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1. P a g e | 1
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 –
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
✓ Hydrolysis reaction: 𝐵4𝑂7
2−
+ 5𝐻2𝑂 ⇌ 2𝐻2𝐵𝑂3
−
+ 2𝐻3𝐵𝑂3
✓ Titration reaction: 2𝐻2𝐵𝑂3
−
+ 2𝐻+
→ 2𝐻3𝐵𝑂3
✓ Net reaction: 𝐵4𝑂7
2−
+ 5𝐻2𝑂 + 2𝐻+
→ 4𝐻3𝐵𝑂3
- This is neutralization reation of weak base 𝐻2𝐵𝑂3
−
by strong acid HCl. The
pH – jump – range is about 6.24 ÷ 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:
𝑁𝐻𝐶𝑙 =
𝑁𝐵𝑜𝑟𝑎𝑥 × 𝑉
̅𝐵𝑜𝑟𝑎𝑥
𝑉
̅𝐻𝐶𝑙
b. Practise:
- Phasing 𝐻𝐶𝑙~0.1𝑁 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 𝑁𝑎2𝐵4𝑂7 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (0.100000 ± 0.000015𝑁) 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:
𝑉𝐻𝐶𝑙(𝑚𝐿)
𝑉
̅𝐻𝐶𝑙 (𝑚𝐿)
1𝑠𝑡
2𝑛𝑑
3𝑟𝑑

2. P a g e | 2
9.75 9.80 9.85 9.80
- Cocentration of HCl:
✓ 𝑁𝐻𝐶𝑙 =
𝑁𝐵𝑜𝑟𝑎𝑥×𝑉
̅𝐵𝑜𝑟𝑎𝑥
𝑉
̅𝐻𝐶𝑙
=
0.100000×10.00
9.8
= 0.102041𝑁
✓ 𝑆𝑉𝐻𝐶𝑙
= √
∑ (𝑉𝑖−𝑉
̅)2
𝑛
𝑖=1
𝑛−1
= √
(9.75−9.80)2+(9.80−9.80)2+(9.85−9.80)2
3−1
= 0.05 (𝑚𝐿)
✓ 𝑢𝑉
̅𝐻𝐶𝑙
= 𝑆̅𝑉𝐻𝐶𝑙
=
𝑆𝑉𝐻𝐶𝑙
√3
=
0.05
√3
= 0.0286675 … 𝑚𝐿
✓ 𝑢𝑉
̅𝑏𝑜𝑟𝑎𝑥
=
𝑢𝑉𝑝𝑖𝑝𝑒𝑡𝑡𝑒
√3
=
𝑎𝑉𝑝𝑖𝑝𝑒𝑡𝑡𝑒
√3 ×√6
=
0.02
√3 ×√6
= 0.004714 (mL)
✓ 𝑢𝑁𝑏𝑜𝑟𝑎𝑥
=
𝑈𝑉𝑏𝑜𝑟𝑎𝑥
2
=
0.000015
2
= 0.0000075(𝑁)
✓
𝑢𝑁𝐻𝐶𝑙
𝑁𝐻𝐶𝑙
= √(
𝑢𝑉
̅
𝐻𝐶𝑙
𝑉
̅𝐻𝐶𝑙
)
2
+ (
𝑢𝑁𝑏𝑜𝑟𝑎𝑥
𝑁𝑏𝑜𝑟𝑎𝑥
)
2
+ (
𝑢𝑉
̅
𝑏𝑜𝑟𝑎𝑥
𝑉
̅𝑏𝑜𝑟𝑎𝑥
)
2
= √(
0.02866675
9.80
)
2
+ (
0.004714
10
)
2
+ (
0.0000075
0.100000
)
2
= 0.0029638 … (𝑁)
→ 𝑢𝑁𝐻𝐶𝑙
= 0.000302436𝑁
𝑈𝑁𝐻𝐶𝑙
= 4.3 × 0.000302436 = 0.00013(𝑁)
 𝑾𝒊𝒕𝒉 𝑷 = 𝟎. 𝟗𝟓, 𝑵𝑯𝑪𝒍 = 𝟎. 𝟏𝟎𝟐𝟎𝟒 ± 𝟎. 𝟎𝟎𝟎𝟏𝟑 𝑵
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 𝐶𝑂3
2−
, spend V1 mL:
𝐶𝑂3
2−
+ 𝐻+
→ 𝐻𝐶𝑂3
−
Lost Va HCl solution
- Second stage: titrate all 𝐶𝑂3
2−
𝑎𝑛𝑑 𝐻𝐶𝑂3
−
, spend 𝑉
̅2 mL:
𝐶𝑂3
2−
+ 2𝐻+
→ 𝐻2𝐶𝑂3 → 𝐶𝑂2 + 𝐻2𝑂 Lost 2Va HCl solution
𝐻𝐶𝑂3
−
+ 𝐻+
→ 𝐻2𝐶𝑂3 → 𝐶𝑂2 + 𝐻2𝑂 Lost Vb HCl solution
 𝑉𝑏 = 𝑉
̅2 − 2𝑉
̅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.
𝑁𝐶𝑂3
2− =
𝑁𝐻𝐶𝑙 × 𝑉1
̅
𝑉𝑠𝑎𝑚𝑝𝑙𝑒
- With second stage, at the end point the titration produce H2CO3. If we don’t heat
system to push CO2, the pH-jump-range is about 4.35 ÷ 3.3, so that we can use
Methyl Orange indicator (pT = 4.0) to determine endpoint.
𝑁𝐻𝐶𝑂3
− =
𝑁𝐻𝐶𝑙 × (𝑉
̅2 − 2𝑉
̅1)
𝑉𝑠𝑎𝑚𝑝𝑙𝑒
b. Data and calculations:
Indicator 1st
2nd
3rd
𝑽
̅𝑯𝑪𝒍 (𝒎𝑳)
pT = 8.3 6.80 6.75 6.75 6.767
Methyl Orange 13.80 13.9 13.95 13.883
❖ pT = 8.3
- Volume of sample solution: Vsample = 10.00 mL
- Concentration of HCl: 𝑁𝐻𝐶𝑙 = 0.10204 ± 0.00013 𝑁
- Concentration of 𝐶𝑂3
2−
:
✓ 𝑁𝐶𝑂3
2− =
𝑁𝐻𝐶𝑙×𝑉1
̅̅̅
𝑉𝑠𝑎𝑚𝑝𝑙𝑒
=
0.10204×6.767
10.00
= 0.06905 (𝑁)
✓ 𝑆𝑉1
= √
∑ (𝑉𝑖−𝑉
̅)2
𝑛
𝑖=1
𝑛−1
= √
(6.80−6.767)2+(6.75−6.767)2+(6.75−6.767)2
3−1
= 0.0288704 (𝑚𝐿)
→ 𝑢𝑉
̅1
= 𝑆̅𝑉1
=
𝑆𝑉1
√3
=
0.0288704
√3
= 0.016668 … (𝑚𝐿)
✓ 𝑢𝑉
̅𝑠𝑎𝑚𝑝𝑙𝑒
=
𝑢𝑉𝑝𝑖𝑝𝑒𝑡𝑡𝑒
√3
=
𝑎𝑉𝑝𝑖𝑝𝑒𝑡𝑡𝑒
√3 ×√6
=
0.02
√3 ×√6
= 0.004714 (mL)
✓
𝑢𝑁
𝐶𝑂3
2−
𝑁𝐶𝑂3
2−
= √(
𝑢𝑉
̅1
𝑉
̅1
)
2
+ (
𝑢𝑁𝐻𝐶𝑙
𝑁𝐻𝐶𝑙
)
2
+ (
𝑢𝑉
̅
𝑠𝑎𝑚𝑝𝑙𝑒
𝑉
̅𝑠𝑎𝑚𝑝𝑙𝑒
)
2
=
√(
0.016668
6.767
)
2
+ (0.0029638)2 + (
0.004714
10.00
)
2
= 0.003882 … (𝑁)

