Experiment 3: Analysis of Carbonate and Bicarbonate
Objective:
The objective of this experiment is to determine the amount of carbonate and bicarbonate in an
unknown sample using direct and indirect titration techniques. The acid-base titrations will involve the
use of primary and secondary standards. The percentages of carbonate and bicarbonate in the
unknown will be analyzed using statistical analysis.
Background:
Acid-base titrations involve an acid or base titrant with a known concentration to measure the
concentration of an unknown solution. When the titration reaches the equivalence point there are
stoichiometrically equivalent amounts of titrant and analyte. Direct titrations involve adding the titrant
directly to the analyte until the equivalence point is reached. Indirect titrations involve adding a
standardized reagent to the analyte and the added reagent is titrated to the equivalence point.
Procedure:
Unknowns A, B, and C were obtained and stored in the desiccator. HCl (1 L, 0.1 M) was
prepared by diluting HCl (17.00 mL, 6 M) with distilled water. NaOH was prepared by dissolving solid
NaOH (4.00 g) in distilled water. KHP (0.500 g) was used to standardize the NaOH. The standardized
NaOH was used to standardize the HCl solution.
An Unknown A solution (250 mL) was made by dissolving Unknown A (2.0000 g) in carbon
dioxide free water. The Unknown A solution (25.00 mL) was pipetted in a 250 mL Erlenmeyer flask and
titrated with HCl and bromocresol green indicator. Unknown A (25.00 mL) and standardized NaOH
(50.00 mL) were pipetted into an Erlenmeyer flask. Barium Chloride (10 mL, 10% wt/v) was added to the
flask and immediately titrated with HCl and phenolphthalein indicator. The aforementioned method
was used to analyze Unknowns B and C.
Data:
Chemical equations
𝐻𝐶𝑂3 − (𝑎𝑞) + 𝐻𝐶𝑙(𝑎𝑞) → 𝐻2 𝐶𝑂3 (𝑎𝑞) → 𝐶𝑙 −
Total Alkalinity
(𝑎𝑞)
+ 𝐻2 𝑂(𝑙) + 𝐶𝑂2 (𝑔)
𝐶𝑂32− (𝑎𝑞) + 2 𝐻𝐶𝑙(𝑎𝑞) → 𝐻2 𝐶𝑂3 (𝑎𝑞) → 2 𝐶𝑙 − (𝑎𝑞) + 𝐻2 𝑂(𝑙) + 𝐶𝑂2 (𝑔)
Bicarbonate
conversion to
Carbonate
𝐻𝐶𝑂3 − (𝑎𝑞) + 𝑁𝑎𝑂𝐻(𝑎𝑞) → 𝑁𝑎𝐶𝑂3 − (𝑎𝑞) + 𝐻2 𝑂(𝑙)
Carbonate
Precipitation
𝐵𝑎𝐶𝑙2 (𝑎𝑞) + 𝐶𝑂32− (𝑎𝑞) → 𝐵𝑎𝐶𝑂3 (𝑠) + 2 𝐶𝑙 − (𝑎𝑞)
Table 1: Reaction equations for the carbonate and bicarbonate analysis
Mass of KHP
Trial 1
Trial 2
Trial 3
Mass KHP
Standardization 1
