Experiment 3: Analysis of a Mixture of Carbonate and Bicarbonate

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Experiment 3: Analysis of a Mixture of Carbonate and Bicarbonate
Objective:
In this lab we will learn about titrating and standards, along with a method to determine
two different carbonate compounds.
Background:
Titrations are a process by which an unknown concentrated solution is measured with a
solution with a known titration called a titrant. This process is done until the reaction is
complete. Acid/base titrations involve an acid or base titrant that is being reacted in an acidic or
basic solution until the equivalence point has been reached. At this point there will be a color
change to indicate that the equivalence point has been reached.
Procedures:
1.) Obtain one unknown containing a mixture solution of sodium bicarbonate and sodium
carbonate – store in desiccator until needed. Prepare 1 liter of 0.1 M HCl and 500.0
mL of 0.1 M NaOH
2.) Use primary standard grade KHP (in oven drying) weigh enough to require at least 25
mL of titrant. Standardize 0.1 M NaOH. Repeat until 3 good trials are obtained.
3.) Standardize the base, use this to standardize the 0.1 M HCl. Repeat until 3 good trials
are obtained
4.) Weigh 2.0 – 2.5 g of unknown into 250 mL volumetric flask. Dilute to the mark with
freshly boiled and cooled triply distilled water
5.) Pipet 25.00 mL aliquot of known into 250 mL Erlenmeyer flask, titrate with
standardized HCl using bromocresol green οƒ  determine the endpoint. Repeat until 3
good trials are obtained.
6.) Pipet 25.00 mL aliquot of unknown and 50.00 mL aliquot of standardized NaOH into
250 mL Erlenmeyer flask. Add 10 mL of 10 wt% BaCl2, swirl to precipitate all
BaCl2. Immediately titrate with standardized HCl using phenolphthalein indicator.
Repeat until 3 good trials are obtained.
Data:
Unknown B
Standardize HCl w/ H2O NaOH
Trial
mL HCl Added
1
28.2
2
25.5
3
24.5
Average
26.07
Standardize NaOH (base)
Trial
mL NaOH
1
26.4
2
24.9
3
26.0
Average
25.77
Mass of KHP:
1. 0.5098 g
2. 0.5109 g
3. 0.05098 g
0.5098𝑔 ∗
1π‘šπ‘œπ‘™
= 𝟎. πŸŽπŸŽπŸπŸ’πŸ—πŸ• π’Žπ’π’
204.2𝑔
0.5109𝑔 ∗
1π‘šπ‘œπ‘™
= 𝟎. πŸŽπŸŽπŸπŸ“πŸŽπŸ π’Žπ’π’
204.2 𝑔
0.5098𝑔 ∗
1π‘šπ‘œπ‘™
= 𝟎. πŸŽπŸŽπŸπŸ’πŸ—πŸ• π’Žπ’π’
204.2𝑔
mL HCl to unknown
Trial
1
2
3
Average
mL
31.5
30.4
29.7
30.53
Trials
1
2
3
mL
36.6
36.3
37.5
Calculations:
Molarity NaOH
0.002497π‘šπ‘œπ‘™
