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Date: 11/3/23
Lab Day:
Lab Start Time: 11:00
Activity B.3
Relationships in Acid-Base Reactions
Lab Report
Collective Analysis
A
variable
is a property of an object that can take on different values. The relationships between
four variables were explored in experiments A (Part I) and B (Part II). The variables were V
HCl
(volume of hydrochloric acid), N
NaOH
(total number of drops of sodium hydroxide) and V
H
2
O
(volume of water), and
T (temperature change). Variables can be categorized into different
classes; this provides a structure to the data set that can help you to identify patterns (i.e.,
relationships between the values of two or more variables). In the sciences, experiments are
designed to explore relationships between independent and dependent variables.
1
. Identify the independent variable and dependent variable in Experiment A.
Type of Variable
Variable Name Experiment A
Independent Variable
Volume of HCl
Dependent Variable
Total # of drops in NaOH
1a)
Briefly explain the reasons why you identified the variables above as independent or
dependent.
The independent variable, which was around the same molarity and volume around a few of
the tests, was altered in color by sodium hydroxide in varied drops, which is a dependent part
of the chemical reaction.
1b)
Describe the relationship between the independent and dependent variable in Experiment A.
Hydrochloric acid, the independent, is not the amount being studied and is a resolute variable,
compared to sodium hydroxide, which for our intents and purposes, is noted for how many
drops alters the reaction. Both increase as the reaction moves on, showing a linear reaction.
1c).
Consider experiment A in which the variable V
H
2
O
changed values between test tubes
containing the same V
HCl
(e.g., between test tubes A1 and A2). Was the relationship between the
two variables V
HCl
and N
NaOH
affected by the variable V
H
2
O
? Support your reasoning using data
from the group file
as evidence. [Hint: you may want to calculate average values of the
dependent variable for each value of V
H
2
O
]
When the amount of hydrochloric acid was watered down by the variable of water, it had a
unique effect of making the interaction more of a much more watered-down base; being more
green or even dark blue. Take for example example A6 from the class data, a volume of 8.0 H2O
and 2.0 HCl.
In sixteen drops, it went from a pH level of 1 to a harsh and deep 10. Whereas with
A3 in the same class data two rows down, a volume of just 1.2 HCL, it took twelve drops, but the
reaction was not as severe, going from a 1 to simply eight. For this, water is an extremely potent
variable in the reaction.
2.
Identify the independent variable and dependent variable in experiment B. Describe the
relationship between these two variables.
Type of Variable
Variable Name Experiment B
Independent Variable
NaOH
Dependent Variable
Temperature change
2a)
Briefly explain the reasons why you identified the variables above as independent or
dependent.
The independent variable is NaOH because it is held at a constant number. The dependent
variable in this case is the change in temperature because it is dependent on the NaOH
solution.
2b)
Describe the relationship between the independent and dependent variable in Experiment B.
The dependent variable (temperature) increases as the independent variable (volume)
increases, demonstrating a linear relationship.
2c)
Consider Experiment B in which the variable V
H
2
O
changed values between test tubes having
the same V
NaOH
(e.g, between test tubes B3 and B4). Was the relationship between the two
variables N
NaOH
and
T affected by the variable V
H
2
O
? Support your reasoning using data from
the group file
as evidence. [Hint: you may want to calculate average values of the dependent
variable for each value of V
H
2
O
]
Both variables are not affected by the variable of water here this time. When comparing B2 and
B4, there was not a significant temperature change. B2, with a water value of 4.4, had a final
temperature of 24.2. B4, with a water value of 4.0, had a final temperature of 24.8.
Gonzalez and Monteyne, Chemistry by Inquiry Activity B.3
2
3.
Two chemical systems, interaction between hydrochloric acid and sodium hydroxide and the
interaction between hydrochloric acid and potassium hydroxide, were investigated. Use the
collection of data from the group file
to plot volume of base (y-axis) versus volume of
hydrochloric acid (x-axis) data from both systems. Use different markers to identify the different
systems.
Use the trend line function in Excel to plot the best-fit line for the class data. Be sure to check the
box for the trend line to go through the origin, and check the box to display the equation of the
trend line on the graph.
3a)
1
7
1
3
1
9
2
5
3
1
3
7
4
3
4
9
5
5
6
1
6
7
7
3
7
9
8
5
9
1
9
7
1
0
3
1
0
9
1
1
5
1
2
1
1
2
7
1
3
3
1
3
9
1
4
5
0.00
10.00
20.00
30.00
40.00
50.00
60.00
Neutralization of HCl With NaOH and KOH (projected)
3b)
How would the value of the slope of the trend line in the previous question change if the
concentration of the hydrochloric acid was increased? Explain your reasoning.
Because the trend lines for potassium hydroxide and sodium hydroxide titrations are precisely
proportional, they would both rise if the hydrochloric acid concentration were to be raised.
3c)
How would the value of the slope of the trend line change if the concentration of base
(sodium hydroxide or potassium hydroxide) was increased? Explain your reasoning.
