Acid/Base Studies
By: Kelli Parr
April 12, 2010
General Chemistry 32-106
Introduction:
This laboratory experiment focuses on acid/base titrations. A titration involves adding a
solution of know concentration to a solution of unknown concentration until the reaction is
complete.1 In this lab the solution with a known concentration is 0.10 M NaOH. The solutions of
unknown concentration are vinegar, solution A and solution B. The goal of the first part of the
lab was to find the molarity of Acetic Acid in vinegar solution through titration. In the later parts
of the lab the goal was to find the identities of solutions A and B.
The equivalence points of titrations were crucial in solving the goals of the experiment.
The equivalence point is when a stoicheometric equivalent amount of base has been added to an
acid.2 In this experiment, when halfway to the equivalence point the ratio of base to acid is one
to one. Using the Henderson-Hasselbalch equation, pH=pKa+log(base/acid)3, at this halfway
point pH will equal pKa. The pKa ia the negative log of the Ka , acid ionization constant. Each
acid has its own Ka value that can be used to differentiate it from other acids. Determining the
pKa graphically will allow for the identification of the unknown solution.
Some acids have one Ka value and two equivalence points, these are called monoprotic
acids. These only give a single H+ per unit acid. Others have two Ka values and two equivalence
points and give two H+ per unit acid.
Experimental Method:
First, 10 mL of vinegar was poured into a graduated cylinder. From the graduated
cylinder, 5.00 mL was transferred into a 150 mL beaker using a 5.00 mL volumetric pipette.
Using a squirt bottle, approximately 15 mL of deionized water was added to the beaker as well.
A 50.00 mL burette was filled with 0.10 M standardized NaOH and set over a Fisher Scientific
stir and hot plate using burette clamps. The beaker was placed on the stir and hot plate and a
stirring bead was placed inside. The plate was set to stir.
A Vernier pH sensor was inserted into the beaker so that it would not be hit by the
stirring bead and was connected by a clamp. The pH sensor was connected to an iMac computer
using a Vernier LabPro interface. The LoggerPro computer program was used to collect the pH
data given by the sensor. The 0.10M NaOH was titrated until the pH was about 12. 5mL
increments were used at all times during this portion of the experiment except for when the graph
showed a steep incline. At this point 1mL was the increment. Between each addition the keep
button was pressed on LoggerPro and the next addition wasn’t added until the pH reading was
stable.
After the first titration was complete, 15mL of solution A was placed into a graduated
cylinder. From the graduated cylinder, 10.00mL of solution A was transferred to a 150mL
beaker using a 10.00mLvolumetric pipette. Using a squirt bottle about 15mL of deionized water
was added to the beaker. The beaker was placed over the stirring plate and the pH sensor was
inserted. The 50mL burette was filled to the top line with standardized 0.10 M NaOH. The
titration was undergone until the pH reached 12. 1mL increments were used except for the steep
portions of the graph were drops were single drops were added.
The previous paragraph was then repeated replacing solution A with solution B.
Results:
Four graphs were made during this experiment. Graphs 1,3 and 4 show the of vinegar,
solution A, or solution B with NaOH. Graph 2 shows the equivalence point of the titration of
vinegar and NaOH. This was done by converting the y axis from graph 1 to (pH/volume). The
point on the graph represented the equivalence point for the solution. Equivalence point graphs
were analyzed for graphs 3 and 4 but are not physically included in the report.
The overall goal from graphs 1 and 2 were to find the molarity of Acetic Acid in vinegar
solution. Graph 1 shows the pH results along with the volume for the titration of vinegar with
NaOH.
Graph 1:
Graph 2, gives a more defined approximation of the equivalence point which was found
to be 45.8mL.
Graph 2
Knowing the equivalence point and the molarity of the NaOH allowed for the moles of OH- to be
calculated.
Calculation 1:
0.0456 L × (0.10 moles NaOH/ 1L) = 0.0046 moles OHAt the equivalence point the moles of OH- equals the moles of H+ so the moles of H+ is also
0.0046. Dividing the moles of H+ by the number of liters of Acetic Acid equaled the molarity of
Acetic Acid in vinegar solution.
Calculation 2:
0.0046 moles H+ / 0.00500 L Acetic Acid = 0.92 M Acetic Acid in vinegar solution
The purposes of graphs 3 and 4 were to use the equivalence points to determine the pKa
values to then identify solutions A and B. Graph 3 represents the titration of solution A with
NaOH. There were two equivalence points for graph 3 so it is a diprotic acid.
Graph 3:
When the y-axis was changed to show (pH/volume) the equivalence point was found to be
9.94mL. The pH value when the equivalence is divided by 2 equals the pKa value.
Calculation 3:
9.94mL /2 = 4.97mL
The pH value at 4.97 mL was 2.84.
Graph 4 shows the titration of solution B with NaOH. There are two equivalence points
for the graph so it is also a diprotic acid.
Graph 4:
The equivalence point for this solution was found to be 10.14mL. Finding the pH at half
of the equivalence point gives the pKa value for solution B.
Calculation 4:
10.14mL/2 = 5.07mL
The pH and therefore pKa for solution B was 3.32.
Discussion:
According to calculation 2 the concentration of Acetic Acid in vinegar was found to be
0.92M. The pKa value for solution A was determined to be 2.84. This number closely relates to
the pKa value for Malonic acid which is 2.83.4 Graph 3 gave two equivalence points which
means it is a diprotic acid. Malonic acid is also diprotic which supports it as the acid for solution
A.
Graph 4 for solution B was also of a diprotic acid. From this graph the pKa was found to
be 3.32. The conclusion for the identity of solution B is Maleic acid. Maleic acid has a pKa value
of 1.83. This conclusion was decided because first equivalence point in graph 4 is very faint and
hardly distinguishable. An error could have been made while using the pipette to measure and
transfer solution B into the beaker. The beaker could have also not been cleaned off well enough
and had traces of other solutions in it. Another source for error could have been having too large
of increments when adding the NaOH. Finally the mixture could have not been mixed
thoroughly enough so an increase in stirring could have resulted in more accurate results.
References:
1. Chang, R. General Chemistry: The Essential Concepts, 5th Edition.; McGraw-Hill: Boston
2003,
2.New Mexico State University. Chemistry Department. Titrations.
http://www.chemistry.nmsu.edu/studntres/chem116/notes/titrations.html (accessed 4-12-10)
3. University of Calgary. Chemistry/Science: Structure and pKa
http://www.chem.ucalgary.ca/courses/351/Carey/Useful/pka.html (accessed 4-12-10)
4. Penn State University. Penn State Department of Chemistry: pKa Data Table Compiled by R.
Williams. http://research.chem.psu.edu/brpgroup/pKa_compilation.pdf (accessed 4-12-10)