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Volumetric pipette
Indicator
Retort stand
• Quantitativeanalysis - determineunknownconc of an analyte
• Add knownconc (std)from buretteto unknownconc in flask.
• Titrant/titrator(added)+ analyte/titrand(to be analyzed)
Conical flask
Pipette filler
Molarity- 0.1M Na2CO3
Titration
TitrationSet up
Burette
Titrant/Titrator
Conical flask
Analyte/Titrand
Notes/ sample titration calculation
White tile
Volumetric flask
Standardizationof (ACID)with standard(BASE)
10.6g Na2CO3 10.6g in 1 L
Burette
Preparation of std (BASE) Preparation of std (ACID)
Titration
Redox Titration Acid Base Titration Complexometric titration
Neutralization
Indicator
pH sensor Conductometric
Changes colour (end point) at equivalent point
Indicator pKa pH
range
Colour
Acid
Colour
Base
Methyl orange 3.46 3.2- 4.4 RED Yellow
Bromophenol Blue 4.10 3.0- 4.6 Yellow Blue
Bromocresol Green 4.90 3.8- 5.4 Yellow Blue
Methyl Red 5.00 4.8- 6.0 Red Yellow
Bromothymol Blue 7.30 6.0- 7.6 Yellow Blue
Phenol Red 8.00 6.6- 8.2 Yellow Red
Phenolphthalein 9.50 8.2-
10.0
Colourless Pink
Video on pH sensor Video on indicator
Video on conductometric
Video on conductometric
3
1
62
4
5
TitrationSteps
Rinse burette – titrant
Bottom fill with titrant NOT air gap Fix it on retort stand
read (Bottom of meniscus)
Record vol to nearest 0.01 ml
9
8 7
Final vol to nearest 0.01 ml
Click here to view video
Rinse burette – deionized water Add dropwise – end point is near
Titration
Redox Titration Acid Base Titration Complexometric titration
Neutralization
Condition for Acid/Alkali Titration
One reactant – must be std (known conc) or capable being standardised
Equivalentpoint – equal amt neutralize each other
End point detectable by colour change, pH change /conductivity
Acid/Base as primary standard
-Stable/solid
-Soluble in water
-Does not decompose over time
Primary std acid
- Potassium hydrogen phthalate
Primary std base
- Anhydrous sodium carbonate
Standard 0.1M Na2CO3
10.6g in 1 L
Volumetric Burette
Unable to prepare accurate conc of NaOH/HCI
•hygroscopic nature NaOH – absorb water vapour
• HCI is in vapour state – diff to measure amt
VolumetricBurette
Standard 0.1M KHP
20.4 g KHP
20.4 g in 1L
Unknown
Conc NaOH
Unknown
Conc HCI
? ?
10.6 g Na2CO3
0.1M KHP – 0.1 mole of KHP in total vol of solution (1L)
Primary standard acid
Mass of KHP → 0.1 mole KHP x M = 0.1 x 204.22gStep 1
Step 2
Pour to 1L volumetricflaskStep 3
Add water until1L mark
Transfer to beaker,add water to dissolve
Step 4
Step 5
20.4 g
Video std solutionpreparation
Molarity = 0. 1 mole
(0.1M) 1 L total vol (solute + solvent)
Primary standard
Preparing std solution – 0.1 M – 0.1 mole KHP in 1 L
0.1M Na2CO3 – 0.1 mole of Na2CO3 in total vol of solution (1L)
Primary standard base
Mass of Na2CO3 → 0.1 mole Na2CO3 x M = 0.1 x 106 gStep 1
Step 2
Pour to 1L volumetricflaskStep 3
Add water until1L mark
Transfer to beaker,add water to dissolve
Step 4
Step 5
10.6 g
Video std solutionpreparation
Molarity = 0. 1 mole
(0.1M) 1 L total vol (solute + solvent)
Primary standard
Preparing std solution – 0.1 M – 0.1 mole Na2CO3 in 1 L
Pipette25 ml NaOH in conical flask
Step 2
Fill burettewith std 0.1M KHP sol
Step 3
Step 4
Step 5
2 dropphenolthalein– colourlessto pink
Ini vol KHP recorded
Titrate until pink colourfadesaway
Final vol KHP recordedStep 6
Step 7 Repeat till consistentresult agree within0.1 cm3 (triplicate)
Standardization (ACID)with standard(BASE)
? Conc NaOH
Standardization(BASE)with std (ACID)
Step 1
Step 2
Step 3
Step 4
Step 5
Step 6
Step 7
Fill burettewith std 0.1M Na2CO3 sol
Ini vol Na2CO3 recorded
Pipette25 ml HCI in conicalflask
2 dropmethyl orangeto HCI – red
Titrate until red changesto pink
Final vol Na2CO3 recorded
Repeat till consistentresult agree within0.1 cm3 (triplicate)
Std 0.1M Na2CO3
? Conc HCI
Step 1 Std 0.1M KHP
Standardizationof (BASE)with std (ACID)
Vol KHP
Fin vol = 29.50 ± 0.05
Ini vol = 3.10 ± 0.05
Vol = 26.4 ± 0.1
Uncertainty invol
Add abs ∆ for final + ini
= (0.05 + 0.05) = ± 0.1
Ave Vol KHP ± uncertainty
= 26.4 + 26.4 + 26.4
3
= (26.4± 0.1) cm3
Conc NaOH = 0.106 ± 0.5%
0.5 x 0.106
100
= 0.001
Conc NaOH = 0.106 ± 0.001M
Lit value - NaOH = 0.100M
Expt value– NaOH = 0.106M
Difference = 0.006
% Error – Difference x 100%
Lit value
0.006 x 100 % = 6%
0.100
Data ProcessingData collection
Vol KHP /cm3
Titration trials
1 2 3
Final
vol/(±0.05)cm3
38.50 29.50 45.90
Ini vol/ (±0.05)cm3 12.10 3.10 19.50
Total vol/ (±0.1)cm
3
26.4 26.4 26.4
NaOH + KHP → NaKP + H2O
M = ? M = 0.1
V = 25ml V = 26.4 ml
M
M
VM
VM
b
aa
bb
106.0
1
1
0264.01.0
025.0
1
1




