# Chemistry 2 Changes in P H Lab Report

BuffersHow does a buffer keep the pH
Example
A buffer solution of acetic acid and sodium
has 1.0 moles of acetic acid and 1.0 moles of
sodium acetate
O
O
C
C
H3C
acetic acid
OH
H3C
ONa
sodium acetate
Mixed together in solution the acetic acid will
remain protonated and the sodium acetate will
dissociate, so the
O
[acetic acid]=
= 1.0 moles/L = [HA]
C
OH
H3C
O
[acetate]=
= 1.0 moles/L=[A-]
C
H3C
O-
The buffer equation is
[ HA]
pH = pK A − log
[ A−]
in this case
[1.0 M ]
pH = 4.87 − log
= 4.87
[1.0 M ]
Now add 10.0 mL of 1.0 M HCl
If acid is added it reacts with base, in this case
the acetate to form the conjugate acid according
to the following reaction equation:
O
O
+ H+
C
H3C
O-
C
H3C
OH
To find the new concentrations of
protonated and deprotonated acid
Let’s make a table
[Acetic acid]
[Acetate]
1.00 moles /liter x 1.00
liters = 1.00 moles
1.00 moles /liter x 1.00
liters = 1.00 moles
Change of moles from
reaction with HCl
ml of 1.00 M HCl
+1.00 moles/liter x
0.0100L=
+0.010 moles
-1.00 moles/liter x 0.0100L
=
-0.010 moles
=1.01
=0.99 moles
•Initial[acetic
acid]
moles
Final concentration of each
after the acid is added (the 1.00 M
new total volume is 1.01 L)
0.98 M
The new pH is:
[ HA]
pH = pK A − log
[ A−]
in this case
[1.0 M ]
pH = 4.87 − log
[0.98M ]
pH = 4.87 − 0.0088 = 4.86
Very little change
Now add the 10.0 mL of HCl to 1.00
liter of water with no buffer
moles H +
1.00
x0.010 liters = 0.010 moles H +
liter
concentration H + is
0.010 moles H +
= 0.0099 M
1.010 liters
pH = 2.00
Wow 5 pH units a really big change!!
Now add 10.0 mL of 1.0 M NaOH
If base is added it reacts with protonated acid, in
this case the acetic acid to form the conjugate
base, acetate, according to the following
reaction equation:
O
O
+ OH-
C
H3C
OH
+ H2O
C
H3C
O-
To find the new concentrations of
protonated and deprotonated acid
Let’s make a table
[Acetic acid]
[Acetate]
1.00 moles /liter x 1.00
liters = 1.00 moles
1.00 moles /liter x 1.00
liters = 1.00 moles
Change of moles from
reaction with NaOH
ml of 1.00 M NaOH
-1.00 moles/liter x
0.0100L=
-0.010 moles
+1.00 moles/liter x 0.0100L
=
+0.010 moles
=0.99
=1.01 moles
•Initial[acetic
acid]
moles
Final concentration of each
0.98 M
(the new total volume is
1.01 L)
1.00 M
The new pH is:
[ HA]
pH = pK A − log
[ A−]
in this case
[0.98M ]
pH = 4.87 − log
[1.00 M ]
pH = 4.87 + 0.0088 = 4.88
Again very little change
Now add the 10.0 mL of NaOH to 1.00
liter of water with no buffer
moles OH −
1.00
x0.010 liters = 0.010 moles OH liter
concentration OH − is
0.010 moles H +
= 0.0099 M
1.010 liters
pOH = 2.00
pH = 12.00
Wow 5 pH units a really big change!!
Lab 9 Buffers
Procedure:
We will measure the pH of several solutions
using a pH meter. First the pH meter is
standardized using solutions with known pH.
Between measurements, the pH meter is rinsed
with RO water. Your instructor will demonstrate
how to use the pH meter.
The 4 solutions are:
0.500 M sodium acetate and 0.500 M acetic acid [acetate buffer]
0.500 M ammonium chloride and 0.500 M ammonia [ammonium
buffer]
HCl and NaCl (hydrochloric buffer) unknown concentrations
NaOH and NaCl (hydroxide buffer) unknown concentrations
Now see if they change pH when acid or base is added to them.
