. Continuously stir in 20 seconds -> Stop stirring and record pH.
• Step 3: Continuously stir in 20 seconds-> Stop stirring and record pH.
• Step 4: Continuously stir in 20 seconds-> Stop stirring and record pH.
• Step 5: Keep doing the same procedure until there is NO SIGNIFICANT CHANGE in pH
value.
2. pH OF STRONG ACID Section 1: Preparation
• Step 1: Take 10mL 0.1M HCl.
• Step 2: Take 20mL 0.1M NaOH.
• Step 3: Prepare 100ml 0.01M NaOH solution (10mL 0.1M NaOH: 90mL H
2
O).
Internaonal University, Vietnam Naonal University - HCMC 5
GENERAL CHEMISTRY LABORATORY
Section 2: pH measurement
• Step 1: Take 10 mL 0.1M HCl –> Record pH.
• Step 2: Add 90 mL distilled water –> Record pH.
• Step 3: Add 10 mL 0.1M NaOH –> Record pH.
• Step 4: Add 90 mL 0.01M NaOH –> Record pH.
3. pH OF WEAK ACID Section 1: Preparation
• Solution A: 0.1M CH
3
COOH.
• Solution B: 0.01M CH
3
COOH (dilute solution A 10 times).
• Solution C: 0.001M CH
3
COOH (dilute solution A 100 times or dilute solution B 10 times).
Section 2: pH measurement
• Step 1: Take 20mL Solution A ->Record pH, K
a.
• Step 2: Take 20mL Solution B ->Record pH, K
a.
• Step 3: Take 20mL Solution C ->Record pH, K
a.
Internaonal University, Vietnam Naonal University - HCMC 6
GENERAL CHEMISTRY LABORATORY
4. pH OF SALTS Section 1: Preparation
Solution A: 0.1M NaCl
Solution B: 0.1M CH
3
COONa
Solution C: 0.1M NH
4
Cl
Section 2: pH measurement
• Step 1: Take 20mL Solution A ->Record pH, K
a.
• Step 2: Take 20mL Solution B ->Record pH, K
a.
• Step 3: Take 20mL Solution C ->Record pH, K
a.
5. pH OF BUFFERS Section 1: Preparation
Section 2: Buffer A
Internaonal University, Vietnam Naonal University - HCMC 7
GENERAL CHEMISTRY LABORATORY
• Step 1: Add 10 mL 0.1M CH3COOH and Add 40 mL 0.1M CH3COONa to have 50 mL Buffer
A-> Record pH (2 times).
• Step 2: Divide equally into 2 parts:
- A1
▪ 25mL Buffer A1-> Record pH.
▪ Add 10 drops 0.1M HCl-> Record pH.
▪ Add more drops 0.1M HCl.
▪ Until the pH CHANGES by one unit from the start, record VHCl (in drops).
- A2
▪ 25mL Buffer A2-> Record pH.
▪ Add 10 drops 0.1M NaOH-> Record pH.
▪ Add more drops 0.1M NaOH.
▪ Until the pH CHANGES by one unit from the start, record VNaOH (in drops).
Section 3: Buffer B
• Step 1: Add 40 mL 0.1M CH
3
COOH and add 10 mL 0.1M CH3COONa to have 50 mL Buffer
B-> Record pH (2 times).
• Step 2: Divide equally into 2 parts:
- B1
▪ 25mL Buffer B1-> Record pH.
▪ Add 10 drops 0.1M HCl-> Record pH.
▪ Add more drops 0.1M HCl.
▪ Until the pH CHANGES by one unit from the start, record VHCl (in drops). -
B2
Internaonal University, Vietnam Naonal University - HCMC 8
GENERAL CHEMISTRY LABORATORY
▪ 25mL Buffer B2-> Record pH.
▪ Add 10 drops 0.1M NaOH-> Record pH.
▪ Add more drops 0.1M NaOH.
▪ Until the pH CHANGES by one unit from the start, record VNaOH (in
drops).
II. Results and discussion 1.
pH OF DEIONIZED WATER
Time
(second)
Observed pH
Explanation
0
According to the theory, the quantity of distilled water is
quite limited, resulting in an equal concentration of H₃O⁺
and H⁺ ions. As a result, the pH level of water is almost 7.