Error Analysis
% ∆ - burette % ∆ - pipette
Abs ∆ vol x 100%
Ave vol
= 0.1 x 100%
26.4
= 0.38%
Abs ∆ vol x 100%
vol
= 0.03 x 100%
25.00
= 0.12 %
Total % ∆ = % ∆ burette+ %∆ pipette
= 0.38% + 0.12%
= 0.5%
Data Processing
% uncertainty
(Abs uncertainty)
Acid (KHP)
Base (NaOH)
Standardizationof (ACID)with std (BASE)
Vol Na2CO3
Fin vol = 29.50 ± 0.05
Ini vol = 3.10 ± 0.05
Vol = 26.4 ± 0.1
Uncertainty invol
Add abs ∆ for final + ini
= (0.05 + 0.05) = ± 0.1
Ave Vol Na2CO3 ± uncertainty
= 26.4 + 26.4 + 26.4
3
= (26.4 ± 0.1) cm3
Conc HCI = 0.2112 ± 0.5%
0.5 x 0.2112
100
= 0.001
Conc HCI = 0.2112 ± 0.001M
Lit value - HCI = 0.200M
Expt value – HCI = 0.2112M
Difference = 0.011
% Error – Difference x 100%
Lit value
0.011 x 100 % = 6%
0.200
Data ProcessingData collection
Vol Na2CO3 /cm3
Titration trials
1 2 3
Final
vol/(±0.05)cm3
38.50 29.50 45.90
Ini vol/ (±0.05)cm3 12.10 3.10 19.50
Total vol/ (±0.1)cm
3
26.4 26.4 26.4
Na2CO3 + 2HCI → 2NaCI + H2O + CO2
M = 0.1 M = ?
V = 26.4 ml V = 25.0 ml
M
M
VM
VM
a
aa
bb
2112.0
2
1
0250.0
0264.01.0
2
1




