Volumetric pipette Indicator Retort stand • Quantitative analysis - determine unknown conc of an analyte • Add known conc (std) from burette to unknown conc in flask. • Titrant/titrator (added) + analyte/titrand (to be analyzed) Conical flask Pipette filler Molarity- 0.1M Na 2 CO 3 Titration Titration Set up Burette Titrant/Titrator Conical flask Analyte/Titrand Notes/ sample titration calculation White tile Volumetric flask Standardization of (ACID) with standard (BASE) 10.6g Na 2 CO 3 10.6g in 1 L Burette Preparation of std (BASE) Preparation of std (ACID)
27
Embed
IB Chemistry Titration Techniques and IA on Titration
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Volumetric pipette
Indicator
Retort stand
• Quantitative analysis - determine unknown conc of an analyte • Add known conc (std) from burette to unknown conc in flask. • Titrant/titrator (added) + analyte/titrand (to be analyzed)
Conical flask
Pipette filler
Molarity- 0.1M Na2CO3
Titration
Titration Set up
Burette Titrant/Titrator
Conical flask
Analyte/Titrand
Notes/ sample titration calculation
White tile
Volumetric flask
Standardization of (ACID) with standard (BASE)
10.6g Na2CO3 10.6g in 1 L
Burette
Preparation of std (BASE) Preparation of std (ACID)
i. Find mass, of KIO3, required to prepare 0.250 dm3 of 0.002M KIO3
ii Titration results shown in table below Find % uncertainty in mean vol of KIO3 used.
Mean vol = (7.20 ± 0.10) cm3
Find amt of KIO3 used
Mol = M x V = 0.002 x 7.20 1000 = 1.44 x 10-5 mol
41000.5
250.0002.0
250.0002.0
.
mol
mol
mol
vol
molacidConc
Convert mole KIO3 → Mass/g
X RMM = 214.00
5.00 x 10-4 x 214.00 = 0.107 g
% ∆ vol = (0.10/7.20) x 100 % = 1.4 %
Find amt, Vit C in sample Find mass of Vit C
Convert mole Vit C → Mass
RMM Vit C – 176.14
M x 0.0292 = 2.5 x 10-3 acid M = 2.5 x 10-3 0.0292 M = 0.0856M
Acid/Base Titration– Ethanoic acid in vinegar
CH3COOH
M = ? V = 29.2ml
NaOH M = 0.1M V = 25.0ml
NaOH + CH3COOH → CH3COONa + H2O M = 0.1M M = ? V = 25ml V = 29.2ml
V = 250ml M = ?
Mole ratio (1 : 1) 1 mole NaOH - 1 mole acid
2.5 x 10-3 mole NaOH - 2.5 x 10-3 acid
Mole ratio – 1: 1
Diluted 10x
V = 25 ml M = ?
25ml of conc vinegar (ethanoic acid) was diluted to total vol 250 ml in a flask. Diluted vinegar was transfer to a burette. 25ml, 0.1M NaOH is pipette into a flask, with indicator added. End point reached when average 29.2 ml of
M1 = Ini molarity M2= Final molarity V1 = Ini vol V2 = Final vol
Mole NaOH = MV = (0.1 x 0.025) = 2.5 x 10-3
0856.01
1
0292.0
025.01.0
1
1
a
aa
bb
M
VM
VM
formula
MM
M
VMVM
856.0
2500856.025
1
1
2211
Acid/Base Titration - Empirical formula Na2CO3. x H2O
HCI M = 0.100 M V = 48.8ml
Na2CO3 M = ? M V = 25 ml
2HCI + Na2CO3 → 2NaCI + CO2 + H2O M = 0.1M M = ? V = 48.8ml V = 25.0ml
V = 1L
M = ?