Nevertheless, following our initial experiment, we (group
4) discovered that the pH level of distilled water actually
drops within the range of 5.80 to 4.88.
20
40
60
80
100
120
140
Internaonal University, Vietnam Naonal University - HCMC 9
GENERAL CHEMISTRY LABORATORY
160
“The possible explanation is that the water container or the
stirring rod might contain a small amount of base, which
consequently causes the pH of the water to undergo a
change."
Comments:
- According to the theory, the pH of distilled water should be around 7. However, our
experimental results diverged from this expectation. This discrepancy may be attributed to
potential contamination from the water container or the stirring rod used by our group,
which could have introduced a base.
- Consequently, our pH value differs from that of the others. In reality, the outcome of the
experiment can be influenced by various factors. To obtain precise results, it is imperative
to exercise caution and ensure that the experimental conditions meet the established
criteria.
2. pH OF STRONG ACID
Solution
Theoretical
pH
Measured pH
Comments/ Explanation
1st
(Group___)
2nd
(Group____)
10 mL of 0.1M HCl
1.00
1.42
1.31
The variation between two
values could be due to the
evaporation of the solution,
which results in a reduction in
hydrogen ions. HCl is a potent
acid that fully dissociates in the
solution, releasing a substantial
amount of H⁺ ions.
Consequently, the
concentration of H⁺ ions is
given by: [H⁺] = [HCl] = 0.1M,
and the calculated pH of the
solution is as follows: pH =
log[H⁺] = -log[0.1] = 1.
Internaonal University, Vietnam Naonal University - HCMC 10
GENERAL CHEMISTRY LABORATORY
Add 90 mL of distilled
water
2.00
2.30
2.27
The measured pH value closely
resembles the theoretical pH.
Upon the addition of distilled
water to the solution, the H⁺
concentration decreased by
0.01 mol/L ([H⁺] = (0.1 mol/L)
x (0.01 L) / (0.1 L) = 0.01
mol/L).
Subsequently, the pH value
becomes: pH = -log[0.01] = 2.
Add 10 mL of
0.1M NaOH
2.04
2.93
2.89
The measured pH value is close
to the theoretical pH.
NaOH is a strong base, and
strong bases can completely
dissociate to release OH⁻ ions
into the solution. OH⁻ ions react
with the H⁺ ions already
present in the solution to form
water molecules.
We calculate the concentrations
of H⁺ and OH⁻ as follows:
[NaOH] = [HCl] = (0.1 mol/L) x
(0.01 L) / (0.11 L) = 9.1 x 10⁻³.
Because [H⁺] = [OH⁻], the
solution is neutral, and the pH is
equal to pOH, which is 7.
Internaonal University, Vietnam Naonal University - HCMC 11
GENERAL CHEMISTRY LABORATORY
Add 90 mL of
0.01M NaOH
11.70
11.11
11.09
The inadequate stirring of the
solution resulted in the disparity
between the theoretical and
measured values.
When additional NaOH is
introduced, the concentration of
OH⁻ in the solution increases. We
have: [OH⁻] = [(0.01 mol/L) x
(0.09 L)] / (0.2 L) = 4.5 x 10⁻³.
Therefore, pOH = -log[OH⁻] =
2.35.
The pH value is then calculated
as: pH = 14 - pOH =
14 - 2.35 = 11.65=11.70
Comments: When adding NaOH solution, the H
+
ions in the acid are neutralized by OH
-
ions, and
the excess OH
-
ultimately results in an increase in pH.
3. pH OF WEAK ACID
Solution
Measured pH
K
a
Comments/ Explanation
1st
(Group___
)
2nd
(Group____
)
0.1M acetic acid
3.24
3.20
3.31x10
-6
Acetic acid is a weak acid, meaning
it only partially dissociates in the
solution. This substance has [H⁺] =
[A⁻] and K
a
=(10
-3.24
)
2
/0.1=3.31x10
-
6
0.01M acetic acid
3.03
2.96
8.71x10
-5
When the solution’s concentration
reduced, the dissociation of H⁺
ions decreased as well K
a
=(10
-
3.03
)
2
/0.01=8.71x10
-5
0.001M acetic acid
4.02
3.93
9.12x10
-6
When the solution’s concentration
lowered, there was a reduction in
the dissociation of H⁺ ions K
a
=(10
-
4.02
)
2
/0.001=9.12x10
-6
Comments:
We are provided with an equation to calculate:
K
a
=[H
+
][A
−
]/[HA].
