# George Mason University General Chemistry Lab Report

Laboratory 8 Standardization of NaOH and Titration of aWeak Acid
The most commonly used strong acids and bases are quite difficult to
prepare precisely. Most common commercial acids (HCL, HNO3 and H2SO4)
vary from manufacturer to manufacturer and even from batch to batch. The
common strong bases, NaOH and KOH, both absorb water from the surrounding
air. Therefore when a solution of strong acid or base is required, as a solution of
strong base is required to titrate the weak acid, it must be standardized. The
easiest method for standardizing a strong base is to titrate it with an acid, which
is a solid at room temperature. The most reliable standard acid is potassium
hydrogen phthalate. Below is the structure of this compound.
H
H
C
C
C
C
C
H
O
O- K+
C
O
C
C
H
O
H
The concentration of acid or base is frequently expressed as the pH of the
solution. The pH is defined as -log [H+]. Therefore a solution of 0.100 M HCl has
a pH of 1.0. When an experiment is performed in water it is found that the
concentration of H+ ions multiplied by the concentration of OH- ions is 10-14. That
is: [H+][OH-] =1.0 x 10-14.
If we take the log of both sides and use the definition -log[H+] = pH and log[OH-] = pOH, then we get: pH + pOH = 14. Any system with pH less than 7
has a greater concentration of hydrogen ions and is acidic. A solution with a pH
greater that 7 is basic.
To find out the real concentration of a prepared solution of NaOH you will
react it with a known amount of potassium hydrogen phthalate (KHP). If you start
with a strong acid in solution, at first the concentration of H+ is high and the pH is
low. As you start to add the base eventually the pH reaches 7 and the initial
moles of acid you equals the moles of base added. A graph of a titration
curve is given in figure 1. If more base is added the pH continues to increase
quickly, and it approaches the pH of base solution.
80
Figure 1
pH
Titration Curve
14
12
10
8
6
4
2
0
0
20
40
60
mL 0.10 M NaOH added
To determine when the pH = 7.0, or moles of H+ = moles of OH- we will
use a pH meter. When the pH is 7.0, we will assume that is the endpoint.
Strong acids (HCl HBr, HI, HNO3, and H2SO4 for example) and strong
bases (NaOH and KOH) dissociate completely in water. Weak acids and bases
may dissolve completely, but do not necessarily dissociate completely. They
exist in solution with undissociated acid in equilibrium with H+ and the conjugate
base. For example acetic acid, CH3COOH does the following in water:
O
O
H
H
H
C
C
OH
+ H2O
H
acid is
H
C
C
O-
+ H3O+
H
The equilibrium constant for a weak acid is designated KA, and for acetic
[CH3COO -] [H3O+]
KA = ——————————[CH3COOH]
= 1.8 x 10-5
81
The equilibrium for a weak acid or base can be solved just like any other
equilibrium problem. For example to calculate the pH of a 0.54 M acetic acid
solution:
species
initial conc.
change
equilibrium
CH3COO –
0.0
+x
x
H3 O +
0.0
+x
x
CH3COOH
0.54
-x
0.54-x
[CH3COO – ] [H3O+]
KA = ——————————[CH3COOH]
= 1.8 x 10-5
x2
= ————-0.54-x
Solving this equation for x gives x = 3.11×10-3, and this is the
concentration of H+. Therefore the pH is -log 3.11×10-3 = 2.5. This is much less
acidic than a solution containing 0.54 M of a strong acid.
To titrate an acid with a base is to dissolve the acid in water, and then add
the base one milliliter or less at a time and measure the pH of the solution after
each addition. A titration curve for a strong acid with a strong base was shown in
Figure 1. The pH changes slowly at first as the base reacts with some of the
acid. It changes very rapidly near the equivalence point. At pH 7, the
equivalence point for a strong acid with a strong base, the titration curve has an
inflection point, and the concentration of base added equals the concentration of
the acid. After the equivalence point, adding base increases the pH of the
solution. The pH asymptotically approaches the pH of the base solution added.
When a weak acid is titrated with a strong base, the equivalence point
occurs at a pH greater than 7. A titration curve for a weak acid is shown in figure
2.