(g)
0.5100
0.5098
0.5097
Mass KHP
Standardization 2
(g)
0.5015
0.5008
0.5012
Table 2: Mass of KHP used in NaOH standardizations
Mass of Unknown Samples
Unknown A
Unknown B
Unknown C
2.0094 g
2.0010 g
2.0038 g
Table 3: Mass of Unknowns A, B, and C
Standardization of NaOH
Standardization 1
Initial Volume
NaOH (mL)
Trial 1
Trial 2
Trial 3
0.72
0.22
0.89
Final
Volume
NaOH (mL)
26.92
26.12
26.82
Table 4: Titration data for NaOH standardization
Standardization 2
Volume
NaOH used
(mL)
26.20
25.90
25.93
Initial Volume
NaOH (mL)
0.00
0.00
0.00
Final
Volume
NaOH (mL)
26.80
26.75
26.85
Volume
NaOH
used (mL)
26.80
26.75
26.85
Standardization of HCl
Standardization 1
Initial Volume
NaOH (mL)
Trial 1
Trial 2
Trial 3
Final
Volume
NaOH (mL)
0.01
10.01
0.11
Standardization 2
Volume
NaOH used
(mL)
10.01
20.02
10.13
Initial Volume
NaOH (mL)
10.00
10.01
10.02
Final
Volume
NaOH (mL)
0.00
10.45
20.92
Volume
NaOH
used (mL)
10.45
20.92
31.35
10.45
10.47
10.43
Table 5: Titration data for HCl standardization
Total Alkalinity Titration
Unknown A
Trial 1
Trial 2
Trial 3
Unknown B
Unknown C
Initial
Volume
HCl (mL)
Final
Volume
HCl (mL)
Volume
HCl used
(mL)
Initial
Volume
HCl (mL)
Final
Volume
HCl (mL)
Volume
HCl
used
(mL)
0.31
0.21
0.67
32.51
32.13
32.71
32.20
31.92
32.04
0.11
0.10
0.22
29.37
29.31
29.50
29.26
29.21
29.28
Initial
Final
Volume
Volume Volume
HCl
HCl
HCl
used
(mL)
(mL)
(mL)
0.32
0.69
0.21
26.79
27.11
26.67
26.47
26.42
26.46
Table 6: Carbonate/Bicarbonate titration with HCl
Moles of Excess NaOH
Unknown A
Trial 1
Trial 2
Trial 3
Unknown B
Unknown C
Initial
Volume
HCl (mL)
Final
Volume
HCl (mL)
Volume
HCl used
(mL)
Initial
Volume
HCl (mL)
Final
Volume
HCl (mL)
Volume
HCl
used
(mL)
0.00
0.00
0.00
42.78
42.80
42.82
42.78
42.80
42.82
0.30
0.00
0.00
37.4
37.15
37.12
37.10
37.15
37.12
Table 7: Indirect titration of NaOH with HCl
Initial
Final
Volume
Volume Volume
HCl
HCl
HCl
used
(mL)
(mL)
(mL)
0.00
0.00
0.00
26.95
27.00
26.90
26.95
27.00
26.90
Calculations:
Mass of NaOH Needed for Solution
Equation
0.1 𝑀 𝑁𝑎𝑂𝐻 39.9971 𝑔 𝑁𝑎𝑂𝐻
1𝐿×
×
= 3.99971 𝑔 𝑁𝑎𝑂𝐻
𝐿
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
Calculation 1: Mass of NaOH needed to make 1 L of a 0.1 M solution
Volume of HCl Needed for Solution
Equation
𝑀1 𝑉1 = 𝑀2 𝑉2
Example