Error Analysis
% ∆ - burette % ∆ - pipette
Abs ∆ vol x 100%
Ave vol
= 0.1 x 100%
26.4
= 0.38%
Abs ∆ vol x 100%
vol
= 0.03 x 100%
25.00
= 0.12 %
Total % ∆ = % ∆ burette+ %∆ pipette
= 0.38% + 0.12%
= 0.5%
Data Processing
% uncertainty
(Abs uncertainty)
Base (Na2CO3)
Acid (HCI)
NaOH
M = ?
V = 25.0ml
KHP
M = 0.100M
V = 26.4 ml
HCI
M = ?
V = 25.0ml
Sample TitrationCalculation
Na2CO3
M = 0.100M
V = 26.4 ml
Standardizationof (BASE)with std (ACID) Standardizationof (ACID)with std (BASE)
KHP + NaOH → NaKP + H2O
M = 0.100 M = ?
V = 26.40 ml V = 25.0 ml
M
M
VM
VM
b
bb
aa
106.0
1
1
0250.0
0264.01.0
1
1




Na2CO3 + 2HCI → 2NaCI + H2O + CO2
M = 0.100 M = ?
V = 26.4 ml V = 25.0 ml
M
M
VM
VM
a
aa
bb
2112.0
2
1
0250.0
0264.01.0
2
1




Click here titration simulation Click here titration simulation Click here titration simulation Click here titration simulation
Simulation on Titration
Sample TitrationCalculation
M
M
VM
VM
b
aa
bb
12.2
1
2
5.026.00.1
0250.0
1
2




3
20
1
2
305.0
5.1
1
2
cm
V
VM
VM
b
aa
bb




25 ml NaOH require 26.5cm3 of 1.0M H2SO4
for neutralization.
Find its molarity of NaOH.
Find vol of 1.5M NH3 required to neutralize
30 ml of 0.5M H2SO4
H2SO4
M = 1 M
V = 26.5 ml
NaOH
M = ?
V = 25 ml
2NaOH + H2SO4 → Na2SO4 + 2H2O
M = ? M = 1M
V = 25.0ml V = 26.5ml
H2SO4
M = 0.5M
V = 30ml
NH4OH
M = 1.5M
V = ? ml
2NH4OH + H2SO4 → (NH4)2SO4 + 2H2O
M = 1.5M M = 0.5M
V = ? ml V = 30.0ml
Simulation on Titration
Click here titration simulation Click here titration simulation Click here titration simulation Click here titration simulation
Sample TitrationCalculation
3
25
2
1
0.2
505.0
2
1
cm
V
VM
VM
a
aa
bb




M
M
VM
VM
a
aa
bb
16.0
2
1
25
102.0
2
1




Simulation on Titration
Click here titration simulation Click here titration simulation Click here titration simulation Click here titration simulation
Find vol of 2.0M HCI needed to neutralize 0.5M
Na2CO3 in 50ml water.
HCI
M = 2.0M
V = ? ml
Na2CO3
M = 0.5M
V = 50ml
Na2CO3 + 2HCI → 2NaCI + CO2 + H2O
M = 0.5M M = 2.0M
V = 50ml V = ? ml
Na2CO3
M = 0.2M
V = 10.0ml
HCI
M = ?
V = 25 ml
10 cm3 of 0.2 M Na2CO3 need 25 cm3 of HCI for
neutralization. Find molarity of HCI.
Na2CO3 + 2HCI → 2NaCI + H2O + CO2
M = 0.2M M = ?
V = 10 ml V = 25 ml
Titration for IA assessment
Acid Base Titration
StandardizationHCI
with std Na2CO3
StandardizationNaOH
with std KHP
Titrationbet NaOH
with std HCI
Titrationbet HCI
with std NaOH
Determinewater crystallization
in hydratedNa2CO3 with std HCI
StandardizationKMnO4
with std ammonium
iron(II)sulphate
Fe 2+ in iron pill
with std KMnO4
Hypochlorite(OCI-) in bleach
with iodine/thiosulphate
Determineethanoicacid
in vinegar
Cu2+ in brass
with iodine/thiosulphate
StandardizationKI/I2
with std KIO3
Determineacetylsalicylic
acid in aspirin
Vit C in fruitswith
iodine/thiosulphate
StandardizationExpt Acid/Base Expt
StandardizationExpt
Redox Expt
Redox Titration
StandardizationKI/I2
with std
sodium thiosulphate
Iodine/thiosulphate(iodometric titration)
Equilibrium established when ethanoic acid and ethanol react together
in strong acid, using propanone as solvent. Eqn given.
CH3COOH + C2H5OH ↔ CH3COOC2H5 + H2O
Density ethanoic acid is 1.05 g cm–3.
i. Find amt, mol, of acid present
ii. Conc acid is 1.748 mol dm–3. Find % uncertainty of conc.
Titration performed on acid using a base. Result shown below
Find absolute uncertainty of titre for Titration 1 (27.60 cm3).
Liquid Vol/cm3
Ethanoic acid 5.00 ± 0.05
Ethanol 5.00 ± 0.05
Hydrochloric acid 1.00 ± 0.02
Propanone 39.0 ± 0.5
gmass
voldenstiymass
25.500.505.1 