82
Titration of barbituric acid with NaOH
14
12
10
pH
8
pH
6
4
2
0
0
5
10
15
20
25
30
mL 0.100 M NaOH added
Using the example of acetic acid, the strong base reacts with the weak
acid according to the following reaction:
HA + OH-
H2O + AAt the equivalence the OH- has reacted with all of the acetic acid, and we
can calculate the pH of the resulting solution by assuming we added the
conjugate base, sodium acetate to water. Remember the total volume of liquid
has changed. When the conjugate base is added to water, the reaction is:
A-(aq) + H2O(l) à HA + -OH.
The equilibrium constant for this reaction is:
[CH3COOH] [-OH]
KB = —————————-[CH3COO-]
Using Kw= 1.0 x 10-14, and some algebra we get KB = Kw/KA.
If we do not know the value of KA for an acid, how can we know the
equivalence point? The answer is the equivalence point for an acid always has
an inflection point. At the inflection point the moles of acid = the moles of base.
83
In this experiment we will measure the titration curve of a weak acid, and use this
to determine its molar mass.
The other important point on the titration curve is the half-equivalence
point. This is the volume of base added that is one half that needed to reach the
equivalence point. At the half equivalence point the concentration of the acid
equals the concentration of its conjugate base. Looking at the first equation you
can see that at that point the pH = pKA.
Procedure:
Directions for calibrating the pH meter are included in appendix C.
instructor will also demonstrate this.
Your
Standardize the NaOH
1. First make approximately 200 mL of a solution which is 0.100 M NaOH.
This is the solution you will standardize. Weigh out the amount of NaOH that you
need to make this solution, and in your largest beaker dissolve this in 200 mL of
distilled water. Make sure all the NaOH dissolves.
2. Weigh 2.5 x 10-3 moles of KHP (molar mass = 204 g/mol.) and place in
a 250 mL beaker. Add 25.0 mL of distilled water. Record the exact mass of the
KHP.
3. Now standardize your NaOH solution. Fill a buret with your NaOH
solution. Be careful and rinse out the buret with about 5 milliliters of your NaOH
solution first. Then add 2-3 drops of the phenolphthalein indicator to the beaker
containing the KHP. Put the calibrated pH probe into the beaker. The instructor
will demonstrate the calibration of the pH meter. Record your initial Volume on
the buret and add NaOH 1.0 mL at a time, recording with each addition the
volume NaOH solution added and the pH. When you near the equivalence point
add smaller amounts of the NaOH solution so that you may observe the inflection
point. Continue titrating until the pH reading begins to level out. Record the pH at
which the pink color is observed. Record the final buret volume. Calculate the
concentration of NaOH in your solution and repeat the titration until two results
agree within 5% or for 4 runs. For your calculations you may consider using a
table like the one below.
Run #
Vi NaOH
Vf NaOH at the
inflection point
1
2
3
4
84
Mole KHP titrated
Titration Curve of a Weak Acid
Obtain a buret and a sample of unknown acid. To three decimal places
weigh about 0.250 grams of your acid and place it in a 250 mL beaker. Dissolve
the acid in 50 mL of distilled water and add a stir bar to beaker. If the acid does
not dissolve totally, it will during the titration. Place the beaker on a stir plate.
Put your standardized NaOH solution in the buret. Mount the pH probe in the
250 mL beaker. Be sure the stir bar does not strike the pH probe. Record the
pH of the acid solution. Add 1.0 mL of the NaOH solution after 2 or 3 minutes
record the pH. Add another mL of NaOH and again record the pH. Plot the pH
versus volume of NaOH added. As you near the equivalence point the pH will
begin to change more rapidly. Add less NaOH to measure more accurately the
equivalence point. Continue to do this until you have recorded the entire titration
curve. That is go several mL past the equivalence point. Label the equivalence
point and the half equivalence point. Remember the concentration of your NaOH
solution was determined last time.
At the end of the experiment return the electrode to the storage solution,
and turn the power off.
Calculations:
At the equivalence point the moles of acid = moles of NaOH added.
Determine the moles of NaOH added at the equivalence point for your unknown
weak acid. The molar mass of the acid is the mass of acid divided by moles of
the acid. Determine the molar mass of the acid. Determine KA for the acid.