As indicated by the previous findings, it’s evident that the solution’s concentration affects the
solution’s pH. The higher the pH, the smaller the equilibrium constant K
a
and vice versa.
Internaonal University, Vietnam Naonal University - HCMC 12
GENERAL CHEMISTRY LABORATORY
4. pH OF SALTS
Solution
Predicted
pH
Measured pH
Explanation
1st
(Group____)
2nd
(Group____)
0.1M NaCl
7
6.25
6.25
While salt typically has a pH
value around 7, it was measured
at approximately 5.0 in this
experiment. This discrepancy
could be attributed to technical
issues with the pH meter.
0.1M CH
3
COONa
>7
7.15
7.18
Sodium acetate (CH₃COONa) is
a salt that results from the
combination of a weak acid and
a strong base. When dissolved in
a solution, it breaks down into
CH₃COO⁻ and Na⁺ ions. The
measured pH closely
approximates the expected pH.
0.1M NH
4
Cl
<7
4.70
4.69
NH₄Cl is a salt resulting from the
combination of a weak base and
a strong acid. When it’s
dissolved in a solution, it breaks
down into NH₄⁺ and Cl⁻ ions.
While the Cl⁻ ion is electrically
neutral, it has acidic properties
that cause the solution to have a
pH below 7. The measured pH
closely aligns with the
anticipated pH.
Comments:
The measured pH is in close proximity to the expected pH. In the first experiment, this similarity
might be attributed to potential technical issues with the pH meter or the possibility of the stirring
process being affected by some remaining impurities of a base.
5. pH OF BUFFERS
Buffer
Volume
(mL) 0.1M
CH
3
COOH
Volume (mL)
0.1M
CH
3
COONa
Acid
Base
Calculated
pH
Measured pH
1st
(Group__
)
2nd
(Group__)
A
10.0
40.0
0.02
0.08
5.34
5.72
5.72
B
40.0
10.0
0.08
0.02
4.13
4.59
4.59
Comments:
The value pH when calculating and measuring is approximately equal.
Internaonal University, Vietnam Naonal University - HCMC 13
GENERAL CHEMISTRY LABORATORY
Applying the same concept for mixture B
Part I: Addition of 10 drops 0.1 M HCl
Buffer
pH from
the start,
pH
o
pH after
adding 10
drops HCl
Total volume HCl
(drops) to change pH by
one unit (pH
o
-1)
Comments/ Explanation
A
5.23
5.65
4.6 (60 drops)
CH₃COOH is a weak acid
and only partially dissociates
in the solution, whereas
CH₃COONa completely
dissociates. *CH₃COOH ⇌
CH₃COO⁻ + H⁺
*CH₃COONa ⇌ CH₃COO⁻ +
Na⁺ * The solution contains
CH₃COOH, CH₃COO⁻, H⁺,
Na⁺ from CH₃COONa, and
CH₃COOH. It is evident that
the acetate ions (CH₃COO⁻)
outnumber the H⁺ ions, yet
the solution maintains an
acidic pH. *When 0.1M HCl
is added, acetate ions
Internaonal University, Vietnam Naonal University - HCMC 14
GENERAL CHEMISTRY LABORATORY
B
5.30
5.83
6.83 (30 drops)
combine with H⁺ from HCl,
reducing the excess H⁺ and
generating more CH₃COOH.
Surprisingly, the addition of
HCl does not significantly
decrease the pH value of the
solution.
Comments: 0.25 pts Part II: Addition of 10 drops 0.1 M NaOH
Buffer
pH from the
start, pH
o
pH after
adding 10
drops NaOH
Total volume NaOH
(drops) to change pH by
one unit (pH
o
+1)
Comments/ Explanation
A
4.59
4.43
3.4 (80 drops)
When NaOH is introduced into
the solution, acetic acid
(CH₃COOH) reacts with
NaOH to produce acetate
(CH₃COO⁻) and form
B
4.59
4.64
5.6 (125 drops)
CH₃COONa. Surprisingly, the
addition of NaOH does not
result in a substantial increase
in the pH value of the solution.
