Analysis of a Mixture of Carbonate and Bicarbonate
Purpose:
The purpose of this lab is to become familiar with the technique of titration. In addition, the
experiment will introduce the use of primary and secondary standards. Finally the amount of
two different carbonate species via indirect titration method will be determined. The
approximate relative percent will be calculated.
Procedure:
Obtain an unknown which contains a mix of the carbonates and store in desiccator. Standardize
the acid and base by preparing approximately 0.1M HCl (1 Liter) and approximately 0.1M
NaOH (500mL). Using KHP, weigh enough KHP to require at least 25mL of titrant. Repeat to
obtain three good trials. Weigh 2.0-2.5g of unknown into a 250mL volumetric flask. Dilute to
mark with freshly boiled and cooled distilled water. Pipet a 25mL aliquot of unknown into a
250mL Erlenmeyer flask and titrate with standardized HCl, using bromocresol green. Repeat for
three good trials. Pipet a 25mL aliquot of unknown and a 50mL aliquot of standard NaOH into a
250 Erlenmeyer. Add 10mL of 10 wt% BaCl2, swirl to precipitate all BaCO3 and then
immediately titrate with standard HCl, use phenolphthalein. Repeat for three good trials.
Reactions:

HCO3  2H   H 2CO3
2
CO3  2H   H 2 CO3

HCO3  OH   CO3
Ba 2  CO3
Ba
3
2
2
 H 2O
 BaCO 3

 2OH  Ba (OH ) 2
Data:
Standardization of NaOH and HCl
Calculations:
Grams of KHP to produce 25mL of titrant:
. 1M =
moles of base
= 0.0025 moles Base
0.025 L
𝑀𝑜𝑙𝑒𝑠 𝑎𝑐𝑖𝑑 = 𝑚𝑜𝑙𝑒𝑠 𝑏𝑎𝑠𝑒
204.2212 g
Example: 0.0025 mol KHP × 1 mol KHP = .51 grams KHP
Preparation of NaOH:
. 1M =
moles of NaOH
39.995 g
= .05 moles NaOH ×
= 1.99975 g NaOH
.5 L
1 mol NaOH
Preparation of HCl:
6M × V1 = 1L(. 1M)
V1 = 0.1667L or 16.67 mL
Standardizing NaOH:
g KHP
1 mol KHP
1 mol NaOH
×
×
= moles NaOH
1
204.221 g KHP 1 mol KHP
. 5704 g KHP
1 mol KHP
1 mol NaOH
×
×
= 0.00248 moles NaOH
1
204.221 g KHP 1 mol KHP
moles NaOH
= Molarity of NaOH
L Titrant (NaOH)
0.00248 moles NaOH
= 0.09616 Molarity of NaOH
0.02579 L Titrant (NaOH)
Standardizing HCl:
Mbase Vbase = Macid Vacid
(0.0963)(0.02405) = Macid (.025)
Macid = 0.0926
Moles of carbonates:
MHCl × mL of HCl added ×
0.0923 × 38.52 ×
Initial moles NaOH:
1
= moles of carbonates
1000 mL
1
= .00356 moles of carbonates
1000
. 05L × M NaOH = Moles of NaOH initial
. 05 L × 0.0942 = 0.00471 moles NaOH initial
Moles NaOH reacted with HCl:
Molarity HCl × mL added ×
0.0923 × 38.49 ×
1
1 mol NaOH
×
= Mol NaOH reacted
1000 mL
1 mol HCl
1
1 mol NaOH
×
= 0.00355 Mol NaOH reacted
1000 mL
1 mol HCl
Mol NaOH reacted with HCO3-:
Mol NaOH initial − mol reacted = moles NaOH reacted with HCO− 3
0.00471 − 0.00349 = 0.00116 moles NaOH = moles HCO− 3
Mol CO32-:
Mole carbonates − Moles HCO− 3 = moles CO2− 3
0.00355 − 0.00124 = 0.00239 moles CO2− 3
Wt% of HCO− 3
%wt HCO− 3 = mole HCO− 3 ×
−
%wt HCO
3
g HCO− 3
1
×
× 100
−
1 mol HCO 3 mass of unknown
61.058 g HCO− 3
1
= 0.00116 ×
×
× 100 = 3.12%
1 mol HCO− 3
2.2660 g
Wt% of CO2− 3
%wt CO2− 3 = mole CO2− 3 ×
2−
%wt CO
3
g CO2− 3
1
×
× 100
2−
1 mol CO 3 mass of unknown
60.008 g CO2− 3
1
= 0.00239 mole CO 3 ×
×
× 100 = 6.33%
1 mol CO2− 3
2.2660 g
2−
Standard deviation:
𝑠 = √∑(𝑥𝑖 − 𝑥𝑚𝑒𝑎𝑛 ÷ 𝑛 − 1
Example: 𝑠 =
√(.0312−.0307)2 +(.0299−.0307)2 +(.0312−.0307)^2
3−1
= 0.000613
Standard Deviation of HCO3-: 3.07 ± 0.00057008
Standard Deviation of CO32-: 6.42 ± 0.00150
Ratio: 50 to 50
Conclusion:
This idea of this lab was to familiarize students with the technique of titration and understands
the components of a titration. The results conclude that the approximate relative percent ratio
between carbonate and bicarbonate is 50 to 50. Error that could have occurred during this lab is
that the titrations could have been slightly overshot.
Post-Lab Questions:
1. What is meant by a primary standard?
A primary standard is a reagent that is pure enough and stable enough to be used directly
after weighing.
2. What is a secondary standard?
A secondary standard is a solution that has been made or standardized from a primary
standard and thus, is not pure like a primary standard.
3. What is an indirect titration?
An indirect titration is one that is used when the analyte cannot be directly titrated. For
example, analyte A may be precipitated with excess reagent R. The product is filtered, and the
excess R washed away. Then AR is dissolved in a new solution, and R can be titrated.
4. Define titrant.
A titrant is the substance that is placed in the buret and added to the analyte.