Questions.
1. Based on your chemical experience, indicate the acidic hydrogen in
potassium hydrogen phthalate.
2. For a solution which is 0.0100 M NaOH, what is its pOH and its pH?
3. What is the pH of your NaOH solution?
4. What is the pH of a 0.54 M HCl solution?
5. What is KB for acetic acid?
Safety and Disposal:
All solutions may be flushed down the drain with excess water.
85
Standardization of NaOH and
Titration of a weak acid
Standardization of NaOH
The reaction is:
HA + OH- → H2O + AThe color change is at the stoichiometric
equivalence. So at that event, moles OH- =
moles HA Initial
To find the initial moles of HA take the mass of
each titration and divide by its molar mass. To
find the concentration of the NaOH, take the
moles of HA and divide by the volume needed
to titrate it. That volume is Vf – Vi
Standardization of NaOH
Mass Acid
(g)
Moles
Vi for
Acid=
each run
mass/ 204 (mL)
g/mol
V f for
each run
(mL)
0.53
2.45
4.72
0.49
4.72
6.88
0.51
6.88
9.10
V tot= VfVi (mL)
V total L
= V total
mL x
1000mL/L
Concentra
tion
NaOH=
moles
acid/V
total in L
Titration of a Weak Acid
You will need to plot pH of the solution versus
volume of the NaOH added. It will look
something like this.
At the equivalence point, which is the inflection
point in the graph (15.0 mL of NaOH added) from
the previous example, the moles of acid initial =
moles of bases added.
Moles base = 0.100 Mol/L x 0.0150 L = 1.50 x 10-3
moles base = moles acid initial
MM acid = 0.264 g / 1.50 x 10-3 moles = 176 g/mol
At the ½ eq. pt. (7.5 moles of base added)
pH = pKa
Ka = 10-4.0 = 1.0 x 10-4
To analyze our data first subtract the initial
volume on the buret from all of the readings on
the buret. That will be volume of NaOH added.
Plot volume NaOH added versus pH. Then find
the volume of inflection point. This is where
moles NaOH added = moles unknown acid in the
original sample. The moles of NaOH added will
be volume (in liters) x the average concentration
from slide 3. The mass of our unknown acid was
0.31 grams. The molar mass is grams
acid/moles NaOH added at the inflection point.
Volume on the burret
pH
23.20
2.31
23.81
2.65
24.62
3.53
25.01
4.02
25.33
4.26
25.68
4.61
26.30
5.24
26.77
5.69
27.11
6.05
27.36
6.31
27.49
6.76
27.58
7.06
28.00
11.67
28.51
11.97
28.95
12.08
30.00
12.24
31.20
12.34
Find pKa by going to the volume ½ of the volume
at the inflection point.
Read the pH at that point and that is the pKa of
the acid.
Ka= 10-pKa.
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.
Weigh filled volumetric flask
Mass of empty flask = 12.032 grams
Mass of full flask = 18.685 grams
Part 2 Solid
Part 2 Unknown #12 Shiny orange color
Fill a graduated cylinder with about 25 mL of
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
The Equilibrium Constant of a Reaction
The purpose of this experiment is to determine the equilibrium constant, Kc, of the reaction Fe3+
+ SCN- ⇔ FeSCN2+.
Observations
Procedure
Absorbance:
Make five standard solutions by adding
1. 0
10.0ml of 0.2M Fe(NO3)3 prepared in 0.1M
2. 0.08
HNO3 to all five 50.0mL flasks as well as
3. 0.15
the following volumes of 2.00*10^(-3)M
4. 0.22
KSCN prepared in 0.1M HNO3 in each:
5. 0.34
1.
2.
3.
4.
5.
0.00 mL
1.00 mL
2.00 mL
3.00 mL
4.00 mL
Then measure the absorbance of those
standard solutions at 447nm.
Label five 10 mL volumetric flasks.
Obtain ~30 mL of 2.00*10^(-3)M solution
of Fe(NO3)3 in 1.0 M HNO3, and ~20 mL of
2.00*10^(-3)M KSCN solution.