0.1 𝑀 × 1 𝐿 = 6 𝑀 × 𝑉2
𝑉2 = 16.667 𝑚𝐿
Calculation 2: Volume of 6 M HCl needed for 0.1 M dilution
Mass of KHP Needed for Titration
Equation
0.1 𝑀 𝑁𝑎𝑂𝐻
1 𝑚𝑜𝑙 𝐾𝐻𝑃 204.22 𝑔 𝐾𝐻𝑃
0.025 𝐿 ×
×
×
= 0.5100 𝑔 𝐾𝐻𝑃
𝐿
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1 𝑚𝑜𝑙 𝐾𝐻𝑃
Calculation 3: Mass of KHP needed for 25.00 mL titration
Moles of NaOH
Equation
𝑔 𝐾𝐻𝑃
𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
×
= 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻
𝑀𝑊 𝐾𝐻𝑃 𝑚𝑜𝑙 𝐾𝐻𝑃
Example Trial 1
0.5100 𝑔 𝐾𝐻𝑃 1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
×
204.22 𝑔 𝐾𝐻𝑃 1 𝑚𝑜𝑙 𝐾𝐻𝑃
= 0.00249731 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻
Calculation 4: Moles of NaOH calculation with Standardization 1 Trial 1 data
Molarity of NaOH
Equation
𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
= 𝑀 𝑁𝑎𝑂𝐻
𝑉
Example Trial 1
0.00249731 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
= 0.09532 𝑀 𝑁𝑎𝑂𝐻
0.02620 𝐿
Calculation 5: Molarity of NaOH calculation with Standardization 1 Trial 1 data
Moles of HCl
Equation
𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
𝑚𝑜𝑙 𝐻𝐶𝑙
𝑉 𝑁𝑎𝑂𝐻 ×
×
𝐿
𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
= 𝑚𝑜𝑙 𝐻𝐶𝑙
Example Trial 1
0.09598 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
1 𝑚𝑜𝑙 𝐻𝐶𝑙
0.0100 𝐿 𝑁𝑎𝑂𝐻 ×
×
𝐿
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
= 0.000959844 𝑚𝑜𝑙 𝐻𝐶𝑙
Calculation 6: Moles of HCl calculation with Standardization 1 Trial 1 data
Molarity of HCl
Equation
𝑚𝑜𝑙 𝐻𝐶𝑙
= 𝑀 𝐻𝐶𝑙
𝑉
Example Trial 1
0.000959844 𝑚𝑜𝑙 𝐻𝐶𝑙
= 0.09608 𝑀 𝐻𝐶𝑙
0.010 𝐿
Calculation 7: Molarity of HCl calculation with Standardization 1 Trial 1 data
Total Alkalinity
Equation
𝑚𝑜𝑙 𝐻𝐶𝑙 𝑚𝑜𝑙 𝐶𝑂3 2− /𝐻𝐶𝑂3 −
𝑉 𝐻𝐶𝑙 ×
×
𝐿
𝑚𝑜𝑙 𝐻𝐶𝑙
= 𝑚𝑜𝑙 𝐶𝑂3 2− /𝐻𝐶𝑂3 −
Unknown A Example Trial 1
0.09608 𝑚𝑜𝑙 𝐻𝐶𝑙 1 𝑚𝑜𝑙𝐶𝑂3 2− /𝐻𝐶𝑂3 −
0.03220 𝐿 𝐻𝐶𝑙 ×
×
𝐿
1 𝑚𝑜𝑙 𝐻𝐶𝑙
= 0.003093776 𝑚𝑜𝑙 𝐶𝑂3 2− /𝐻𝐶𝑂3 −
Calculation 8: Total alkalinity calculation of Unknown A
Moles of NaOH to react with Bicarbonate
Equation
𝑉 × 𝑀 = 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁𝑎𝑂𝐻
Unknown A Example Trial 1
0.050 𝐿 𝑁𝑎𝑂𝐻 × 0.09598 𝑀
= 0.004799 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁𝑎𝑂𝐻
Calculation 9: Total moles of NaOH to react with Bicarbonate before the titration. Unknown C used Standardization 2 data.
