User Guide: Orionβ’ Weather Station (Columbia Weather Systems)
Β
Acid-Base.pdf
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Date:22/06/2022 Group:21 Sample test: 10
Name: Pham Thi Tuong Vy Student ID: 20147113 Class: 20HOH_CLC2
EXPERIMENT REPORT QUANTITATIVE ANALYSIS TITRATION OF ACID β
BASE
1. Standarddization of HCl 0.1 N solution:
a. Principle:
- Titrating concentration of HCl solution using borax solution approximately 0.1N with
5 significant digits with indicator pT 5.1
β Hydrolysis reaction: π΅4π7
2β
+ 5π»2π β 2π»2π΅π3
β
+ 2π»3π΅π3
β Titration reaction: 2π»2π΅π3
β
+ 2π»+
β 2π»3π΅π3
β Net reaction: π΅4π7
2β
+ 5π»2π + 2π»+
β 4π»3π΅π3
- This is neutralization reation of weak base π»2π΅π3
β
by strong acid HCl. The
pH β jump β range is about 6.24 Γ· 4.3 (with accuracy 99.9%) so that we
could use pT 5.1 indicator to determine endpoint of titration.
- Concentration of HCl is by below equation:
ππ»πΆπ =
ππ΅ππππ₯ Γ π
Μ π΅ππππ₯
π
Μ π»πΆπ
b. Practise:
- Phasing π»πΆπ~0.1π solution from HCl 6N by taking 4mL HCl 6N by pipette line,
water norms up to 250mL in a 500mL container, shake evenly. Coat the burret once
with distilled water and recoat with HCl. Lock rhe burret, bring HCl to the level line.
- Take 10.00mL ππ2π΅4π7 π πππ’π‘πππ (0.100000 Β± 0.000015π) by pipette elected to
erlen, for 1 drop of pT 5.1. Slowly lower the HCl from the top of ther burret, until the
solution turns from green to grape red. Record the volume of the used HCl and repeat
3 times, recording the average value.
c. Data and caculations:
ππ»πΆπ(ππΏ)
π
Μ π»πΆπ (ππΏ)
1π π‘
2ππ
3ππ
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9.75 9.80 9.85 9.80
- Cocentration of HCl:
β ππ»πΆπ =
ππ΅ππππ₯Γπ
Μ π΅ππππ₯
π
Μ π»πΆπ
=
0.100000Γ10.00
9.8
= 0.102041π
β πππ»πΆπ
= β
β (ππβπ
Μ )2
π
π=1
πβ1
= β
(9.75β9.80)2+(9.80β9.80)2+(9.85β9.80)2
3β1
= 0.05 (ππΏ)
β π’π
Μ π»πΆπ
= πΜ ππ»πΆπ
=
πππ»πΆπ
β3
=
0.05
β3
= 0.0286675 β¦ ππΏ
β π’π
Μ πππππ₯
=
π’ππππππ‘π‘π
β3
=
πππππππ‘π‘π
β3 Γβ6
=
0.02
β3 Γβ6
= 0.004714 (mL)
β π’ππππππ₯
=
πππππππ₯
2
=
0.000015
2
= 0.0000075(π)
β
π’ππ»πΆπ
ππ»πΆπ
= β(
π’π
Μ
π»πΆπ
π
Μ π»πΆπ
)
2
+ (
π’ππππππ₯
ππππππ₯
)
2
+ (
π’π
Μ
πππππ₯
π
Μ πππππ₯
)
2
= β(
0.02866675
9.80
)
2
+ (
0.004714
10
)
2
+ (
0.0000075
0.100000
)
2
= 0.0029638 β¦ (π)
β π’ππ»πΆπ
= 0.000302436π
πππ»πΆπ
= 4.3 Γ 0.000302436 = 0.00013(π)
ο° πΎπππ π· = π. ππ, π΅π―πͺπ = π. πππππ Β± π. πππππ π΅
2. Determine concentration of exam sample of Na2CO3 and NaHCO3 mixture
a. Principle:
- Titration a exact of solution Na2CO3 and NaHCO3 2 times by using standard solution
HCl 0.1N. One titration reach 1st
stager to determine Na2CO3. The second one reach
2nd
stage to determine sum of Na2CO3 and NaHCO3.
- First stage: Only titrate πΆπ3
2β
, spend V1 mL:
πΆπ3
2β
+ π»+
β π»πΆπ3
β
Lost Va HCl solution
- Second stage: titrate all πΆπ3
2β
πππ π»πΆπ3
β
, spend π
Μ 2 mL:
πΆπ3
2β
+ 2π»+
β π»2πΆπ3 β πΆπ2 + π»2π Lost 2Va HCl solution
π»πΆπ3
β
+ π»+
β π»2πΆπ3 β πΆπ2 + π»2π Lost Vb HCl solution
ο° ππ = π
Μ 2 β 2π
Μ 1
3. P a g e | 3
- H2CO3 is a weak acid which have pKa1 = 6.35, pKa2 = 10.30. First stage titration only
have accuracy about 95%. With very short pH-jump-range, this titration should use
pT = 8.3 indicator.
ππΆπ3
2β =
ππ»πΆπ Γ π1
Μ
ππ πππππ
- With second stage, at the end point the titration produce H2CO3. If we donβt heat
system to push CO2, the pH-jump-range is about 4.35 Γ· 3.3, so that we can use
Methyl Orange indicator (pT = 4.0) to determine endpoint.
ππ»πΆπ3
β =
ππ»πΆπ Γ (π
Μ 2 β 2π
Μ 1)
ππ πππππ
b. Data and calculations:
Indicator 1st
2nd
3rd
π½
Μ π―πͺπ (ππ³)
pT = 8.3 6.80 6.75 6.75 6.767
Methyl Orange 13.80 13.9 13.95 13.883
β pT = 8.3
- Volume of sample solution: Vsample = 10.00 mL
- Concentration of HCl: ππ»πΆπ = 0.10204 Β± 0.00013 π
- Concentration of πΆπ3
2β
:
β ππΆπ3
2β =
ππ»πΆπΓπ1
Μ Μ Μ
ππ πππππ
=
0.10204Γ6.767
10.00
= 0.06905 (π)
β ππ1
= β
β (ππβπ
Μ )2
π
π=1
πβ1
= β
(6.80β6.767)2+(6.75β6.767)2+(6.75β6.767)2
3β1
= 0.0288704 (ππΏ)
β π’π
Μ 1
= πΜ π1
=
ππ1
β3
=
0.0288704
β3
= 0.016668 β¦ (ππΏ)
β π’π
Μ π πππππ
=
π’ππππππ‘π‘π
β3
=
πππππππ‘π‘π
β3 Γβ6
=
0.02
β3 Γβ6
= 0.004714 (mL)
β
π’π
πΆπ3
2β
ππΆπ3
2β
= β(
π’π
Μ 1
π
Μ 1
)
2
+ (
π’ππ»πΆπ
ππ»πΆπ
)
2
+ (
π’π
Μ
π πππππ
π
Μ π πππππ
)
2
=
β(
0.016668
6.767
)
2
+ (0.0029638)2 + (
0.004714
10.00
)
2
= 0.003882 β¦ (π)