vol
mass
Density 
molMol
RMM
mass
Mol
0874.0
60
25.5


RMM acid = 60
% uncertaintyacid conc = % uncertaintyin vol acid + % uncertaintyin total vol
vol
mol
acidConc .
(0.05/5.00) x 100 %
= 1 %
(0.62/50) x 100%
= 1.24 %
Total % uncertainty = (1 + 1.24) %
= 2.24%
Uncertaintyfinal – initial vol
(28.80 ±0.05 – 1.20 ±0.05 )
= (27.60 ± 0.1)
Add absolute uncertaintytogether
Two rxn kinetic investigated using iodine clock rxn.
Reaction A: H2O2 + 2I− + 2H+ → I2 + 2H2 O
Reaction B: I2 + 2S2O3
2− → 2I− + S4O6
2-
i. Find total uncertainty, in vol of rxn mixture
Mixture contained:
5.0 ± 0.1 cm3 of 2M H2O2
5.0 ± 0.1 cm3 of 1 % starch
20.0 ± 0.1 cm3 of 1M H2SO4
20.0 ± 0.1 cm3 of 0.01 M Na2S2O3
50.0 ± 0.1 cm3 of water with 0.0200 ± 0.0001 g KI
i. Add all vol together:Add all absolute uncertaintytogether.
(5.0 ± 0.1)+ (5.0 ± 0.1) + (20.0 ± 0.1)+ (20.0 ± 0.1 ) + (50.0 ± 0.1)
= (100 ± 0.5) cm 3
ii. Conc KI =Mass/ vol
% uncertainty conc KI = % uncertaintymass + % uncertainty vol KI
% ∆ mass = (0.0001/0.02) x 100% = 0.5 %
% ∆ vol = (0.1/50) x 100% = 0.2 %
% conc KI = (0.5 + 0.2)% = 0.7 %
iii. Final Conc KI = Conc KI in total mixture
ii. Find % uncertainty for KI conc in final rxn sol.
iii. Find % uncertainty for KI conc in overall rxn mixture
% ∆ final conc KI = % ∆ conc KI + % ∆ total vol KI
% ∆ conc KI
= 0.7 %
% ∆ total vol
= (0.5/100) x 100 %
= 0.5%
% conc KI
= (0.5 + 0.7)
= 1.2 %
Mixture contained:
5.0 ± 0.1 cm3 of 2M H2O2
5.0 ± 0.1 cm3 of 1 % starch
20.0 ± 0.1 cm3 of 1M H2SO4
20.0 ± 0.1 cm3 of 0.01 M Na2S2O3
50.0 ± 0.1 cm3 of water with 0.0200 ± 0.0001 g KI
total vol
Only vol/mass/conc KI
4.32 x 10-5 x 176.14
= 7.61 x10-3 g Vit C
KIO3 + 5KI + 6H+ → 3I2 + 6K+ + 3H2O
3C6H8O6 + 3I2 → 3C6H6O6 + 6I- + 6H+
Iodometric titration on Vit C, (C6H8O6).
Vit C titrated with 0.002M KIO3 , using excess KI and starch.
Redox Titration – Vit C quantification
KIO3
M = 0.002M
Vit C
Amt = ?
Mole ratio (1 :3)
1 mol KIO3 : 3 mol I2 : 3 mol C6H8O6
1 mol KIO3 3 mol C6H8O6
Amt = ?
transfer
1g KI excess + starch
titrated
Vit C
5
3
1032.4)..(
3
1
)..(
0072.0002.0
3
1
).(
)(