25 ml transfer
Mole ratio – 2: 1
Mass Na2CO3 . x H2O = 27.82 g Mass Na2CO3 = 10.36 g Mass of water = (27.82 – 10.36) g = 17.46 g
Diuted to 1L
27.82g
Na2CO3. xH2O
27.82 g of hydrated (Na2CO3 . x H2O) dissolved in water, making up to 1L. 25 ml of sol was neutralized by 48.8ml of 0.1 M HCI. Cal molarity and mass of Na2CO3 present in 1L of sol. Find x
Convert mol dm-3 → g dm-3
Empirical formula
Na2CO3 H2O
Mass/g 10.36 17.46
RMM 106 18.02
Mole 10.34/106 = 0.09773
17.46/18.02 = 0.9689
Lowest ratio
0.09773/0.09733 1
0.9689/0.09733 10
Empirical formula Na2CO3 . 10 H2O
MM
VM
VM
b
bb
aa
0976.01
2
0250.0
0488.01.0
1
2
0.0976 x 106 = 10.36g/dm3
X RMM
Redox Titration - % Fe in iron tablet
Iron tablet contain FeSO4.7H2O. One tablet weigh 1.863 g crushed, dissolved in water and sol made up to 250 ml. 10ml of this sol added to 20 ml of H2SO4 and titrated with
0.002M KMnO4. Ave 24.5ml need to reach end point. Cal % Fe in iron tablet.
1.863 g
250ml
KMnO4 M = 0.002M V = 24.5 ml
Fe2+
M = ? V = 30ml
MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe2+ + 4H2O
M = 0.002M M = ? V = 24.5ml
Mole ratio – 1: 5
Mass (expt yield) = 1.703g Mass (Actual ) = 1.863g % Fe = 1.703 x 100% 1.863 = 91.4%
6.125 x 10-3 x 278.05 = 1.703 g FeSO4
10ml sol contain - 2.45 x 10-4 Fe2+ 250ml sol contain - 250 x 2.45 x 10-4 Fe2+ 10 = 6.125 x 10-3 mole Fe2+
42
2
1045.2.
5
1
.
0245.0002.0
5
1
FeMole
FeMole
VM
VM
bb
aa
Convert mole → Mass
X RMM
Mole bef dil = Mole aft dil M1 V1 = M2V2
M1 x 10 = 1.78 x 10-2 x 250 M1 = 1.78 x 10-2 x 250 10 M1 = 0.445M
2CIO- + 2I- + 2H+ → I2 + 2CI- + H2O
I2 + 2S2O32- → S4O6
2- + 2I-
10ml bleach (CIO-) diluted to a vol of 250 ml. 20ml is added to 1g of KI (excess) and iodine produced is titrated with 0.0206 M Na2S2O3.Using starch indicator, end point was 17.3ml.
M x V = Mol CIO- M x V = 3.56 x 10-4 M x 0.02 = 3.56 x 10-4 M = 3.56 x 10-4 002 M = 1.78 x 10-2 M diluted 25x
Diuted 25x
V = 10 M = ?
titrated
Water added
till 250ml
4
32
1056.3)..(
2
2
0173.00206.0
).(
2
2
)(
)(
CIOMole
CIOMole
OSMV
CIOMV
KIO3 + 5KI + 6H+ → 3I2 + 6K+ + 3H2O 3C6H8O6 + 3I2
→ 3C6H6O6 + 6I- + 6H+
Iodometric titration on Vit C, (C6H8O6). 25 ml Vit C titrated with 0.002M KIO3 from burette, using excess KI and starch. Ave vol KIO3 is 25.5ml. Cal molarity of Vit C.
M x V = 1.53 x 10-4 M x 0.025 = 3.56 x 10-4 M = 3.56 x 10-4 0025 M = 6.12 x 10-3 M
2.82 x 10-3 x 63.5 = 0.179 g Cu in 25ml 1.79 g Cu in 250ml
% Cu = mass Cu x 100% mass brass = 1.79 x 100% 2.5 = 71.8%
2Cu2+ + 4I- → I2 + 2CuI I2 + 2S2O3
2- → S4O62- + 2I-
2.5g brass react with 10ml HNO3 producing Cu2+ ion. Sol made up to 250ml using water. Pipette 25ml of sol to flask. Na2CO3 add to neutralize excess acid. 1g KI (excess) and starch added. Titrate with 0.1M S2O3
2- and end point, is 28.2 ml. Find molarity Cu 2+ and % Cu found in brass.