Using a pipet of graduated cylinder, add 5.0
mL of Fe(NO3)3 to all 5 flasks and the
following volumes of KSCN to the
following:
1. 1.0 mL
2. 2.0 mL
3. 3.0 mL
4. 4.0 mL
5. 5.0 mL
And distilled water (to the line):
1. 4.0 mL
2. 3.0 mL
3. 2.0 mL
4. 1.0 mL
5. 0.0 mL
Measure the absorbance of the five solutions
Data and Calculations
2.00*10-3M Fe(NO3)3 observation: clear
yellow solution
2.00*10-3M KSCN observation: clear
colorless solution
When mixing the two solutions a clear
orange solution is made.
Test tube one has the lightest tone of orange
and increases in orange tone as more KSCN
Absorbance:
1) 0.03
2) 0.07
3) 0.17
4) 0.25
5) 0.33
[Fe3+]init = 2.00*10-3M * VFe init/Vtot
[SCN-] init = 2.00*10-3M * VSCN init/Vtot
[Fe3+]eq = [Fe3+]init – [FeSCN2+]eq
[SCN-] eq = [SCN-]init – [FeSCN2+]eq
[FeSCN2+]eq (from abs)
Keq = [FeSCN2+]eq/[Fe3+]eq*[SCN-]eq
Calculations and table on excel spreadsheet.
Results and Conclusions
In this experiment, we determined the equilibrium constant Kc for the reaction Fe3+ + SCN- ⇔
FeSCN2+. Initial solutions are clear light yellow and colorless and once added together, they
form FeSCN2+ which is an orange-colored compound. As a higher volume of SCN- is present
going up to test tubes 1-5, the solution color darkens. In test tube 1 a clear orange solution is
formed and the color darkens to a darker orange in the following test tubes. This is because more
SCN- is available to react with Fe3+ in order to form FeSCN2+. This tells us that more FeSCN2+ is
formed in each. Kc was achieved first by determining [Fe3+]init which was 0.0010M for all five
runs, [SCN-]init, [FeSCN2+]eq, [Fe3+]eq, [SCN-]eq and Keq. The Kc found was 167. Some
possible errors that could have been made in this experiment were, not perfect cuvette cleaning
for absorbance measures, using the opposite side of the cuvette, as well as inaccurate readings of
volumes. If the cuvette was not perfectly cleaned, some impurities in the cuvette could affect the
concentration as well as absorbance. If an inaccurate volume reading was made, the calculated
concentration would be affected and so affecting the Kc found.
Questions:
1. Is the equilibrium constant the same for all five test tubes?
No.
2. If a reaction is exothermic, how will lowering the temperature affect Kc?
Lowering the temperature of an exothermic reaction will increase Kc. This is because the
position of the equilibrium will shift in a way to increase temperature again/release heat.
3. For the reaction: H2(g) + I2(g) → 2HI(g), Kc = 49.5. If 0.500 moles of H2 are mixed
with 0.500 moles of I2 in a 10.0 L flask, what are the concentrations of all species at
equilibrium?
[H2] = moles / volume = 0.500 moles / 10.0 L = 0.0500 M
[I2] = moles / volume = 0.500 moles / 10.0 L = 0.0500 M
[HI] = 0 M
Start with the concentrations mentioned above, 0.0500 M and no HI but as it proceeds,
the reactants decrease (change recorded as x) as they make more product.
H2: 0.500 M – x
I2: 0.500 M – x
HI: 0 M + 2x
Kc = 49.5
(2x)2 / (0.0500 – x) (0.0500 – x) = 49.5 = Kc
4×2 / 0.00250 – 0.1000x + x2 = 49.5
4×2 = 49.5(0.0025 – 0.1000x + x2)
4x = 49.5×2 – 4.95 + 0.124
0 = 45.5×2 – 4.95x + 0.124
X = (4.95 +/- squareroot((-4.95)2 – 4(45.5)(0.124)))/2(45.5)
X = 0.0697 or 0.0391 M
Can’t be 0.0.697 M because then H2 and I2 will have a negative concentration so x =
0.0391 M.
H2: 0.0500 M – 0.0391 M = 0.0109 M
I2: 0.0500 M – 0.0391 M = 0.109 M
HI: 0 M + 2(0.0391 M) = 0.0782M

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