If the base's concentration (potassium hydroxide or sodium hydroxide) were to rise, the trend
line's slope would fall. More concentration of NaOH would result in a faster reach of the
equivalency point.
Gonzalez and Monteyne, Chemistry by Inquiry Activity B.3
Volume of Base (aq) (+/- 0.05mL)
3
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3d)
. Use the drawing tools in Microsoft Word to sketch a line in each set of axes below that
illustrates the patterns between moles of acid and moles of base for each chemical system.
What is the numerical value of the slope (i.e., the
proportionality constant
) in the equations for
these graphs? Explain your reasoning.
Gonzalez and Monteyne, Chemistry by Inquiry Activity B.3
Amount of Hydrochloric Acid
(moles)
Amount of Potassium Hydroxide
(moles)
Amount of Hydrochloric Acid
(moles)
Amount of Sodium Hydroxide
(moles)
The graph is linear because the acid and
base have a mole ratio of one to one.
Because it uses less basal energy, the
line's slope is less steep than that of
KOH.
m = 1
The graph is linear because the acid and
base have a mole ratio of one to one.
Because it needs more base than NaOH,
the line's slope is higher. There will be a
straight proportionality between the two
variables when comparing the mole to
mole ratios of KOH and HCl. This is
because of the same rationale as a reaction
between HCl and NaOH. The rationale is
that since the solution will be neutralized,
moles of acid and base will always be
equal.
m = 1
4
4.
Determination of Concentration of Base from Titration Data Part III
Complete the table below using data that you
personally
collected for two volumes of HCl in
Act B3 Part III. Identify the base as sodium hydroxide or potassium hydroxide. Assume that
the concentration of HCl solution is 0.100 M.
4a)
Concentration of Base - NaOH
Base Used:
Trial 1
Trial 2
Trial 3
Volume HCl (+/-
0.01 mL)
10
10
10
Initial Vol Base
(+/-0.01 mL)
5
3
4
Final Vol Base (+/-
0.01 mL)
4.9
6.8
5.4
Volume of Base at
Equivalence (+/-
0.02 mL)
15
13
14
Concentration of
Base (M)
0.124
0.126
0.125
Average
Concentration of
Base for All Trials
0.125
Show your calculations for the values reported in the table for Trial 1.
Balance Equation: NaOH (aq) + HCl (aq)
H2O (l) + NaCl (aq)
Volume of Base at Equivalence (+/-0.02 mL): Final Vol – Initial Vol = Vol at Equivalence
15-4.9=10.1 x (10^-3 L/ 1 mL) = 0.0101
Concentration of Base (M):
Concentration of HCl = 0.100 M
Volume of HCl = 10.00 mL x (10-3 L / 1 mL) = 1.000 x 10-2 L - M = moles/L
0.100 M =
unknown moles HCl / 1.000 x 10-2 L
moles of HCl = 1.000 x 10-3 mols
M = moles/L
M = 1.000 x 10-3 mol NaOH / 0.00101 =
1.24
Concentration of Base - KOH
Base Used:
Trial 1
Trial 2
Trial 3
Volume HCl (+/-
0.01 mL)
10
10
10
Initial Vol Base
(+/-0.01 mL)
4
3
4
Final Vol Base (+/-
0.01 mL)
6
7
6
Volume of Base at
Equivalence (+/-
0.02 mL)
14
13
14
Concentration of
.128
.129
.128
Gonzalez and Monteyne, Chemistry by Inquiry Activity B.3
5
Base (M)
Average
Concentration for
All Trials
.128
Show your
calculations for the
average
concentration of
base for all trials.
.128+.129.128/3 =
.128
.124/.126/.125=
.125
Expand Your Thinking
4.
A student completed two experiments in which several samples from two hydrochloric acid
solutions, Solution #1 and Solution #2, were titrated with the same solution of sodium hydroxide
until each sample of acid was completely neutralized.
4a.
Use the particulate-level representations of the two HCl solutions to describe the data the
student would have recorded in his laboratory notebook during the titration experiments (i.e.,
what did he observe? what measurements did he make? Did the data differ between the two
experiments?).
4b.)
Use MicroSoft Word drawing tools to sketch the pattern the student would observe in his
data in the axes below. Plot the data from both experiments on the same graph; be sure to
properly label your graph. Use different markers to identify the data for each solution. Explain
your reasoning.
Gonzalez and Monteyne, Chemistry by Inquiry Activity B.3
Reasoning:
Reasoning:
Reasoning:
Reasoning:
The student was supposed to record the molarity of each solution as well as the amount
of HCl and NaOH used in the titration experiment. It was the student's responsibility to
notice the hue shifts. The HCl solution should have become a light pink color when a
few drops of NaOH were added to it to neutralize it. Because Solution 1 has less water
molecules and more hydrogen ions than chloride ions, the data from the two trials differ.
Each one is linear. Because Solution 2 has less
acid and requires fewer drops of NaOH to
complete the titration, it is shown further down
the graph.
6
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1
2
Gonzalez and Monteyne, Chemistry by Inquiry Activity B.3
7
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