= 𝟎. πŸŽπŸ—πŸ”πŸ— 𝑴
0.02577 𝐿
0.002502π‘šπ‘œπ‘™
= 𝟎. πŸŽπŸ—πŸ•πŸ 𝑴
0.02577𝐿
0.002497π‘šπ‘œπ‘™
= 𝟎. πŸŽπŸ—πŸ”πŸ— 𝑴
0.02577𝐿
Average: 0.097 M
Molarity HCl
0.002497π‘šπ‘œπ‘™
= 𝟎. πŸŽπŸ—πŸ“πŸ•πŸ–π‘΄
0.02607𝐿
0.002502π‘šπ‘œπ‘™
= 𝟎. πŸŽπŸ—πŸ“πŸ—πŸ• 𝑴
0.02607𝐿
0.002497π‘šπ‘œπ‘™
= 𝟎. πŸŽπŸ—πŸ“πŸ•πŸ– 𝑴
0.02607𝐿
Moles carbonate
𝑀𝐻𝐢𝑙 ∗ 𝑉𝐻𝐢𝑙
Trial 1 0.0969 ∗ 0.0566 = 𝟎. πŸŽπŸŽπŸ‘πŸ“πŸ“
Trial 2 0.0971 ∗ 0.0363 = 𝟎. πŸŽπŸŽπŸ‘πŸ“πŸ
Trial 3 0.0969 ∗ 0.0375 = 𝟎. πŸŽπŸŽπŸ‘πŸ”πŸ‘
Moles Bicarbonate
π‘€π‘π‘Žπ‘‚π» ∗ π‘‰π‘π‘Žπ‘‚π» π‘Ÿπ‘’π‘Žπ‘π‘‘π‘–π‘›π‘”
Trial 1 0.00355 − 0.00256 = πŸ—. πŸ—π’™πŸπŸŽ^ − πŸ’
Trial 2 0.00352 − 0.00242 = πŸ•. πŸ–π’™πŸπŸŽ^ − πŸ’
Trial 3 0.0363 − 0.00252 = 𝟎. 𝟎𝟎𝟏𝟏𝟏
Moles NaOH
𝑀 ∗ 0.0500 𝐿
0.097 ∗ 0.0500 = 𝟎. πŸŽπŸŽπŸ’πŸ–πŸ“
Moles NaOH reacting with HCl
𝑀 ∗ 𝑉𝐻𝐢𝑙
Trial 1 0.097 ∗ 0.0264 = 𝟎. πŸŽπŸŽπŸπŸ“πŸ”
Trial 2 0.097 ∗ 0.0294 = 𝟎. πŸŽπŸŽπŸπŸ’πŸ
Trial 3 0.097 ∗ 0.0260 = 𝟎. πŸŽπŸπŸ“πŸ
Moles Carbonate
π‘šπ‘œπ‘™ π‘π‘Žπ‘Ÿπ‘π‘œπ‘›π‘Žπ‘‘π‘’ − π‘šπ‘œπ‘™π‘’ π‘π‘–π‘π‘Žπ‘Ÿπ‘π‘œπ‘›π‘Žπ‘‘π‘’
Trial 1 0.00355 − 9.9π‘₯10−4 = 𝟎. πŸŽπŸŽπŸ“πŸ”
Trial 2 0.00352 − 7.8π‘₯10−4 = 𝟎. πŸŽπŸŽπŸπŸ•πŸ’
Trial 3 0.00363 − 0.00111 = 𝟎. πŸŽπŸŽπŸπŸ“πŸ
Mass % Bicarbonate
π‘šπ‘œπ‘™ ∗ π‘€π‘Š
∗ 100
π‘šπ‘Žπ‘ π‘  π‘’π‘›π‘˜π‘›π‘œπ‘€π‘›
Trial 1
Trial 2
Trial 3
9.9π‘₯10−4 ∗61.01069
0.20988
7.8π‘₯10−4 ∗61.01069
0.20988
0.0011∗61.01069
0.20988
∗ 100 = πŸπŸ–. πŸ•πŸ–
∗ 100 = 𝟐𝟐. πŸ”πŸ•
∗ 100 = πŸ‘πŸ. πŸπŸ•
Mass % of Carbonate
π‘šπ‘œπ‘™ ∗ π‘€π‘Š
∗ 100
π‘šπ‘Žπ‘ π‘  π‘’π‘›π‘˜π‘›π‘œπ‘€π‘›
Trial 1
Trial 2
Trial 3
0.0056∗60.008
2.0988
∗ 100 = πŸπŸ”. πŸŽπŸ•
0.00274∗60.00∗
2.0988
∗ 100 = πŸ•. πŸ–πŸ‘
0.00252−60.008
2.0988
∗ 100 = πŸ•. 𝟐𝟏
Conclusion:
The NaOH and HCl were standardized to calculate the molarity of the NaOH and the HCl
and KHP used. Then it was used to titrate an unknown. Indirect titration allowed for bicarbonate
to be brought out of solution, allowed us to determine its concentration.
Discussion:
1.) Primary standard is at least 99.9% pure, weighed and used directly. KHP was ours
2.) 2° standard is prepared in the lab and standardized against 1° standard. NaOH was
ours.
3.) Indirect titration / back titration: reagent added to analyte to determine concentration
of unknown analyte. Bicarbonate and carbonate were indirectly tirated with NaOH
4.) Titrant is a substance of known concentration used in a titration to determine the
unknown concentration.
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