M x V = 2.85 x 10-3 M x 0.025 = 2.85 x 10-3 M = 2.85 x 10-3 0025 M = 1.14 x 10-3 M
Convert mole Cu → Mass Cu
2.85 x 10-3 x 63.5 = 0.18 g Cu
X RMM
% Cu = mass Cu x 100% mass brass = 0.18 x 100% 0.456 = 39.7 %
% Calcium carbonate in egg shell - Back Titration
250ml,
2M HNO3
Amt of HNO3 added
Amt of base (egg)
Amt of HNO3 left
Titrate NaOH M = 1.0 V = 17.0ml
Amt HNO3 react = Amt HNO3 – Amt HNO3 add left
HNO3 left
Transfer to flask
Left overnight in acid
added
25 g of egg shell (CaCO3) dissolved in 250 ml, 2 M HNO3. Sol titrated with NaOH. 17 ml, 1M NaOH needed to neutralize excess acid. Cal % CaCO3 by mass in egg shell.
NaOH + HNO3 → NaNO3 + H2O
M = 1.00M mol = ? V = 17 ml
Amt HNO3 add = M x V = 2.0 x 0.250 = 0.50 mol
Amt HNO3 react = Amt HNO3 add – Amt HNO3 left = 0.50 – 1.7 x 10-2 = 0.483 mol
% CaCO3 = mass CaCO3 x 100% mass egg = 0.153 x 100% 0.188 = 81.4 %
0.188g of egg shell (CaCO3) dissolved in 27.2 ml, 0.2 M HCI. Sol was titrated with NaOH. 23.8ml, 0.10M NaOH need to neutralize excess acid.
Cal mol, mass and % of CaCO3 by mass in egg shell.
0.188g impure CaCO3 in egg shell
27.20ml, 0.2M HCI
Amt of HCI added
Amt of base
Amt of HCI left
Titrate NaOH M = 0.1108 V = 33.64 ml
Amt HCI react = Amt HCI – Amt HCI add left
HCI left
Transfer
to flask
Left overnight in acid
added
NaOH + HCI → NaCI + H2O M = 0.1108 M mol = ? V = 33.64 ml
Amt HCI add = M x V = 0.250 x 0.05 = 0.0125 mol
Amt HCI react = Amt HCI add – Amt HCI left = 0.0125 – 3.727 x 10-3 = 0.008773 mol
2HCI + Ca(OH)3 → CaCI3 + H2O Mole Mole 0.008773 ?
Mole ratio (2 : 1) 2 mol HCI - 1 mol Ca(OH)2
0.008773 mol HCI - o.004386 mol Ca(OH)2
310727.3..
1
1
).(
03364.01108.0
1
1
acidMole
acidMole
VM
VM
aa
bb
Convert mole Ca(OH)2 → Mass /g
X RMM
0.004386 x 74.1 = 0.325g Ca(OH)2
% Ca(OH)2 = mass Ca(OH)2 x 100% mass impure = 0.325 x 100% 0.5214 = 62.3 %
50 ml, 0.250M HCI
% Calcium hydroxide in antacid tablet - Back Titration
0.5214 g of impure Ca(OH)2 from antacid was dissolved in 50 ml, 0.250M HCI. 33.64ml, 0.1108 M NaOH need to neutralize excess acid. Cal % Ca(OH)2 in tablet.
0.5214g impure Ca(OH)2
Amt of NaOH added
Amt of acid
Amt of NaOH left
Titrate HCI M = 0.5
V = 17.6 ml
Amt NaOH react = Amt NaOH – Amt NaOH add left
NaOH left
Transfer
to flask
Left overnight in acid
added
HCI + NaOH → NaCI + H2O M = 0.5 M mol = ? V = 17.6 ml
Amt NaOH add = M x V = 2 x 0.02 = 0.04 mol
Amt NaOH react = Amt NaOH add – Amt NaOH left = 0.04 – 8.8 x 10-3 = 0.0312 mol
2NaOH + H2A → Na3 A+ 2H2O Mole Mole 0.0312 ?
Mole ratio (2 : 1) 2 mol NaOH - 1 mol acid
0.0312 mol NaOH - 0.0156 mol acid
3108.8.
1
1
).(
0176.05.0
1
1
baseMole
acidMole
VM
VM
bb
aa
Molar mass of insoluble acid in tablet -Back Titration
2.04 g insoluble acid dissolve in 20 ml, 2M NaOH. Excess NaOH require 17.6 ml, 0.5M HCI to neutralize it. Find molar mass acid
2.04 g impure acid H2A
20 ml, 2M NaOH
Molar Mass Acid 0.0156 mol acid - 2.04 g 1 mol acid - 2.04 0.0156 = 131