Comments: When we add the NaOH base solution to the buffer, the pH value increases
because the concentration of H⁺ ions decreases. The observed volume in the experiment differs
from the theoretical volume due to variations in laboratory conditions and potential errors
during the experimental procedure.
Internaonal University, Vietnam Naonal University - HCMC 15
GENERAL CHEMISTRY LABORATORY
III. Data sheet
Internaonal University, Vietnam Naonal University - HCMC 16
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Internaonal University, Vietnam Naonal University - HCMC 17
GENERAL CHEMISTRY LABORATORY
Internaonal University, Vietnam Naonal University - HCMC 18
GENERAL CHEMISTRY LABORATORY
IV. Conclusion
When we introduce the NaOH base into the buffer solution, the pH level goes up. This occurs because
the concentration of H⁺ ions decreases in the solution. The deviation in the observed volume compared
to the theoretical volume may stem from differences in laboratory conditions and potential errors
during the experimental process.
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International University, Vietnam National University - HCMC 1
GENERAL CHEMISTRY LABORATORY REPORT2
EXPERIMENT 2: pH AND BUFFERS
Group: 04 Submission date: 10/14/2023 Score: _________ No. Full name Student ID Contribution (%) 1 Le Thi Thu Ha IELSIU21194 100% 2 Le Thanh Thai IELSIU20412 100% 3 Le Huu Phat IELSIU20386 100% 4 Le Phuong Thao IELSIU20418 100% 5 Dinh Nam Dan BEBEIU21199 100%
International University, Vietnam National University - HCMC 2
GENERAL CHEMISTRY LABORATORY I. Introduction Theories
According to the Arrhenius theory, an acid is a substance that dissociates in water to form
hydronium ion (H3O+), and a base is a substance that dissociates in water to form hydroxide
(OH-) ions. For the Lewis-Brłnsted theory, an acid is a proton donor, and a base is a proton
acceptor. In an aqueous solution, the H+ from an acid is associated with water to form H3O+ (a
hydronium ion), while a base accepts a proton from water to form OH- (a hydroxide ion).
Strong acid/strong base is completely dissociated in water to produce hydronium ion/hydroxide
ion, respectively. Weak acid/base dissociates only partially in an aqueous solution and forms
little or very little H3O+ /OH-.
Acid: HA (aq) + H2O ⇌ H3O+(aq) + A-(aq) Ka Base: A-(aq) + H2O
⇌ HA (aq) + OH– (aq) Kb
We have: Ka x Kb = Kwater = 1.0 x 10-14 at 25oC
The pH scale is a compact way to specify the acidity of a solution: pH = - log[H3O+] Therefore:
• Acidic solution: pH < 7 or [H3O+] > [OH-]
• Basic solution: pH > 7 or [H3O+] < [OH-]
• Neutral solution: pH = 7 or [H3O+] = [OH-]
Strong acids and strong bases are completely dissociated in water to produce hydrogen ions or
hydroxide ions, respectively. Weak acids dissociate only partially and form little or very little
H+ . A buffer is a solution of a weak acid or weak base and its conjugate weak base or weak
acid, respectively. Buffers have the function that resists a large change in pH on the addition of
H+ or OH- . This is because the weak base, A-, will react with added H+ and the weak acid, HA,
will react with added OH- . Changes in pH of buffer solutions can be determined using the
Henderson-Hasselbach equation:
International University, Vietnam National University - HCMC 3
GENERAL CHEMISTRY LABORATORY
A pH meter can be used to measure the pH of prepared solutions. Different classes of chemicals
behave differently when dissolved in water. By doing this experiment, you will gain a better
understanding of strong acids and strong bases, weak acids and weak bases, salts and buffers.
Dilution is the process of reducing the concentration of a solution by adding solvent into that
solution. In fact, the moles of solute after being diluted in solution are equal to the moles of
solute in the initial solution.
Furthermore, based on the concentration formula, we can know that the moles of solute = the
solution volume x the concentration of the solution. Therefore:
 The protocol to make a standard solution-solution dilution:
a) Calculation: First, determine the volume of initial solvent needed for dilution by
substituting the given values into the formula Mi x Vi =Mf x Vf. Finally, Vi can be obtained. b) Equipment:
+ 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).