CVitMole
CVitMole
CVitMV
KIOMV
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
4
1000.5
250.0002.0
250.0
002.0
.





mol
mol
mol
vol
mol
acidConc
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
diluted vinegar added. Find its molarity.
mole ratio
Moles bef dilution = Moles aft dilution
M1 V1 = M2V2
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
0856.0
1
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
M
M
VM
VM
b
bb
aa
0976.0
1
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 + 2S2O3
2- → 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.
Cal molarity of CIO in bleach.
Redox Titration – CIO- in Bleach
Na2S2O3
M = 0.0206M
V = 17.3ml
I2
M = ?
Mole ratio ( 1 : 1)
2 mole CIO- : 1 mole I2 : 2 mole S2O3
2-
2 mole CIO- 2 mole S2O3
2-
10.0ml CIO-
transfer
V = 250ml
M = 1.78 x 10-2 M
20ml transfer
1g KI excess
added
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.
Redox Titration – Vit C quantification
KIO3
M = 0.002M
V = 25.5ml
Vit C
M = ?
V = 25ml
Mole ratio (1 :3)
1 mol KIO3 : 3 mol I2 : 3 mol C6H8O6
1 mol KIO3 3 mol C6H8O6
V = 25ml
M = ?
25ml transfer
1g KI excess + starch
titrated
Vit C
4
3
1053.1)..(
3
1
)..(
0255.0002.0
3
1
).(
)(





CVitMole
CVitMole
CVitMV
KIOMV M x V = Mol 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- → S4O6
2- + 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.
Redox Titration - % Cu in Brass
Na2S2O3
M = 0.1M
V = 28.2ml
I2
M = ?
Mole ratio (1 : 1)
2 mol Cu2+ : 1 mol I2 : 2 mol S2O3
2-
2 mol Cu2+ 2 mol S2O3
2-
Pour into
Volumetric flask
V = 250ml
M = ?
25ml transfer
1g KI excess/ starch
10 ml HNO3
titrated
Water added 250ml
2.5g brass
32
2
2
32
2
1082.2).(
2
2
0282.01.0
).(
2
2
)(
)(








CuMole
CuMole
OSMV
CuMV
M x V = Mol Cu 2+
M x V = 2.82 x 10-3
M x 0.025 = 2.82 x 10-3
M = 2.82 x 10-3
0025
M = 1.13 x 10-3 M
Convert mole Cu→ Mass Cu
X RMM
X 10
2Cu2+ + 4I- → I2 + 2CuI
I2 + 2S2O3
2- → S4O6
2- + 2I-
0.456 g brass react with 25ml HNO3 producing Cu2+ ions. Sol was titrate with 0.1M S2O3
2-
and end point, reached when 28.5ml added. Cal mol, mass, molarity of Cu 2+ and % Cu in brass.
Redox Titration - % Cu in Brass
Na2S2O3
M = 0.1M
V = 28.5ml
I2
M = ?
Mole ratio (1 : 1)
2 mol Cu2+ : 1 mol I2 : 2 mol S2O3
2-
2 mol Cu2+ 2 mol S2O3
2-
transfer
1g KI excess starch
25ml HNO3
titrated
0.456g
brass
32
2
2
32
2
1085.2).(
2
2
0285.01.0
).(
2
2
)(
)(








CuMole
CuMole
OSMV
CuMV
M x V = Mol Cu 2+
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
2HNO3 + CaCO3 → (CaNO3)2 + H2O + 2CO2
Mole Mole
0.483 ?
Mole ratio (2 : 1)
2 mol HNO3 - 1 mol CaCO3
0.483 mol HNO3 - o.242 molCaCO3
2
107.1..
1
1
).(
017.000.1
1
1





