Naomi Bryner
Experiment 3: Analysis of a Mixture of Carbonate and Bicarbonate
Introduction: This lab is designed to exercise the technique of titration. Standards will be made using
serial dilutions. Data collected will determine the amount of carbonate and bicarbonate via indirect and
direct titration.
Procedure:
One liter of 0.1 M HCL and 0.5 L of 0.1 M NaOH were prepared. NaOH solution was standardized
using KHP (x 3) and HCL solution was standardized using standardized NaOH solution (x 3). Unknown C
(2.2141 g) was diluted with degassed H2O. 25.00 mL unknown solution was titrated using HCl solution,
with bromocresol green as indicator (x 3). BaCl2 was precipitated using 25.00 mL unknown solution,
50.00 mL standardized NaOH solution, and 10% BaCl2. This was immediately titrated using standardized
HCl solution with phenolphthalein indicator.
Reactions:
HCO3- + H+ → H2CO3
CO32- + 2 H+ → H2CO3
HCO3- + OH- → CO32- + H2O
Ba2+ + CO32- → BaCO3
Ba3+ + 2 OH- → Ba(OH)2
Data:
Standardization of NaOH
Trial 1
Trial 2
Trial 3
Average
KHP (g)
0.5141
0.5127
0.5127
0.5132
Volume (L)
0.02640
0.02680
0.02661
0.07981
Molarity (M)
0.095355
0.093676
0.094345
0.093857
Volume NaOH (L)
0.01058
0.01063
0.01065
0.01062
Volume HCl (L)
0.010
0.010
0.010
0.010
Molarity HCl
0.099301
0.099770
0.999577
0.099676
Standardization of HCl
Trial 1
Trial 3
Trial 3
Average
Unknown titrated with HCL
Trial 1
Trial 2
Trial 3
Volume HCl (L)
0.02845
0.02840
0.02838
Unknown and NaOH titrated with HCl
Volume HCl (L)
Trial 1
0.02542
Trial 2
0.02570
Trial 3
0.02555
Mass NaOH used: 2.0294 g (500 mL)
Mass Unknown C used: 2.2141 g (250 mL)
Volume unknown solvent (L)
0.025
0.025
0.025
Volume NaOH (L)
0.050
0.050
0.050
Volume unknown solvent (L)
0.025
0.025
0.025
Calculations:
in lab
How much KHP for 25 mL 0.1 M NaOH?
0.1 molNaOH
1 LNaOH
×
0.025 LNaOH
molNaOH
1 mol
× 1 mol KHP ×
NaOH
204.2212 g
=
1 molKHP
0.51 g KHP
Mass NaOH for dilution
39.995 g
[(0.1 MNaOH) (0.5 LNaOH)]× 1 mol NaOH = 1.99975 gNaOH
NaOH
Dilution of HCl
0.1M×1 L
= 16.67 mL
6M
post-lab
Molarity of NaOH
(mass of KHP ×
1 mol KHP
1 mol NaOH
×
)
molar mass KHP
1 mol KHP
1 mol
(0.5141KHP g × 204.2212KHP
g
×
KHP
1 molNaOH
1molKHP
÷ Volume NaOH = Molarity NaOH
) ÷ 0.02640 L = 0.095355 M
Molarity of HCl
MNaOH VNaOH = MHCl VHCl
(0.093857 M) (0.01058 L) = (MHCl) (0.010 L)
-> MHCl= 0.099301 M
Total moles carbonate and bicarbonate
MHCl VHCl = mole𝑠𝐻2 𝐶𝑂3 +𝐶𝑂3
(0.099676 M) (0.02845 L) = 0.0028308 mole𝑠𝐻2 𝐶𝑂3 +𝐶𝑂3
Initial moles NaOH
MNaOH VNaOH = molesNaOH
(0.093857 M) (0.050 L) = 0.00469285 molesNaOH
Excess Moles NaOH
1 𝑚𝑜𝑙
MHCl VHCl ( 1 𝑚𝑜𝑙𝑁𝑎𝑂𝐻 ) = molesNaOH (initial)
𝐻𝐶𝑙
1 𝑚𝑜𝑙𝑁𝑎𝑂𝐻
)
1 𝑚𝑜𝑙𝐻𝐶𝑙
(0.099676 M) (0.02542 L ) (
= 0.0025338 molesNaOH (excess)
Moles HCO3(Molesinitial NaOH – molesexcess NaOH) ×
1 𝑚𝑜𝑙𝐻𝐶𝑂3−
1 𝑚𝑜𝑙𝑁𝑎𝑂𝐻
= moles HCO3-
(0.00469285 molesNaOH - 0.0025338 molesNaOH (excess)) = 0.0025474 moles HCO3Moles CO3-2
mole𝑠𝐻𝐶𝑂3 −+𝐶𝑂3 2− – moles HCO3- = mole𝑠𝐶𝑂3 2−
0.0028308 mole𝑠𝐻2 𝐶𝑂3 +𝐶𝑂3 - 0.0021455 moles HCO3- = 0.0006853 mole𝑠𝐶𝑂3 2−
Relative Percent HCO3-
moles HCO3- MolarMassHCO3- (MassUnknown-1) (100%) = Relative Percent
𝑔
(0.0021455 moles HCO3-)(61.058 𝑚𝑜𝑙)(2.2345 g)-1(100%) = 6.96%
Relative Percent CO32mole𝑠𝐶𝑂3 2− - MolarMas𝑠𝐶𝑂3 2− - (MassUnknown-1) (100%) = Relative Percent
(0.0006846 mole𝑠𝐶𝑂3 2− )(60.008
Calculated Data
Trial
1
2
3
Average
Standard deviation
𝑔
)
𝑚𝑜𝑙
Moles HCO30.0025474
0.0021312
0.0025617
0.0024134
0.000221
(2.2345 g)-1 (100%) = 1.84%
% HCO36.96
5.82
7.00
6.59
0.670037
Moles CO320.0006853
0.0006996
0.0002671
0.0005507
0.000246
% CO321.84
1.88
0.72
1.48
0.65848
Conclusions:
All titrations were performed by one person to minimize error, which worked well as seen by the
standard deviation of the moles of HCO3-. Unknown C had an average composition of 6.59% HCO3- and
1.48% CO32-. These compositions are likely false due to a high standard deviation of percent
composition.
Post-Lab Questions:
1. A primary standard is a very pure and can be used to determine the concentrations of
subsequent solutions.
2. A solution that is prepared using a primary standard and is used for a specific analysis.
3. An indirect titration is usually the first titration performed. It is fast and inaccurate, and yields an
approximation of the real amount of titrant needed.
4. A titrant is the reagent solution added to the analyte for titration.