International University, Vietnam National University - HCMC 4
GENERAL CHEMISTRY LABORATORY c) Dilution process
(1) First take the needed volume of initial solution (Vi) by pipette
(2) Transfer the needed volume into volumetric flask
(3) Add solvent into the volumetric flask until the solution reaches the marked level of flask (meniscus)
(4) Close the cap and shake the flask gently
Experimental 1. pH OF DEIONIZED WATER
• Step 1: Pour 50mL deionized water.
• Step 2: Stirring rod. Continuously stir in 20 seconds -> Stop stirring and record pH.
• Step 3: Continuously stir in 20 seconds-> Stop stirring and record pH.
• Step 4: Continuously stir in 20 seconds-> Stop stirring and record pH.
• Step 5: Keep doing the same procedure until there is NO SIGNIFICANT CHANGE in pH value.
2. pH OF STRONG ACID Section 1: Preparation
• Step 1: Take 10mL 0.1M HCl.
• Step 2: Take 20mL 0.1M NaOH.
• Step 3: Prepare 100ml 0.01M NaOH solution (10mL 0.1M NaOH: 90mL H2O).
International University, Vietnam National University - HCMC 5
GENERAL CHEMISTRY LABORATORY
Section 2: pH measurement
• Step 1: Take 10 mL 0.1M HCl –> Record pH.
• Step 2: Add 90 mL distilled water –> Record pH.
• Step 3: Add 10 mL 0.1M NaOH –> Record pH.
• Step 4: Add 90 mL 0.01M NaOH –> Record pH.
3. pH OF WEAK ACID Section 1: Preparation • Solution A: 0.1M CH3COOH.
• Solution B: 0.01M CH3COOH (dilute solution A 10 times).
• Solution C: 0.001M CH3COOH (dilute solution A 100 times or dilute solution B 10 times). Section 2: pH measurement
• Step 1: Take 20mL Solution A ->Record pH, Ka.
• Step 2: Take 20mL Solution B ->Record pH, Ka.
• Step 3: Take 20mL Solution C ->Record pH, Ka.
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GENERAL CHEMISTRY LABORATORY
4. pH OF SALTS Section 1: Preparation Solution A: 0.1M NaCl Solution B: 0.1M CH3COONa Solution C: 0.1M NH4Cl Section 2: pH measurement
• Step 1: Take 20mL Solution A ->Record pH, Ka.
• Step 2: Take 20mL Solution B ->Record pH, Ka.
• Step 3: Take 20mL Solution C ->Record pH, Ka.
5. pH OF BUFFERS Section 1: Preparation Section 2: Buffer A
International University, Vietnam National University - HCMC 7
GENERAL CHEMISTRY LABORATORY
• Step 1: Add 10 mL 0.1M CH3COOH and Add 40 mL 0.1M CH3COONa to have 50 mL Buffer
A-> Record pH (2 times).
• Step 2: Divide equally into 2 parts: - A1
▪ 25mL Buffer A1-> Record pH.
▪ Add 10 drops 0.1M HCl-> Record pH.
▪ Add more drops 0.1M HCl.
▪ Until the pH CHANGES by one unit from the start, record VHCl (in drops). - A2
▪ 25mL Buffer A2-> Record pH.
▪ Add 10 drops 0.1M NaOH-> Record pH.
▪ Add more drops 0.1M NaOH.
▪ Until the pH CHANGES by one unit from the start, record VNaOH (in drops). Section 3: Buffer B
• Step 1: Add 40 mL 0.1M CH3COOH and add 10 mL 0.1M CH3COONa to have 50 mL Buffer
B-> Record pH (2 times).
• Step 2: Divide equally into 2 parts: - B1 ▪
25mL Buffer B1-> Record pH. ▪
Add 10 drops 0.1M HCl-> Record pH. ▪
Add more drops 0.1M HCl. ▪
Until the pH CHANGES by one unit from the start, record VHCl (in drops). - B2
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GENERAL CHEMISTRY LABORATORY ▪
25mL Buffer B2-> Record pH. ▪
Add 10 drops 0.1M NaOH-> Record pH. ▪
Add more drops 0.1M NaOH. ▪
Until the pH CHANGES by one unit from the start, record VNaOH (in drops).
