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Michael Oliver
Katarina Ong
Connor Wilson
18 & 25 February 2014
Investigation 40 : What is the Acid Dissociation Constant
Experimental Background
A small chemical company has misidentified a batch of an unknown weak acid
...
The company is not only looking for the identity of the
unknown acid, but also for the acid dissociation constant K​a​, and the concentration of the
unknown acid
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Furthermore, the company is interested in how the pH of the
weak acid reacts when titrated with 0
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Materials

Equipment

Unknown Weak Acid Solution (#3)

50 mL Buret

0
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00 and 10
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The NaOH solution was titrated into a known quantity of oxalic acid dihydrate in order to
determine the concentration of the NaOH solution to three significant figures
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2

Method
In order to accomplish the goal laid out in this laboratory investigation, the group had several
intermediate goals, primarily to standardize the given 0
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From these
data, the molarity of weak acid can be determined, as well as the value of Ka
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Using stoichiometry, the mass of
oxalic acid required to neutralize approximately 25 mL of the 0
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Three samples were measured using an analytic balance and each transferred into
a 250mL Erlenmeyer Flask
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Three drops of phenolphthalein were added to provide a visual indication of the neutralization
point
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The average of the
three trials gave a value of 0
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100M NaOH added
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In this case,
the entry was the volume of NaOH delivered, measured in mL, vs the pH probe reading
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with a pH of 4
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00 respectively
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A 50mL buret was set up on a buret stand,
using a burette clamp, and filled with 50
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A 250mL beaker
was filled with 25
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The magnetic stir rod was positioned in the solution such that the pH probe could
remain in the solution while the titration continued
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At the beginning
of the titration, NaOH was delivered in increments of 2mL
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after each
delivery, the reading on the 50mL buret was taken, and the pH of the solution recorded
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At this point in the procedure, all of the data required to address the goals of this laboratory
investigation have been collected
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Using the balanced chemical equation:
2NaOH + H 2 C 2 O 4 • 2H 2 O ⇔ 2H 2 O + Na 2 C 2 O 4
approximately 0
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3 drops
of phenolphthalein was added as an indicator for when the acid was neutralized
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​This table shows the data collected during the standardization of the NaOH solution
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Trial
Number

Initial
Volume (mL)

Final
Volume (mL)

Delivered
Volume (mL)

Mass of
Oxalic Acid
(g)

Molarity of
NaOH

1

0
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45

24
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153


...
00

23
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67


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0997

3

24
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60

24
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153


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​ Trial 2
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155 grams
NaOH used:​ 24
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155 g oxalic acid ×

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02467 L N aOH

1 mol ox
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07 g ox
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acid

=
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0997 M N aOH

The NaOH standardization was carried out three times to ensure an accurate and precise value
for the concentration of NaOH
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100 M
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00 mL of the Unknown
Acid #3
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Figure 3
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The point toward the middle of the initial, flat portion of the graph
where concavity changes is the half-titration point, and the point toward the middle of the
near-vertical section of the graph where concavity changes again is the equivalence point
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58mL, which was calculated by averaging the 2
points on either side of the near-vertical portion of the graph
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00mL), N B is the normality of NaOH (which is equal to the
concentration determined from the standardization), and V B is the volume of the NaOH used at
the equivalence point
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00) = (
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58)
N A = 0
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114 M
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Using initial pH:
The pH when no base was added to the acid was 2
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The concentration of H + ions was found to be 10−2
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9 × 10−3 M
When the acid dissociates, it becomes:
HA ⇔ H + + A−
Using an ICE chart and the information found above, the Ka was found
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HA
I
C
E

H+

⇔

0
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114-x

+

0
+x
x

A−
0
+x
x

x is the concentration of H + ions found from the initial pH
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114−x)
(1
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114−(1
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2 × 10

At the half- equivalence point:
The half- equivalence point was the point at which 14
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The pH
of this point was found using linear interpolation using the 2 points on either side of the
half-equivalence point
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​This table shows the pH and delivered NaOH data for the two points on either side of
the half-equivalence point
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22

4
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29

x

14
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63
6

ex
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29−14
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62
14
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22 = 4
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62

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62

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01

0
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51 = 7
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63
the pH at the half-equivalence point is 4
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At the half-equivalence point, pH = pKa
Therefore,
4
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63 = Ka
Ka = 2
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98, 4
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98 is the mL of NaOH added and 4
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98 mL of NaOH is added
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ex​:
mmol HA = 0
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00
mmol HA = 2
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85

+

OH
0
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77

0

0
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Using the Henderson-Hasselbach equation:
pH = pKa + log( [B]
[A] )

Where [B] is the millimoles of A− after, and [A] is the millimoles of HA after
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21 = pKa + log( [
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77] )
4
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54)
pKa = 5
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75
Ka = 1
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9 × 10−5
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The first part of this process
was to standardize the given 0
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This was accomplished by titrating the NaOH solution into a known
quantity Oxalic acid until the two were neutralized, then calculating the concentration of the
NaOH based on stoichiometry and the quantity of NaOH added
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This was accomplished by slowly titrating the standardized NaOH
solution into 25mL of unknown acid, recording the change in pH with each addition of NaOH
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The K​a​ value was calculated at three points along the titration curve shown in Figure 3, at a
point within the buffer zone, at the half-equivalence point, and at the equivalence point
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9 × 10​-5​
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It
is, however, unlikely that this is the identity of unknown acid #3, as hydrazoic acid at anything
above very dilute concentrations is highly volatile at standard temperature and pressure
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7 ×10​-5​
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A more likely candidate for
the identity of unknown acid #3 is acetic acid, chemical formula HC​2​H​3​O​2​, which has a K​a​ value
of 1
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One of the most prominent issues experienced during this lab was that some of the 50mL burets
provided let air leak into the buret near the nozzle
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Having access to a burette with a more complete seal would likely
completely resolve this issue
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This problem is easily addressed by allowing the system time to
stabilize and reach equilibrium before the pH readings are taken
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Performing several trials
of this standardization ensures the accuracy of the final calculated value

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