acidMole
acidMole
VM
VM
aa
bb
25 g impure
CaCO3 in egg shell
Convert mole CaCO3 → Mass /g
X RMM
0.242 x 100 = 24.2 g CaCO3
% CaCO3 = mass CaCO3 x 100%
mass egg
= 24.2 x 100%
25.0
= 96.8 %
% Calcium carbonate in egg shell - Back Titration
Amt of HCI added
Amt of base (egg)
Amt of
HCI left
Titrate NaOH
M = 0.10
V = 23.8 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 = 0.1 M mol = ?
V = 23.8 ml
Amt HCI add = M x V
= 0.2 x 0.272
= 0.0544 mol
Amt HCI react = Amt HCI add – Amt HCI left
= 0.0544 – 2.38 x 10-3
= 0.00306 mol
2HCI + CaCO3 → CaCI3 + H2O + CO2
Mole Mole
0.00306 ?
Mole ratio (2 : 1)
2 mol HCI - 1 mol CaCO3
0.00306 mol HCI - o.00153 molCaCO3
3
1038.2..
1
1
).(
238.01.0
1
1





acidMole
acidMole
VM
VM
aa
bb
Convert mole CaCO3 → Mass /g
X RMM
0.00153 x 100 = 0.153 g CaCO3
% 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 molCa(OH)2
3
10727.3..
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
3
108.8.
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

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IB Chemistry Titration Techniques and IA on Titration