II. Results and discussion 1. pH OF DEIONIZED WATER Observed pH Time 1st 2nd Explanation (second)
(Group___) (Group___) 0 5.80 5.75
According to the theory, the quantity of distilled water is 20
quite limited, resulting in an equal concentration of H₃O⁺ 5.60 5.52
and H⁺ ions. As a result, the pH level of water is almost 7. 40 5.30 5.28 60 5.16 5.01
Nevertheless, following our initial experiment, we (group 80 4.96 4.97
4) discovered that the pH level of distilled water actually 100 4.93 4.89
drops within the range of 5.80 to 4.88. 120 4.91 4.80 140 4.90 4.79
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GENERAL CHEMISTRY LABORATORY 160 4.88 4.77
“The possible explanation is that the water container or the
stirring rod might contain a small amount of base, which
consequently causes the pH of the water to undergo a change." Comments:
- According to the theory, the pH of distilled water should be around 7. However, our
experimental results diverged from this expectation. This discrepancy may be attributed to
potential contamination from the water container or the stirring rod used by our group,
which could have introduced a base.
- Consequently, our pH value differs from that of the others. In reality, the outcome of the
experiment can be influenced by various factors. To obtain precise results, it is imperative
to exercise caution and ensure that the experimental conditions meet the established criteria. 2. pH OF STRONG ACID Measured pH Theoretical Solution pH Comments/ Explanation 1st 2nd (Group___) (Group____) The variation between two values could be due to the evaporation of the solution,
which results in a reduction in
hydrogen ions. HCl is a potent
acid that fully dissociates in the
solution, releasing a substantial 1.00 amount of H⁺ ions. 1.42 1.31 10 mL of 0.1M HCl Consequently, the concentration of H⁺ ions is
given by: [H⁺] = [HCl] = 0.1M, and the calculated pH of the solution is as follows: pH = log[H⁺] = -log[0.1] = 1.
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GENERAL CHEMISTRY LABORATORY The measured pH value closely resembles the theoretical pH.
Upon the addition of distilled
water to the solution, the H⁺ Add 90 mL of distilled 2.00 concentration decreased by 2.30 2.27
0.01 mol/L ([H⁺] = (0.1 mol/L) water x (0.01 L) / (0.1 L) = 0.01 mol/L). Subsequently, the pH value becomes: pH = -log[0.01] = 2.
The measured pH value is close to the theoretical pH. Add 10 mL of 2.04 2.93 2.89 NaOH is a strong base, and 0.1M NaOH strong bases can completely
dissociate to release OH⁻ ions
into the solution. OH⁻ ions react with the H⁺ ions already
present in the solution to form water molecules.
We calculate the concentrations of H⁺ and OH⁻ as follows:
[NaOH] = [HCl] = (0.1 mol/L) x
(0.01 L) / (0.11 L) = 9.1 x 10⁻³. Because [H⁺] = [OH⁻], the
solution is neutral, and the pH is equal to pOH, which is 7.
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The inadequate stirring of the
solution resulted in the disparity between the theoretical and measured values. When additional NaOH is
introduced, the concentration of
OH⁻ in the solution increases. We Add 90 mL of
have: [OH⁻] = [(0.01 mol/L) x 11.70 11.11 11.09
(0.09 L)] / (0.2 L) = 4.5 x 10⁻³. 0.01M NaOH
Therefore, pOH = -log[OH⁻] = 2.35.
The pH value is then calculated as: pH = 14 - pOH = 14 - 2.35 = 11.65=11.70
Comments: When adding NaOH solution, the H+ ions in the acid are neutralized by OH- ions, and
the excess OH- ultimately results in an increase in pH. 3. pH OF WEAK ACID Measured pH Solution 1st 2nd Ka Comments/ Explanation (Group___ (Group____ ) )
Acetic acid is a weak acid, meaning
it only partially dissociates in the 0.1M acetic acid 3.24 3.20
3.31x10-6 solution. This substance has [H⁺] =
[A⁻] and Ka=(10-3.24)2/0.1=3.31x10- 6
When the solution’s concentration 3.03 2.96
reduced, the dissociation of H⁺ 0.01M acetic acid 8.71x10-5 ions decreased as well Ka=(10- 3.03)2/0.01=8.71x10-5
When the solution’s concentration 9.12x10-6
lowered, there was a reduction in 0.001M acetic acid 4.02 3.93
the dissociation of H⁺ ions Ka=(10- 4.02)2/0.001=9.12x10-6 Comments:
We are provided with an equation to calculate:
Ka=[H+][A−]/[HA].