HCl/NaCl buffer
3.22
pH of 10.0 mL
buffer
+ 2 mL 1.0 M
HCl
1.62
NaOH/NaCl
buffer
Acetate buffer
10.55
1.56
12.86
4.83
4.46
5.12
Ammonium
buffer
9.25
8.88
9.62
Buffer solution
Initial pH
pH of 10.0 mL
buffer + 2 mL
1.0 M NaOH
12.89
Solution
Initial pH
pH of 10.0 mL
pH of 10.0 mL
solution
solution+ 2 mL
+ 2 mL 1.0 M HCl 1.0 M NaOH
0.100 M acetic
acid
2.88
0.77
12.92
0.100 M sodium
acetate
8.87
0.82
13.10
0.100 M
ammonia
11.12
0.82
13.10
0.100 M
ammonium
chloride
5.13
0.75
12.99
0.100 M HCl
.98
0.60
12.92
0.100 M NaOH
13.00
0.78
13.36
Laboratory 9 Buffers
Acids and Bases are important classes of compounds in chemistry. Acids
and Bases may react with each other. The reaction between the common strong
acid, HCl and the common strong base, NaOH, is:
HCl + NaOH
H2O + NaCl
The weak base, ammonia, NH3, reacts with HCl in the following reaction.
NH4+ + Cl-
HCl + NH 3
Strong acids and strong bases are strong electrolytes. They dissociate
100%. Therefore for a strong acid like HCl, the concentration of H+ equals the
concentration of the HCl solution. For a strong base like NaOH, the
concentration of OH- is equal to the concentration of NaOH. The number of
strong acids and strong bases is small Table 1 lists the important strong acids
and bases.
Strong acids
HCl hydrochloric acid
HBr Hydrobromic acid
HI hydroiodic acid
H2SO4 sulfuric acid
HNO3 nitric acid
HClO4 perchloric acid
Strong bases
NaOH sodium hydroxide
LiOH lithium hydroxide
KOH potassium hydroxide
Ba(OH)2 barium hydroxide
Weak acids and weak bases do dissociate or react 100%. The
concentration of H+ for a weak acid, or OH- for a weak base, is not equal to the
initial concentration, but is much lower. Only a few acids or bases are strong.
Most are weak, acetic acid is a weak acid. When it dissolves in water, it only
partially dissociates. It is best expressed by the equilibrium reaction:
O
C
H3C
O
+ H2O
+ H3O+
C
OH
H3C
86
O-
HCl dissolved in water is entirely H+ and Cl- ions, but acetic acid is
primarily undissociated CH3COOH.
The conjugate base of a weak acid is the acid minus an H+. Because it is
not as stable as the conjugate base of a strong acid as an ion in solution, it
actually behaves like a base in water and accepts an H+ from water.
O
O
+ H2O
C
H3C
O-
+ OH-
C
H3C
OH
Similarly the conjugate acids of weak bases, like ammonia, behave like
weak acids when they dissolve in water.
NH3 + H3O+
NH4+ + H2O
An interesting feature of weak acids and bases is the ability they have to
buffer the pH. A mixture of a weak acid and its conjugate base modulates the pH
and prevents it from changing by much. In this experiment we will determine the
pH changes in buffered and nonbuffered solutions.
The Henderson-Hasselbach equation may be used to determine the pH of
a buffer solution:
pH = pK A − log
[ HA]
[ A−]
Procedure:
We will measure the pH of several solutions using a pH meter. First the
pH meter is standardized using solutions with known pH. Between
measurements, the pH meter is rinsed with RO water. Your instructor will
demonstrate how to use the pH meter.
Test the following solutions to see if they resist a change in pH with
addition of an acid or a base? The four solutions are:
87
0.500 M sodium acetate and 0.500 M acetic acid [acetate buffer]
0.500 M ammonium chloride and 0.500 M ammonia [ammonium buffer]
HCl and NaCl (hydrochloric buffer) unknown concentrations
NaOH and NaCl (hydroxide buffer) unknown concentrations
Now see if they change pH when acid or base is added to them.
Buffer solution
Initial pH
pH of 10.0 mL
buffer
+ 2 mL 1.0 M HCl
pH of 10.0 mL
buffer + 2 mL 1.0
M NaOH
HCl/NaCl buffer
NaOH/NaCl buffer
Acetate buffer
Ammonium buffer
Part 2: Determine how one component solutions change with addition of
acid or base.
Solution
Initial pH
pH of 10.0 mL
solution
+ 2 mL 1.0 M HCl
0.100 M acetic
acid
0.100 M sodium
acetate
0.100 M ammonia
0.100 M
ammonium
chloride
0.100 M HCl
0.100 M NaOH
88
pH of 10.0 mL
solution+ 2 mL 1.0
M NaOH
Questions:
1. Using the measured pH of the solution of HCl in part 1, what is the
concentration of HCl in the solution.
2. Using the Henderson-Hasselbach equation and the measured pH in part 1,
determine KA for acetic acid and KB for ammonia.
