David Millard
Naomi Bryner (Partner)
Experiment 3: Analysis of a Mixture of Carbonate and Bicarbonate
Introduction:
Titration techniques will be practiced. There will be an introduction to standards,
secondary standards, which originate from a primary standard, will be used. An acidic and basic
standard will be prepared and used for the titration. An indirect titration will be used to find the
amount of two different carbonate species.
Procedure:

Obtain an unknown and store it in your dessicator until it is needed.

Standardize your acid and base by preparing approximately 0.1M HCl (1L) and
approximately 0.1M NaOH (500mL)

Use primary grade KHP, which is drying in the oven; accurately weigh enough to require
at least 25mL of titrant.

Repeat this procedure until three good trials are obtained.

Now that you have a standardized base, you can standardize the acid.

Accurately weigh 2.0-2.5g of unknown into a 250mL volumetric flask.


Dilute to the mark with freshly boiled and cooled triply distilled water.
Pipet a 25.00mL aliquot of your unknown into a 250mL Erlenmeyer and titrate with
standardized HCl using bromocresol green to determine the end point.

Repeat until at least three good trials are obtained.

Pipet a 25.00mL aliquot of unknown and a 50.00mL aliquot of standard NaOH into a
250mL Erlenmeyer flask.

Add 10mL of 10 wt % BaCl2, swirl to precipitate all BaCO3, then immediately titrate
with standard HCl using phenolphthalein as an indicator.


Repeat until at least three good trials are obtained.
Report relative percent carbonate and bicarbonate in your unknown with appropriate
standard values.
Reaction:
+
HCO−
3 + 2H → H2 CO3
+
CO2−
3 + 2H → H2 CO3
2−
−
HCO−
3 + OH → CO3 + H2 O
Ba2+ + CO2−
3 → BaCO3
3+
Ba + 2OH − → Ba(OH)2
Data:
Standardizing the base (NaOH)
Sample
1
2
3
Average
Mass KHP (g)
0.5141
0.5127
0.5127
0.5132
Volume NaOH (mL)
26.40
26.80
26.61
26.603
Molarity NaOH (M)
0.09368
0.09368
0.09434
0.09390
Volume HCl (mL)
10.58
10.63
10.65
10.62
Molarity HCl (M)
0.09935
0.09982
0.09905
0.099407
Volume HCl (mL)
Moles of HCO3-
28.45
28.40
28.38
28.41
0.002828
0.002823
0.002821
0.002824
Volume HCl (mL)
Moles of CO32-
25.42
25.70
25.55
25.56
0.0009580
0.0009748
0.0009271
0.0009533
Standardizing acid (HCl) using base (NaOH)
Sample
1
2
3
Average
Volume NaOH (mL)
10.00
10.00
10.00
10.00
Determining unknown with standardized acid
Sample
1
2
3
Average
Volume of unknown
(mL)
25.00
25.00
25.00
25.00
Determining unknown with standardized acid and BaCl2
Sample
1
2
3
Average
Volume of unknown
(mL)
25.00
25.00
25.00
25.00
Wt%
Weight percent HCO37.80%
7.78%
7.78%
7.79%
0.00988
Sample
1
2
3
Average
Standard Deviation
Weight percent CO322.60%
2.64%
2.51%
2.58%
0.0656
Calculations:
Finding amount of KHP to use
Molarity NaOH ∗ Volume NaOH = mol NaOH ∗
0.1M NaOH ∗ 0.025L NaOH = 0.0025mol NaOH ∗
mol KHP
g KHP
∗
= g KHP
mol NaOH mol KHP
1mol KHP 204.22g KHP
∗
= 0.51g KHP
1mol NaOH
1mol KHP
Finding molarity for NaOH
g KHP ∗
0.5141g KHP ∗
mol KHP mol NaOH mol NaOH
∗
=
= Molarity NaOH
g KHP
mol KHP
L NaOH
1mol KHP
1mol NaOH 0.0025174mol NaOH
∗
=
= 0.0938M NaOH
204.22g KHP 1mol KHP
0.02040L NaOH
Finding molarity of HCL
Mbase Vbase
= Macid
Vacid
(0.0939M)(0.01058L)
= 0.09935M HCl
0.010L
Finding moles of HCO3Molarity HCl ∗ Volume HCl = mol HCO−
3
0.09935M ∗ 0.02845L = 0.002828mol HCO−
3
Finding moles of initial NaOH
Volume NaOH ∗ Molarity NaOH = inital moles NaOH
0.05L ∗ 0.09390M = 0.004695mol NaOH
Finding moles of NaOH which reacted with HCl
Molarity HCL ∗ Volume HCL ∗
0.09940M ∗ 0.02845L ∗
mol NaOH
= mol NaOH
mol HCl
1mol NaOH
= 0.002828mol NaOH
1mol HCl
Finding moles of NaOH that reacted with HCO3mol NaOH inital − mol NaOH reacted = mol HCO−
3
0.004695mol − 0.002828mol = 0.001866mol HCO−
3
Finding moles of CO32−
2−
mol HCO−
3 − mol HCO3 reacted = mol CO3
0.002828mol − 0.001866mol = 0.000962mol CO2−
3
Finding wt% of HCO3mol
HCO−
3
0.002828mol HCO−
3 ∗
g HCO3−
1
∗
∗
∗ 100 = wt% HCO3−
mol HCO3− g unknown
61.055g HCO−
1
3
∗ 100 = 7.80 wt% HCO−
3
− ∗
1mol HCO3
2.2141g unknown
Finding wt% of CO32mol CO2−
3 ∗
0.000962mol
g CO2−
1
3
2−
2− ∗ g unknown ∗ 100 = wt% CO3
mol CO3
60.008g CO2−
1
3
∗
∗
∗ 100 = 2.61 wt% CO2−
3
2−
2.2141g unknown
1mol CO3
CO2−
3
Finding an average
X1 + X 2 … X n
n
0.0264L + 0.0268L + 0.0261L
= 0.02660
3
Finding standard deviation
∑(xi − x̅)2
√
= standard deviation
n−1
√
(7.80 − 7.79)2 + (7.78 − 7.79)2 + (7.78 − 7.79)2
= 0.00988
2
Conclusion:
The main objective of this lab was to determine the amount of a carbonate species and a
bicarbonate species. Using the titration technique, we were able to determine the amounts of
carbonate and bicarbonate in the unknown samples. On average, the unknown contained 7.79%
(±0.00988) carbonate and 2.58% (±0.0656) bicarbonate.
Discussion:
Using the textbook as a reference;
A primary standard is a sample that is pure, and upon massing, it can be assumed that the entire
mass is what is labeled. A secondary standard is a sample that may have some type of
contamination and the mass recorded may not be the true mass of the sample. An indirect
titration is a type of titration when the analyte cannot be directly titrated. A titrant is something
added to the analyte in the titration process.