4. P a g e | 4
→ 𝑢𝑁𝐶𝑂3
2−
= 0.0002681 𝑁
✓ 𝑈𝑁𝐶𝑂3
2− = 4.3 × 0.0002681 = 0.001153 (𝑁)
 𝑾𝒊𝒕𝒉 𝑷 = 𝟎. 𝟗𝟓, 𝑵𝑪𝑶𝟑
𝟐− = 𝟎. 𝟎𝟔𝟗𝟎𝟓 ± 𝟎. 𝟎𝟎𝟏𝟏𝟓 𝑵
❖ Methyl Orange:
- Volume of sample solution: Vsample = 10.00 mL
- Concentration of HCl: 𝑁𝐻𝐶𝑙 = 0.10204 ± 0.00013 𝑁
- Concentration of 𝐻𝐶𝑂3
−
:
✓ 𝑁𝐻𝐶𝑂3
− =
𝑁𝐻𝐶𝑙×(𝑉
̅2−2𝑉
̅1)
𝑉𝑠𝑎𝑚𝑝𝑙𝑒
=
0.10204×(13.883−2×6.767)
10.00
= 0.003561 (𝑁)
✓ 𝑆𝑉2
= √
∑ (𝑉𝑖−𝑉
̅)2
𝑛
𝑖=1
𝑛−1
= √
(13.80−13.883)2+(13.9−13.883)2+(13.95−13.883)2
3−1
=
0.076377 (𝑚𝐿)
 𝑢𝑉
̅2
= 𝑆̅𝑉2
=
𝑆𝑉2
√3
=
0.076377
√3
= 0.044096 … (𝑚𝐿)
 √(𝑢𝑉
̅1
)
2
+ (2𝑢𝑉
̅2
)
2
= √0.0166682 + 0.0440962 = 0.047141 … (𝑚𝐿)

𝑢𝑁𝐻𝐶𝑂3
−
𝑁𝐻𝐶𝑂3
−
= √(
√(𝑢𝑉
̅1
)
2
+(2𝑢𝑉
̅2
)
2
𝑉
̅2−2𝑉
̅1
)
2
+ (
𝑢𝑁𝐻𝐶𝑙
𝑁𝐻𝐶𝑙
)
2
+ (
𝑢𝑉
̅
𝑠𝑎𝑚𝑝𝑙𝑒
𝑉
̅𝑠𝑎𝑚𝑝𝑙𝑒
)
2
=
√(
0.047141
13.883−2×6.767
)
2
+ (0.0029638)2 + (
0.004714
10.00
)
2
= 0.08906 … (𝑁)
→ 𝑢𝑁𝐻𝐶𝑂3
− = 0.0003172 𝑁
✓ 𝑈𝑁𝐻𝐶𝑂3
− = 4.3 × 0.0003172 = 0.00136 (𝑁)
 𝑾𝒊𝒕𝒉 𝑷 = 𝟎. 𝟗𝟓, 𝑵𝑯𝑪𝑶𝟑
− = 𝟎. 𝟎𝟎𝟑𝟓𝟔 ± 𝟎. 𝟎𝟎𝟏𝟑𝟔 𝑵