Experiment 3
Analysis of a Mixture of Carbonate and Bicarbonate
Purpose:
In this experiment, the method of titration will be introduced in the lab. This technique is very common
but important in determining a wide variety of analytes. Students will also be introduced to the difference
between a primary standard and a secondary standard and the importance of each in an experiment.
Procedure:
1. An unknown was obtained which contains a mix of carbonates.
2. The acid and base were standardized to approximately 0.1M. This is accomplished by using primary
standard grade KHP that would produce at least 25 mL of titrant. This process was repeated until 3 good
trials were obtained.
3. The unknown was weighed between 2.0-2.5g and placed into a 150 mL volumetric flask. This was then
diluted with freshly boiled distilled water.
4. Next, 25.0 mL of the unknown was placed in a 250 mL Erlenmeyer flask and titrated with standardized
HCl and bromocresol green. Three trials were performed.
5. 25.0 mL of the unknown was placed into a 250 mL Erlenmeyer flask with 50 mL of standardized NaOH.
10.0 mL of 10% wt BaCl2 was added to the flask to precipitate all BaCO3 asnd immediately titrated with
standardized HCl. Phenolphthalein indicator was used to determine an endpoint. This was repeated until
three good trials were performed.
Reactions:

HCO3  2H   H 2CO3
2
CO3  2H   H 2 CO3

HCO3  OH   CO3
Ba 2  CO3
2
2
 H 2O
 BaCO 3
Ba 3  2OH   Ba (OH ) 2
Data:
Grams of NaOH: 1.9988g
Grams of Unknown B: 2.2660g
Trial 1
Trial 2
Trial 3
Trial 4
Ave.
KHP (g)
NaOH (mL)
0.5074g
0.5165g
0.5139g
0.5141g
0.5129g
25.79 mL
26.97 mL
26.97 mL
26.95 mL
26.67 mL
Molarity
NaOH
0.0963 M
0.0937 M
0.0933 M
0.0934 M
0.0942 M
Vol. of
HCl (mL)
25
25
25
25
25
NaOH
Added
24.05 mL
24.75 mL
24.90 mL
24.35 mL
24.51 mL
Molarity of
HCl
0.0926 M
0.0928 M
0.0929 M
0.0910 M
0.0923 M
mL HCl (Bromocresol)
38.52 mL
38.48 mL
38.47 mL
38.50 mL
Trial 1
Trial 2
Trial 3
Trial 4
Initial Mol NaOH
Trial 1
Trial 2
Trial 3
Trial 4
0.004817 moles
0.004689 moles
0.004665 moles
0.004671 moles
Trial 1
Trial 2
Trial 3
Trial 4
Average
mL HCl (Phenolpthalein)
38.49 mL
38.89 mL
38.51 mL
38. 62 mL
NaOH reacted with
HCl
0.00387 moles
0.00391 moles
0.00388 moles
0.00388 moles
Moles of CO32- & HCO30.00389 moles
0.00387 moles
0.00387 moles
0.00388 moles
Moles HCO3-
Moles CO32-
0.000947 moles
0.000779 moles
0.000785 moles
0.000791 moles
0.002943 moles
0.003091 moles
0.003085 moles
0.003089 moles
Wt % HCO3-
Wt % CO32-
Ratio of HCO3- to CO32-
2.55%
2.10%
2.12%
2.13%
2.23%
7.79%
8.82%
8.17%
8.18%
8.24%
20:80
20:80
20:80
20:80
20:80
Calculations:
Grams of KHP
. 1M =
moles of base
= 0.0025 moles Base
0.025 L
Moles of acid=Moles of Base
0.0025 mol KHP ×
204.2212 g
= .51 grams KHP
1 mol KHP
Preparation of NaOH
. 1M =
moles of NaOH
39.995 g
= .05 moles NaOH ×
= 1.9998 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
. 5074 g KHP
1 mol KHP
1 mol NaOH
×
×
= 0.002485 moles NaOH
1
204.221 g KHP 1 mol KHP
moles NaOH
= Molarity of NaOH
Vol. NaOH
0.002485 moles NaOH
= 0.09636 M
0.02579 L Titrant (NaOH)
Standardizing HCl
Mbase Vbase = Macid Vacid
(. 09634)(.02405) = Macid (.025)
Macid = 0.0926 M
Moles of Carbonates
MHCl × mL of HCl added ×
. 0926 × 38.57 ×
1
= moles of carbonates
1000 mL
1
= .00357 moles of carbonates
1000
Moles Initial NaOH
. 05L × M NaOH = Moles of NaOH initial
. 05 L × 0.09634 = 0.004817 molesNaOH initial
Mol NaOH reacted with HCl
Molarity HCl × mL added ×
1
1 mol NaOH
×
= Mol NaOH reacted
1000 mL
1 mol HCl
0.1005 × 38.49 ×
1
1 mol NaOH
×
= .004817 Mol NaOH reacted
1000 mL
1 mol HCl
Mol NaOH reacted with HCO3Mol NaOH initial − mol reacted = moles NaOH reacted with HCO− 3
0.004817 − 0.00387 = 0.000947 moles NaOH = moles HCO− 3
Mol CO32Mole carbonates − Moles HCO− 3 = moles CO2− 3
0.00389 − 0.000947 = 0.002943 moles CO2− 3
Wt% of 𝐇𝐂𝐎− 𝟑
%𝐰𝐭 HCO− 3 = mole HCO− 3 ×
%𝐰𝐭 HCO− 3 = 0.000947 ×
g HCO− 3
1
×
× 100
−
1 mol HCO 3 mass of unknown
61.058 g HCO− 3
1
×
× 100 = 2.55%
−
1 mol HCO 3
2.2660 g
Wt% of CO2− 3
2−
%𝐰𝐭 CO
3
g CO2− 3
1
= mole CO 3 ×
×
× 100
1 mol CO2− 3 mass of unknown
2−
%𝐰𝐭 CO2− 3 = 0.002943 mole CO2− 3 ×
60.008 g CO2− 3
1
×
100 = 7.79%
2−
1 mol CO 3
2.2660 g
̅ )2
∑(di −d
S=√
n−1
Std Dev of Wt % HCO3-: 2.225 ± 0.000678
Std Dev of Wt % CO32-: 18.315 ± 0.00378
Conclusion:
After comparison of our trials, we determined that trials 2, 3 and 4 were the most accurate. The first
trial however did not support the other data. We were able to calculate a standard deviation and our
results did support a high success rate. As far as improvements to the lab, there are so many variables
and that can often lead to a source of error. Titrating also cause a larger level of error because of how
quickly the solution reaches its endpoint. The endpoint is difficult to grasp and more practice will lead
to less error.
After-Lab Questions:
1. What is meant by a primary standard?
A substance of a known high degree of purity that undergoes one invariable reaction with the other
reactant of interest is a primary standard.
2. What is a secondary standard?
A secondary standard is a solution that has been titrated against a primary standard.
3. What is an indirect titration?
An indirect titration determines 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. The concentration of
the analyte in the original solution is then related to the amount of reagent consumed.
4. Define titrant:
Titrant is a solution of known concentration, which is titrated to another solution to determine the
concentration of a second chemical species.