As indicated by the previous findings, it’s evident that the solution’s concentration affects the
solution’s pH. The higher the pH, the smaller the equilibrium constant Ka and vice versa.
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GENERAL CHEMISTRY LABORATORY 4. pH OF SALTS Measured pH Explanation Predicted Solution 1 2 pH st nd (Group____) (Group____) While salt typically has a pH
value around 7, it was measured 7 6.25 6.25 at approximately 5.0 in this 0.1M NaCl experiment. This discrepancy
could be attributed to technical issues with the pH meter.
Sodium acetate (CH₃COONa) is a salt that results from the
combination of a weak acid and >7 7.15 7.18
a strong base. When dissolved in 0.1M CH 3COONa
a solution, it breaks down into
CH₃COO⁻ and Na⁺ ions. The measured pH closely approximates the expected pH.
NH₄Cl is a salt resulting from the
combination of a weak base and a strong acid. When it’s
dissolved in a solution, it breaks
down into NH₄⁺ and Cl⁻ ions. 0.1M NH4Cl <7 4.70 4.69
While the Cl⁻ ion is electrically
neutral, it has acidic properties
that cause the solution to have a pH below 7. The measured pH closely aligns with the anticipated pH. Comments:
The measured pH is in close proximity to the expected pH. In the first experiment, this similarity
might be attributed to potential technical issues with the pH meter or the possibility of the stirring
process being affected by some remaining impurities of a base. 5. pH OF BUFFERS Measured pH Volume Volume (mL) Calculated
Buffer (mL) 0.1M 0.1M Acid Base 1st pH 2nd CH 3COOH CH3COONa (Group__ ) (Group__) A 10.0 40.0 0.02 0.08 5.34 5.72 5.72 B 40.0 10.0 0.08 0.02 4.13 4.59 4.59 Comments:
The value pH when calculating and measuring is approximately equal.
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Applying the same concept for mixture B
Part I: Addition of 10 drops 0.1 M HCl pH from pH after Total volume HCl Buffer the start, adding 10
(drops) to change pH by Comments/ Explanation pHo drops HCl one unit (pHo-1) CH₃COOH is a weak acid
and only partially dissociates in the solution, whereas CH₃COONa completely dissociates. *CH₃COOH ⇌ CH₃COO⁻ + H⁺ *CH₃COONa ⇌ CH₃COO⁻ + Na⁺ * The solution contains CH₃COOH, CH₃COO⁻, H⁺, Na⁺ from CH₃COONa, and CH₃COOH. It is evident that
the acetate ions (CH₃COO⁻) outnumber the H⁺ ions, yet the solution maintains an acidic pH. *When 0.1M HCl A 5.23 5.65 4.6 (60 drops) is added, acetate ions
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GENERAL CHEMISTRY LABORATORY combine with H⁺ from HCl, reducing the excess H⁺ and generating more CH₃COOH. Surprisingly, the addition of HCl does not significantly decrease the pH value of the solution. B 5.30 5.83 6.83 (30 drops)
Comments: 0.25 pts Part II: Addition of 10 drops 0.1 M NaOH pH from the pH after Total volume NaOH
Buffer start, pHo adding 10
(drops) to change pH by Comments/ Explanation drops NaOH one unit (pHo+1) A 4.59 4.43 3.4 (80 drops) When NaOH is introduced into the solution, acetic acid (CH₃COOH) reacts with NaOH to produce acetate (CH₃COO⁻) and form B 4.59 4.64 5.6 (125 drops) CH₃COONa. Surprisingly, the addition of NaOH does not
result in a substantial increase
in the pH value of the solution.
Comments: When we add the NaOH base solution to the buffer, the pH value increases
because the concentration of H⁺ ions decreases. The observed volume in the experiment differs
from the theoretical volume due to variations in laboratory conditions and potential errors
during the experimental procedure.
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GENERAL CHEMISTRY LABORATORY III. Data sheet
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GENERAL CHEMISTRY LABORATORY
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GENERAL CHEMISTRY LABORATORY IV. Conclusion
When we introduce the NaOH base into the buffer solution, the pH level goes up. This occurs because
the concentration of H⁺ ions decreases in the solution. The deviation in the observed volume compared
to the theoretical volume may stem from differences in laboratory conditions and potential errors
during the experimental process.