Archer G11
Group member: Jack, Judy
9 February 2012
Determination of Ka of Weak Acids
Purpose: The purpose of this lab is to identify three unknown acids. This is done by using the process
called half neutralization, which neutralized everything but H+ ions to determine the pKa. The
significance of this lab is that it can be used to identify the acid in rain. From the composition of the acid,
the source of acid-waste production can be traced.
Hypothesis: The hypothesis is that the pKa can be calculated and used to identify the unknown acids.
With the half neutralization technique, the acid and conjugate base will neutralize each other, leaving
the pH only basing on the concentration of H+. Because the concentrations of the acid and conjugate
base cancel out, the Ka depends solely on the concentration of H+ and thus, Ka = [H+] and pKa = pH. A list
of acids and their Ka is given so the pKa could be calculated and compared to identify the unknown acids.
Materials:
Materials
Unknown acid B
Unknown acid C
Unknown acid D
pH 7 buffer
pH 4 buffer
Distilled water
1 M Sodium hydroxide solution (NaOH)
1% Phenolphthalein solution
Wash glass
Pipets
Erlenmeyer flasks
pH meter
250-mL graduated cylinder
50-mL graduated cylinder
Stirring rod
Spatula
0.0000-g precision balance
Small screw driver
50-mL beaker
Size 12 stopper
Magnetic Bar
Magnetic Stirrer
Quantities
About 0.5 g
About 0.5 g
About 0.5 g
Procedures:
1.) Pour about 5 mL of pH 7 buffer into a 50-mL beaker
2.) Pour about 15 mL of distilled water into another 50-mL beaker
About 0.5 g
About 0.5 g
About 0.5 g
About 5 mL
About 5 mL
315 mL
50 mL
12 drops
3 wash glasses
5 pipets
3 flasks
1 pH meter
1 cylinder
4 cylinders
3 stirring rods
3 spatulas
1 balance
1 screw driver
3 beakers
1 stopper
1 bar
1 stirrer
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3.) Dip the pH meter into the buffer
4.) Use a small screw driver to calibrate the pH meter
5.) Wash the tip of the pH meter with the distilled water in the beaker
6.) Repeat step 1 to 4 for pH 4 buffer
7.) Label three wash glass: B, C, and D
8.) Mass out about 0.15-0.20 g of unknown B
9.) Transfer the unknown B from the wash glass to a beaker
10.) Label the beaker Unknown B
11.) Add 50 mL of distilled water to the beaker
12.) Stir the solution until the unknown acid completely dissolves
13.) Transfer 25 mL of the unknown B solution to an Erlenmeyer flask
14.) Label the flask Unknown B
15.) Add 4 drops of phenolphthalein into the flask
16.) Titrate the unknown B with 0.1 M NaOH
17.) Transfer the solution in the flask back to the labeled beaker
18.) Use the pH meter to measure the pH of the solution
19.) Repeat step 8 to 18 for 2 more trials
20.) Repeat step 8 to 19 for the other two unknown acids
Results: When calibrating the pH meter, water, which was supposed to have the pH of 7, turns out to
have the pH of 8 instead. During the titration, once the solution reaches the equivalent point, the
solution would slowly turns lighter as time pass, although the time it takes to turn light is much longer
than before it reaches equilibrium.
