Experiment 7: Comparison of Unknowns: Weak Acid Identification
Introduction:
The objective of this experiment is to familiarize the student with acid-base chemistry using a guided, seven-step procedure. After all
seven steps have been completed, the student should be able to correctly identify his/her unknown acid.
Procedure:
Sodium hydroxide solution (0.1M) was prepared and standardized using 0.511g potassium hydrogen phthalate (KHP) dissolved in
50mL distilled water. The titration was monitored using a calibrated pH meter. An unknown weak acid was then dried and
approximately 0.306g of the unknown was used in each trial. Again, the unknown was dissolved in 50mL distilled water and the
titration with sodium hydroxide was monitored using a calibrated pH meter. Sufficient data were recorded to produce titration
graphs for each trial.
Data:
See Excel document for data points and titration graphs.
Equivalence Points
Half Eq.Points
Standardization
Standardization
Trial
Vol. NaOH (mL)
pH
Trial
Vol. NaOH (mL)
pH
1
26.8
8.80
1
13.4
5.05
2
26.2
8.58
2
13.1
4.99
3
26.3
8.98
3
13.2
5.03
Average
26.4
8.79
Average
13.2
5.02
Trial
Vol. NaOH (mL)
pH
Trial
Vol. NaOH (mL)
pH
1
18.2
8.00
1
9.1
4.00
2
18.1
8.42
2
9.1
4.08
3
18.1
8.40
3
9.1
4.10
Average
18.1
8.27
Average
9.1
4.06
Unknown C
Unknown C
Formulas for Calculations:
Mass of KHP Required Equation
204.23𝑔
= 𝑔𝐾𝐻𝑃
1𝑚𝑜𝑙
(assume ~0.1M NaOH, ~25mL titration)
0.1𝑀 ∗ 0.02500𝐿 = 𝑚𝑜𝑙𝑁𝑎𝑂𝐻/𝐾𝐻𝑃 ×
Mass of Unknown C Required Equation
204.23𝑔
= 𝑔𝑢𝑛𝑘
1𝑚𝑜𝑙
(assume ~0.1M NaOH, ~15mL titration, MW = KHP)
0.1𝑀 ∗ 0.01500𝐿 = 𝑚𝑜𝑙𝑁𝑎𝑂𝐻/𝑢𝑛𝑘 ×
Molarity of NaOH Equation
1𝑚𝑜𝑙
÷ 𝑉𝑒 (𝐿) = 𝑀𝑁𝑎𝑂𝐻
204.23𝑔
(because moles of NaOH = moles of KHP)
𝑔𝐾𝐻𝑃 ×
Moles of Unknown C Equation
̅𝑁𝑎𝑂𝐻 × 𝑉𝑒 (𝐿) = 𝑚𝑜𝑙𝑁𝑎𝑂𝐻/𝑢𝑛𝑘
𝑀
Molar Mass of Unknown C Equation
𝑔𝑢𝑛𝑘
= 𝑀𝑊𝑢𝑛𝑘
𝑚𝑜𝑙𝑢𝑛𝑘
Ka of Unknown C Equation
10−𝑝𝐻1/2𝑒𝑞 = 𝑘𝑎
(because pH=pKa at ½ eq. pt.)
Mass of KHP Required Example
0.1𝑀 ∗ 0.02500𝐿 = 0.0025𝑚𝑜𝑙𝑁𝑎𝑂𝐻/𝐾𝐻𝑃 ×
= 0.511𝑔𝐾𝐻𝑃
204.23𝑔
1𝑚𝑜𝑙
Mass of Unknown C Required Example
0.1𝑀 ∗ 0.01500𝐿 = 0.0015𝑚𝑜𝑙𝑁𝑎𝑂𝐻/𝑢𝑛𝑘 ×
= 0.306𝑔𝑢𝑛𝑘
204.23𝑔
1𝑚𝑜𝑙
Molarity of NaOH Example
0.5126𝑔𝐾𝐻𝑃 ×
1𝑚𝑜𝑙
÷ 0.0268𝐿(𝑁𝑎𝑂𝐻) = 0.0937𝑀𝑁𝑎𝑂𝐻
204.23𝑔
Moles of Unknown C Example
̅𝑁𝑎𝑂𝐻 × 0.0182𝑉𝑒 (𝐿) = 0.00173𝑚𝑜𝑙𝑁𝑎𝑂𝐻/𝑢𝑛𝑘
0.0950𝑀
Molar Mass of Unknown C Example
0.3072𝑔𝑢𝑛𝑘
= 177.65𝑀𝑊𝑢𝑛𝑘
0.00173𝑚𝑜𝑙𝑢𝑛𝑘
Ka of Unknown C Example
10−4.00 = 1.00 × 10−4
Summary of Calculations:
Trial
Molarity NaOH
Moles Unk. C
Molar Mass Unk. C
pH @ V1/2 (pka)
Ka Unk. C
1
0.093653556
0.001729212
177.6532001
4.00
0.0001
2
0.095816982
0.001719711
178.9836054
4.08
8.31764E-05
3
0.095564365
0.001719711
178.8673067
4.10
7.94328E-05
Average
0.095011634
0.001722878
178.5013707
4.06
8.75364E-05
Mass KHP Req’d : 0.511g
Mass Unk. C Req’d : 0.306g
Conclusions:
The calculated molar mass of unknown C was 178.50 g/mol. The pka value was 4.06, and the ka value was 8.754E-5. The
unknown was known to be a monoprotic organic acid. Two monoprotic organic acids with similar molecular weights, pka and ka
values to those that were calculated from this experiment are 1-naphthoic acid (172.18g/mol, 3.70, 2.1E-4) and acetylsalicylic acid
(180.157g/mol, 3.50, 3.3E-4). Further experimentation would be necessary to determine which of the two weak acids is unknown C.
The first fits more closely to the pka value whereas the second fits more closely to the molecular weight. One possible reason for not
obtaining a clear-cut identity for the acid is that using a titration curve is inexact unless the first and second derivatives are utilized
to determine the exact equivalence point. As this was not the case with this experiment, the determinations of the equivalence and
half-equivalence points are very inexact, leading to an unclear identity of the unknown. Further experimentation may include NMR
or IR analysis to determine which of the two possible compounds is, in fact, unknown C.