Moles of NaOH in Excess
Equation
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
𝑉×𝑀×
= 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁𝑎𝑂𝐻
1 𝑚𝑜𝑙 𝐻𝐶𝑙
Unknown A Example Trial 1
1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
0.04278 𝐿 𝐻𝐶𝑙 × 0.09608𝑀 𝐻𝐶𝑙 ×
1 𝑚𝑜𝑙 𝐻𝐶𝑙
= 0.0041103 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁𝑎𝑂𝐻
Calculation 10: Moles of NaOH in excess. Unkown C used standardization 2 data
Moles of Bicarbonate in Titration
Equation
𝑇𝑜𝑡𝑎𝑙 𝑀𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻 − 𝑀𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻 𝑖𝑛 𝐸𝑥𝑐𝑒𝑠𝑠
= 𝑚𝑜𝑙 𝐻𝐶𝑂3 −
Unknown A Example
0.004799 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
− 0.0041103 𝑁𝑎𝑂𝐻 𝑖𝑛 𝐸𝑥𝑐𝑒𝑠𝑠
= 0.0006887 𝑚𝑜𝑙 𝐻𝐶𝑂3 −
Calculation 11: Moles of bicarbonate reacted with NaOH
Moles of Bicarbonate in Unknown Solution
Equation
𝑚𝑜𝑙 𝐻𝐶𝑂3 −
× 𝑉𝑓𝑙𝑎𝑠𝑘 = 𝑀 𝐻𝐶𝑂3 −
𝑉𝑝𝑖𝑝𝑒𝑡
Calculation 12: Moles of bicarbonate in Unknown A solution
Unknown A Example
0.0006887 𝑚𝑜𝑙 𝐻𝐶𝑂3 −
× 0.250
0.025 𝐿
= 0.006887 𝑀 𝐻𝐶𝑂3 −
Mass of Bicarbonate in Unknown A
Equation
𝑀𝑊
𝐻𝐶𝑂3 −
−
𝑚𝑜𝑙 𝐻𝐶𝑂3 ×
= 𝑚𝑎𝑠𝑠 𝐻𝐶𝑂3 −
𝑚𝑜𝑙 𝐻𝐶𝑂3 −
Unknown A Example
84.0066 𝑔 𝐻𝐶𝑂3 −
−
0.006687 𝑚𝑜𝑙 𝐻𝐶𝑂3 ×
1 𝑚𝑜𝑙 𝐻𝐶𝑂3 −
= 0.5786 𝑔 𝐻𝐶𝑂3 −
Calculation 13: Mass of bicarbonate in Unknown A
Mass Percent of Bicarbonate in Unknown A
Equation
𝑚𝑎𝑠𝑠 𝐻𝐶𝑂3 −−
× 100 = % 𝐻𝐶𝑂3 −
𝑔 𝑠𝑎𝑚𝑝𝑙𝑒
Unknown A Example
0.5786 𝑔 𝐻𝐶𝑂3 −
× 100 = 28.91 % 𝐻𝐶𝑂3 −
2.0015 𝑔 𝑠𝑎𝑚𝑝𝑙𝑒
Calculation 14: Mass percent of bicarbonate in Unknown A
Moles of Carbonate in Titration
𝑚𝑜𝑙 𝐶𝑂3
2−
/𝐻𝐶𝑂3
2
−
Equation
− 𝑚𝑜𝑙 𝐻𝐶𝑂3 −
= 𝑚𝑜𝑙 𝐶𝑂3 2−
Unknown A Example
𝐶𝑂3 2−
−
0.00307968 𝑚𝑜𝑙 𝐻𝐶𝑂
− − 0.0006887𝑚𝑜𝑙 𝐻𝐶𝑂3
3
2
= 0.0011955 𝑚𝑜𝑙 𝐶𝑂3 2−
Calculation 15: Moles of carbonate in Unknown A titration
Moles of Carbonate in Unknown Solution
Equation
𝑚𝑜𝑙 𝐶𝑂3 2−
× 𝑉𝑓𝑙𝑎𝑠𝑘 = 𝑀 𝐶𝑂3 2−
𝑉𝑝𝑖𝑝𝑒𝑡
Unknown A Example
0.0011955 𝑚𝑜𝑙 𝐻𝐶𝑂3 −
× 0.250
0.025 𝐿
= 0.0119549 𝑀 𝐶𝑂3 2−
Calculation 16: Moles of carbonate in Unknown A solution
Mass of Carbonate in Unknown A
𝑚𝑜𝑙 𝐶𝑂3 2− ×
Equation
𝑀𝑊 𝐶𝑂3 2−
𝑚𝑜𝑙 𝐶𝑂3
2−
= 𝑚𝑎𝑠𝑠 𝐶𝑂3 2−
Unknown A Example
105.988 𝑔 𝐶𝑂3 2−
0.0119549 𝑚𝑜𝑙 𝐶𝑂3 2− ×
1 𝑚𝑜𝑙 𝐶𝑂3 2−
= 1.2671 𝑔 𝐶𝑂3 2−
Calculation 17: Mass of carbonate in Unknown A
Mass Percent of Carbonate in Unknown A
Equation
2− −
𝑚𝑎𝑠𝑠 𝐶𝑂3
𝑔 𝑠𝑎𝑚𝑝𝑙𝑒
× 100 = % 𝐶𝑂3 2−
Calculation 18: Mass percent of carbonate in Unknown A
Unknown A Example
−
1.2671 𝑔 𝐶𝑂3 2−
× 100 = 63.31 % 𝐶𝑂3 2−
2.0015 𝑔 𝑠𝑎𝑚𝑝𝑙𝑒
Mean (x̅)
Example NaOH Molarity
Equation
x
0.09532 M  0.09638 M  0.09625 M
3
x  0.09595 M
x 1  x 2  x 3 ...  x n
n
x
Calculation 19: Mean calculation
Standard Deviation (S)
Equation
n
S
 (x
i 1
1
 x) 2
n 1
Example EDTA Molarity
S=√
(0.09532 M-0.09598 M)2 + (0.09638 M-0.09598 M)2 + (0.09625 M-0.09598 M)2
2
S = 4.749×10−4
Calculation 20: Standard deviation calculation
Results:
NaOH Standardization
Average NaOH Molarity
Standardization1
0.09598 M±0.0004749
Average NaOH Molarity
Standardization 2
0.09157 M±0.0001177
Table 8: Average molarity for NaOH standardization
HCl Standardization
Average HCl Molarity
Standardization1
0.09608 M±0.00010
Average HCl Molarity
Standardization 2
0.10030 M±0.00019
Table 9: Average molarity for HCl standardization
Total Alkalinity
Unknown A
Unknown B
Unknown C
Average Moles of CO32-/HCO30.00308 mol ±1.102E-5
0.00281 mol ±2.829E-6
0.00265 mol ±2.167E-6
Table 10: Average moles of carbonate and bicarbonate from HCl titration.
Mass Percent of Carbonate and Bicarbonate in Unknown Sample
Unknown A
Unknown B
Unknown C
Average Mass % HCO328.91%
50.42%
78.61%
Average Mass % CO363.31%
42.58%
20.57%
Table 11: Mass percent of carbonate and bicarbonate in unknown samples
Discussion:
Primary standards like KHP are extremely pure (≥99.9%) can be weighed and directly used to
standardize the titrant. Secondary standards like the NaOH and HCl solutions are prepared in the
laboratory and standardized against the primary standard. A titrant is a solution used to react with the
analyte during the titration. The NaOH is the titrant in the HCl standardization. HCl is the titrant in the
unknown titrations. Indirect titrations or back titrations involve the excess reagent being titrated after
the reaction with the analyte. In this experiment the excess NaOH is determined after the carbonate is
precipitated with barium chloride. The percentage of bicarbonate and carbonate found in Unknown A
was 28.91% and 63.31%, respectively. There is a discrepancy in the data, the percentages should add up
to 100%; however, the calculated percentage adds up to 92.22%. In Unknown B, the experimental
percentage of bicarbonate and carbonate were 50.42% and 42.58% respectively. Again there is a
discrepancy in the data; the total percentage is 93%. Unknown C has experimental bicarbonate and
carbonate percentages of 78.61% and 20.57% respectively (Table 11), and a total percentage of 99.18%.
Unknown C yielded the most accurate results because the overall percentage is closer to 100%.
Conclusion:
The indirect titration to determine the excess NaOH is essential to the experiment because you
cannot determine the individual amount of sodium bicarbonate using a direct titration. In the future a
more distinctive indicator would be helpful to determine the end point of the indirect titration. The
phenolphthalein indicator used for the indirect titration has a color change from bright pink to a
translucent light pink. Determining the actual end point of the titration was very ambiguous. Overall,
the mass percentage of sodium bicarbonate and sodium carbonate were able to be determined for all
three unknown samples.