Weak Acids
Potassium dihydrogen phosphate
Potassium hydrogen sulfate
Potassium hydrogen phthalate
Potassium hydrogen tartrate
Acetylsalicylic acid
Given Weak Acids
Formula
Ka
KH2PO4
Ka2 of H3PO4 = 6.2 × 10-8
KHSO4
Ka2 of H2SO4 = 1.0 × 10-2
KHC8H4O4
Ka2 of H2C8H4O4 = 3.9 × 10-6
KHC4H4O6
Ka2 of H2C4H4O6 = 4.6 × 10-5
2-CH3CO2C6H4COOH
Ka = 3.2 × 10-4
pKa = -log(Ka)
pKa of potassium dihydrogen phosphate = -log(6.2 × 10-8) = 7.207
pKa of potassium hydrogen sulfate = -log(1.0 × 10-2) = 2.000
pKa of potassium hydrogen phthalate = -log(3.9 × 10-6) = 5.408
pKa of potassium hydrogen tartrate = -log(4.6 × 10-5) = 4.337
pKa of acetylsalicylic acid = -log(3.2 × 10-4) =3.494
pKa
7.21
2.00
5.41
4.34
3.49
Archer G11
X denotes the unknown acids (B, C, and D)
Unknown
Acids
B
C
D
Trial
pH
#1
#2
#1
#2
#1
#2
6.99
6.93
2.16
2.20
5.30
5.21
Literature pKa (given)
Calculated pKa
Percent Error (%)
Identity of Weak Acids
pH
pKa
(average)
Ka
6.96
6.96
1.10 × 10-7
2.18
2.18
6.61 × 10-3
5.26
5.26
5.56 × 10-6
Percent Error Calculation
Acid B
Acid C
7.21
2.00
6.96
2.18
3.47
9.00
(pH average) = [( Trial 1 pH) + (Trial 2 pH)] ÷ 2
Unknown B: (6.99 + 6.93) ÷ 2 = 6.96
Unknown C: (2.16 + 2.20) ÷ 2 = 2.18
Unknown D: (5.30 + 5.21) ÷ 2 = 5.255
(pH average) = pKa
Unknown B: 6.96 = 6.96
Unknown C: 2.18 = 2.18
Unknown D: 5.255 = 5.255
Ka = 10-pKa
Unknown B: 10-6.96 = 1.096 × 10-7
Unknown C: 10-2.18 = 6.607 × 10-3
Unknown D: 10-5.255 = 5.559 × 10-6
Unknown B has similar pKa to potassium dihydrogen phosphate
Unknown C has similar pKa to potassium hydrogen sulfate
Identity of the
unknown
Potassium dihydrogen
phosphate
Potassium hydrogen
sulfate
Potassium hydrogen
phthalate
Acid D
5.41
5.26
2.87
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Unknown D has similar pKa to potassium hydrogen phthalate
(Percent Error) = |[(Calculated pKa) – (Literature pKa)] ÷ (Literature pKa)| × 100%
Acid B: |(6.96 – 7.21) ÷ 7.21| × 100% = 3.467%
Acid C: |(2.18 – 2.00) ÷ 2.00| × 100% = 9.000%
Acid D: |(5.255 – 5.41) ÷ 5.41| × 100% = 2.865%
Analysis: The hypothesis can be verified. The pKa of the acids were not close together enough to need
an extremely accurate pKa to compare. Thus, although the results of pKa calculated may have been
slightly altered, it was close enough to figure out the identity of the acid. It was not necessary to know
the exact mass of the unknown acids when making its solution. This is because with the method of half
neutralization, the acid and conjugate base will cancel out and thus, does not affect the pH of the
solution. Also, the volume of NaOH needed to neutralize the acid would change accordingly as well, so
the mass of the unknown acids is not important to the pH. The exact concentration of NaOH was also
unnecessary because it was just used to do half neutralization. The volume needed to neutralize the acid
changes according to the concentration anyway so the exact concentration did not contribute to the pH
determination. It was necessary, however, to know the exact volume of distilled water used to dissolve
the acid and the exact volume of solution transferred from the beaker to the Erlenmeyer flask because
the half neutralization needed to perfectly neutralize half of the solution. If more or less than half was
neutralized, there would also be other factor that affected the pH such as the acid and conjugate base
left in the solution. If that is the case, pH would not equal to pKa and so the pKa calculated would not be
accurate. Strangely, the Ka calculated and Ka given are somewhat different from each other.
Conclusion: The hypothesis can be verified true. There is about 0-10% differences between literature
and the calculated pKa which shows that there are some errors during the course of the experiment,
slightly altering the data. An error could be that the titration was not done perfectly. The titration uses
the slowly drop the NaOH solution into the acid solution until the color changes. This means that the
accuracy of titration cannot be more accurate than the volume of each drop. Therefore, even if the
titration only needed one-fourth of a drop more, a drop must be added to the solution otherwise the
color will not change. This could have caused the pH to be unequal to pKa and thus causes the data to be
altered. Another error could be that the pH meter was not calibrated accurately. The buffer in the
solution, although resist change, still changes slightly depending on the temperature and could cause
the pH meter calibrated overlooking this to be slightly wrong. This could also cause the pH meter to
Archer G11
show the wrong pH for the solution. These errors can be prevented in the future by using some
technique such as making a capillary head for the pipet so that each drop will be smaller, allowing a
more accurate titration than what was done. Also, the drop could be smear on the side of the flask so
that the adhesive force of the NaOH would pull the drop from the tip of the pipet to the side of the flask.
The other error could be prevented by checking the room temperature of the buffer then calibrate the
pH meter according to the pH at a certain temperature listed on the buffer’s bottle.
Archer G11