Matthew Marthaler
Experiment Three
Analysis of a mixture of Carbonate and Bicarbonate
Purpose:
In this lab, the students will learn the technique of titration. They will also learn the concepts of
standards and how to use them. Lastly, they will determine the amount of the two different carbonate
species via an indirect titration method.
Procedure:
1) Obtain an unknown of carbonate and store it in the desiccator until it is needed. Standardize
the acid and base now by preparing approximately 0.1 M HCl (1 L) and approximately 1 M
NaOH (500 mL) using primary grade KHP which is drying in the oven. Accurately weigh
enough KHP to require at least 25 mL of titrant. Repeat until three good trials are done. Now
that the base is standardized, use this to standardize the acid.
2) Weigh 2.0-2.5g of unknown into a 250 mL volumetric flask. Dilute to mark with freshly
boiled and cooled triply distilled water.
3) Pipet a 25.00 mL aliquot of unknown into a 250 mL Erlenmeyer and titrate with standardized
HCl using bromocresol green to determine the endpoint. Repeat until at least three good
trials are obtained.
4) Pipet 25.00 mL aliquot of unknown and a 50.00 mL aliquot of unknown into a 250 mL
Erlenmeyer flask. Add 10 mL of 10% weight BaCl2, swirl to precipitate all BaCO3, then titrate
with standard HCl using phenolphthalein indicator. Repeat until at least three good trials are
done.
5) Report relative percent carbonate and bicarbonate into unknown with appropriate standard
deviation values. Include all appropriate chemical equations into the lab notebook.
Reactions:
HCO3- + 2H+
H2CO3
CO32- + 2H+
H2CO3
HCO3- + OH-
CO32- + H2O
Ba2+ + CO32-
BaCO3
Ba3+ + 2 OH-
Ba(OH)2
Data for Unknown C:
Standardization of NaOH using KHP
KHP Trial Number
KHP (g)
1
2
3
Average:
0.5108
0.5102
0.5102
0.5104
Volume of 0.1 M
NaOH (mL) Trial
One
26.62
26.09
21.90
24.87
Volume of 0.1 M
NaOH (mL) Trial
Two
26.62
26.63
26.62
26.62
M NaOH
0.09396
0.09381
0.09384
0.09387
Standardization of HCl using NaOH
Trial Number
1
2
3
Average:
Volume HCL (mL)
25.00
25.00
25.00
25.00
Volume NaOH (mL)
26.62
26.39
26.49
26.50
M HCl
0.10049
0.09903
0.09943
0.09965
Mass of Unknown C
1: 2.1035 g
2: 2.1135 g
3: 2.1134 g
Titration of 25.00 mL Unknown C by HCl
Trial Number (25.00 mL of unknown with the
same number)
1
2
3
Average:
Volume of HCl (mL)
Bromocresol
29.13
29.01
28.31
28.82
Titration of 25.00 mL of Unknown C in 50.00 mL of NaOH and 10 mL of 10% weight BaCl2 by HCl
Trial Number
Volume of HCl (mL) First Trial
Phenolphthalein
28.90
27.51
28.01
28.14
1
2
3
Average:
Volume of HCl (mL) Second Trial
Phenolphthalein
27.40
27.51
27.90
27.60
Calculations:
Average mL for each titration set:
Σxi
n
Ex (average mL for titration of HCl in NaOH):
26.62+26.39+26.49
3
Standard Deviation and Percent Relative Deviation:
Ex (titration of HCl in NaOH):
S=
√(26.62−26.50)2 + (26.29−26.50)2 + (26.49+26.50)2
√2
% relative deviation:
0.115
26.50
x 100% = 0.444%
Molarity of NaOH:
Ex:
0.5108 g KHP x 1 mol KHP
204.2212 g KHP x 0.02662
Molarity of HCl :
Ex :
= 0.09396 M
M NaOH x V NaOH
V HCl
0.09396 x 0.02662
0.02500
=0.100049 M
Mass of NaOH (used in step one):
0.1 mols NaOH x 0.5 L x 39.997 g KHP
1 L NaOH x 1 mol NaOH
= 1.999 g NaOH
Mass of KHP (used in step one):
0.1 mol HCl x 0.025 L x 204.2212 g KHP
1 L HCl x 1 mol KHP
= 0.5106 g KHP
= 0.115
= 26.50 mL
Moles of Carbonates (CO32-) and Bicarbonates (HCO3-):
Ex: 0.1000486 M HCl x 0.02913 L HCl (bromocresol) = 0.00291 moles
Moles of NaOH initial:
Ex: 0.09396 M NaOH x 0.0500 L NaOH (added) = 0.00470 mol NaOH initial
Moles of NaOH excess:
Ex: 0.1000486 M HCl x 0.02740 L HCl (Phenolphthalein) x
1 mol NaOH
1 mol HCl
= 0.00274 mol NaOH
Moles of NaOH reacted with HCO3-:
Ex: 0.00470 mol initial – 0.00274 mol excess = 0.00196 mol NaOH reacted
Moles of Bicarbonate (HCO3-):
Ex: 0.00196 moles NaOH x
1 mol Bicarbonate
=
1 mol NaOH
0.00196 mol Bicarbonate
Moles of Carbonate (CO32-):
Ex: 0.00291 mol carbonate/bicarbonate – 0.00196 mol bicarbonate = 0.00095 mol carbonate
Relative Percent of Carbonate:
Ex:
0.00095 mol carbonate x 60.007 g Carbonate x 1
1 mol carbonate x 2.1035 g Unknown C
x 100% = 2.71 %
Average: 2.59%
Relative Percent Bicarbonate:
Ex:
0.00196 mol bicarbonate x 61.014 g bicarbonat x 1
1 mol bicarbonate x 2.10365 g Unknown C
x 100% = 5.69%
Average: 5.65%
Standard deviation of percent Carbonate:
√
(0..0271− .0259)2 + (0.0253−0.0259)2 + (0.0253−0.0259)2
2
= 0.00104
Standard deviation of percent Bicarbonate
√
(0.0569−0.0565)2 + (0.0572−0.0565)2 + (0.0554−0.0565)2
2
= 0.000964
Conclusion:
The purpose of this lab was to determine the amount of carbonate and bicarbonate in unknown
C. Upon the completion of this lab, the calculations show an average of 5.65% bicarbonate and 2.59%
carbonate. The standard deviation of the percent carbonate was 0.00104 and the standard deviation of
the percent bicarbonate was 0.000964. This shows that our calculations are precise, because the
standard deviation is very small, so we did not make any huge errors that would make our trials differ
significantly. I have no idea, however, if those numbers are accurate.
Post-Lab Questions:
1) A primary standard is a reagent that is extremely pure, stable, has no waters of hydration, and
has a high molecular weight. I got this definition from the Department of Chemistry at Adelnide
University’s website. This primary standard also can calibrate other standards.
2) A secondary solution is a standard that has been titrated against a primary standard. A standard
solution is a secondary standard. I got this definition from chemistry-Dictionary.com.
3) An indirect titration is the process of determining the concentration of an analyte by reacting it
with a known number of moles of excess reagent. The excess reagent is then titrated with a
second reagent. I got this definition from Chemistry-Dictionary.com.
4) Titrant is a solution of known concentration which is added (titrated) to another solution to
determine the concentration of a second chemical species. I got this definition from
Chemistry.about.com.