3. Calculate the expected change in pH for addition of the HCl and NaOH to the
acetate buffer in part 1.
89
Measurement of Physical Properties
In any measurement it is important to know the precision of the
measurement and also its accuracy. All physical measurements should be made
as precisely and accurately as possible. Maximizing the precision of a
measurement is accomplished by using the most precise equipment available,
and using it properly. If possible it is also wise to compare your result with either
a known or theoretical value.
Significant Figures
When calculating a result from more than one measurement it important to
retain the uncertainty information from all the measurements. There is an entire
field of mathematics devoted to this topic. In this course we use the relatively
simple method of significant figures. A summary of the rules with examples:
Addition and subtraction: line up the numbers to be added or subtracted;
the answer is truncated to the decimal place of the least precise number.
Ex. 12.1 + 2.345 = 14.4
15.678 – 2.2 = 13.5 (notice I rounded up)
Multiplication and Division: Significant Figures in the answer are equal to
the number of significant figures in the least precise number.
15.6 x 2.1 = 31
16.789 ⎟ 25.67432 = 0.65392
25.1 x 3.00 = 75.0
Note zeroes before another number as in 0.65392 do not count. In the
middle and the end they count.
Laboratory Notebooks
In this course you will be required to keep a laboratory notebook. A good
laboratory notebook is an accurate record of everything, which occurred in the
lab. In patent disputes a good lab book versus an inaccurate lab book can mean
millions of dollars. In this course it may mean hundreds of points. Before the lab
you will be required to prepare a lab report outline to be completed during the lab
session. Each lab report will contain:
7
Title and Purpose
1. Procedure and Observations
2. Data and Calculations
3. Results and Conclusions
4. Answers to questions in manual
An example of a lab report is shown below:
1. Title: Density of liquid and a solid.
Purpose to measure the density of a liquid and an unknown solid.
2. Procedure:
Observations
Part 1 Liquid
Unknown # 5 smells like gasoline
Weigh an empty 10.0 mL volumetric flask
Fill with unknown liquid.
Mass of empty flask = 12.032 grams
Mass of full flask = 18.685 grams
Part 2 Solid
Part 2 Unknown #12 Shiny orange color
water
Measure precisely volume of water.
Volume of water = 24.83 mL
Volume of water + metal = 28.53 mL
Mass of Dry metal = 46.409 grams
weigh dry solid sample
Data and Calculations:
Part 1 Liquid:
Density = Mass / volume
Mass of liquid = mass of liquid + flask – mass of flask = 18.685 grams – 12.032 grams = 6.652
grams.
Density = 6.652 grams / 10.000 mL = 0.6652 g/mL
Part 2 solid
volume of solid = volume of solid + water – volume water = 28.53 mL – 24.83 mL = 3.70 mL
density = 46.409 g/3.70 mL = 12.5 g/mL
Results and Conclusions:
The density of the liquid was determined to be 0.6652 g/mL by comparison with the
density table in the CRC it appears the sample could be hexane, which has a density of 0.660
g/mL
The density of solid was 12.5 grams / mL. The solid looked like copper, but the density of
copper from the CRC is: 8.94 g/mL, which is significantly less than my unknown sample.
Therefore although the sample looks like copper it must be something else.
8
In this example, the data is recorded in the section with the observations,
and the procedure is recorded in one column and the observations are recorded
in an adjoining column. This allows you to record your observations with the
correct section of the procedure. In some experiments, the type and volume of
data is better recorded in a table. In this case it should follow the procedure
section. You should still leave room in the procedure section for observations.
One of the objectives of this course is for students to learn how to determine
what data they need to collect, and how to organize it. For some experiments
explicit instructions for organizing the data and calculations will be given, but for
other experiments you will need to determine this for yourself before class. In the
case of repetitive calculations tables are necessary. A spreadsheet such as
Excel can be used, and instructions are included for the Reaction Rate
experiment. All your calculations must follow the rules for significant figures and
every value must have a correct unit. A spreadsheet or calculator will not
determine the correct number of significant figures; it is up to you.
When determining the results and conclusions, there are some things to
keep in mind. The results should relate back to the purpose. Address directly if
the purpose was fulfilled. If the result is a number clearly restate what it is and
the unit for the number. If possible compare your result with a literature value. If
you received no result or an unexpected result, give some scientific explanation
of this. Human error is not a good explanation, because the experiment or
section, which was in error, should be repeated. Thoroughness is important but it
is not necessary to write everything you know about density or volume etc.
To be ready to use all the lab time efficiently, before lab class you should
have completed the purpose, procedure and arranged the data table or written
down what you need to measure.
Lab Instructors may have additional report requirements.
9

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