  • 1. Volumetric pipette Indicator Retort stand • Quantitativeanalysis - determineunknownconc of an analyte • Add knownconc (std)from buretteto unknownconc in flask. • Titrant/titrator(added)+ analyte/titrand(to be analyzed) Conical flask Pipette filler Molarity- 0.1M Na2CO3 Titration TitrationSet up Burette Titrant/Titrator Conical flask Analyte/Titrand Notes/ sample titration calculation White tile Volumetric flask Standardizationof (ACID)with standard(BASE) 10.6g Na2CO3 10.6g in 1 L Burette Preparation of std (BASE) Preparation of std (ACID)
  • 2. Titration Redox Titration Acid Base Titration Complexometric titration Neutralization Indicator pH sensor Conductometric Changes colour (end point) at equivalent point Indicator pKa pH range Colour Acid Colour Base Methyl orange 3.46 3.2- 4.4 RED Yellow Bromophenol Blue 4.10 3.0- 4.6 Yellow Blue Bromocresol Green 4.90 3.8- 5.4 Yellow Blue Methyl Red 5.00 4.8- 6.0 Red Yellow Bromothymol Blue 7.30 6.0- 7.6 Yellow Blue Phenol Red 8.00 6.6- 8.2 Yellow Red Phenolphthalein 9.50 8.2- 10.0 Colourless Pink Video on pH sensor Video on indicator Video on conductometric Video on conductometric
  • 3. 3 1 62 4 5 TitrationSteps Rinse burette – titrant Bottom fill with titrant NOT air gap Fix it on retort stand read (Bottom of meniscus) Record vol to nearest 0.01 ml 9 8 7 Final vol to nearest 0.01 ml Click here to view video Rinse burette – deionized water Add dropwise – end point is near
  • 4. Titration Redox Titration Acid Base Titration Complexometric titration Neutralization Condition for Acid/Alkali Titration One reactant – must be std (known conc) or capable being standardised Equivalentpoint – equal amt neutralize each other End point detectable by colour change, pH change /conductivity Acid/Base as primary standard -Stable/solid -Soluble in water -Does not decompose over time Primary std acid - Potassium hydrogen phthalate Primary std base - Anhydrous sodium carbonate Standard 0.1M Na2CO3 10.6g in 1 L Volumetric Burette Unable to prepare accurate conc of NaOH/HCI •hygroscopic nature NaOH – absorb water vapour • HCI is in vapour state – diff to measure amt VolumetricBurette Standard 0.1M KHP 20.4 g KHP 20.4 g in 1L Unknown Conc NaOH Unknown Conc HCI ? ? 10.6 g Na2CO3
  • 5. 0.1M KHP – 0.1 mole of KHP in total vol of solution (1L) Primary standard acid Mass of KHP → 0.1 mole KHP x M = 0.1 x 204.22gStep 1 Step 2 Pour to 1L volumetricflaskStep 3 Add water until1L mark Transfer to beaker,add water to dissolve Step 4 Step 5 20.4 g Video std solutionpreparation Molarity = 0. 1 mole (0.1M) 1 L total vol (solute + solvent) Primary standard Preparing std solution – 0.1 M – 0.1 mole KHP in 1 L
  • 6. 0.1M Na2CO3 – 0.1 mole of Na2CO3 in total vol of solution (1L) Primary standard base Mass of Na2CO3 → 0.1 mole Na2CO3 x M = 0.1 x 106 gStep 1 Step 2 Pour to 1L volumetricflaskStep 3 Add water until1L mark Transfer to beaker,add water to dissolve Step 4 Step 5 10.6 g Video std solutionpreparation Molarity = 0. 1 mole (0.1M) 1 L total vol (solute + solvent) Primary standard Preparing std solution – 0.1 M – 0.1 mole Na2CO3 in 1 L
  • 7. Pipette25 ml NaOH in conical flask Step 2 Fill burettewith std 0.1M KHP sol Step 3 Step 4 Step 5 2 dropphenolthalein– colourlessto pink Ini vol KHP recorded Titrate until pink colourfadesaway Final vol KHP recordedStep 6 Step 7 Repeat till consistentresult agree within0.1 cm3 (triplicate) Standardization (ACID)with standard(BASE) ? Conc NaOH Standardization(BASE)with std (ACID) Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Fill burettewith std 0.1M Na2CO3 sol Ini vol Na2CO3 recorded Pipette25 ml HCI in conicalflask 2 dropmethyl orangeto HCI – red Titrate until red changesto pink Final vol Na2CO3 recorded Repeat till consistentresult agree within0.1 cm3 (triplicate) Std 0.1M Na2CO3 ? Conc HCI Step 1 Std 0.1M KHP
  • 8. Standardizationof (BASE)with std (ACID) Vol KHP Fin vol = 29.50 ± 0.05 Ini vol = 3.10 ± 0.05 Vol = 26.4 ± 0.1 Uncertainty invol Add abs ∆ for final + ini = (0.05 + 0.05) = ± 0.1 Ave Vol KHP ± uncertainty = 26.4 + 26.4 + 26.4 3 = (26.4± 0.1) cm3 Conc NaOH = 0.106 ± 0.5% 0.5 x 0.106 100 = 0.001 Conc NaOH = 0.106 ± 0.001M Lit value - NaOH = 0.100M Expt value– NaOH = 0.106M Difference = 0.006 % Error – Difference x 100% Lit value 0.006 x 100 % = 6% 0.100 Data ProcessingData collection Vol KHP /cm3 Titration trials 1 2 3 Final vol/(±0.05)cm3 38.50 29.50 45.90 Ini vol/ (±0.05)cm3 12.10 3.10 19.50 Total vol/ (±0.1)cm 3 26.4 26.4 26.4 NaOH + KHP → NaKP + H2O M = ? M = 0.1 V = 25ml V = 26.4 ml M M VM VM b aa bb 106.0 1 1 0264.01.0 025.0 1 1     Error Analysis % ∆ - burette % ∆ - pipette Abs ∆ vol x 100% Ave vol = 0.1 x 100% 26.4 = 0.38% Abs ∆ vol x 100% vol = 0.03 x 100% 25.00 = 0.12 % Total % ∆ = % ∆ burette+ %∆ pipette = 0.38% + 0.12% = 0.5% Data Processing % uncertainty (Abs uncertainty) Acid (KHP) Base (NaOH)
  • 9. Standardizationof (ACID)with std (BASE) Vol Na2CO3 Fin vol = 29.50 ± 0.05 Ini vol = 3.10 ± 0.05 Vol = 26.4 ± 0.1 Uncertainty invol Add abs ∆ for final + ini = (0.05 + 0.05) = ± 0.1 Ave Vol Na2CO3 ± uncertainty = 26.4 + 26.4 + 26.4 3 = (26.4 ± 0.1) cm3 Conc HCI = 0.2112 ± 0.5% 0.5 x 0.2112 100 = 0.001 Conc HCI = 0.2112 ± 0.001M Lit value - HCI = 0.200M Expt value – HCI = 0.2112M Difference = 0.011 % Error – Difference x 100% Lit value 0.011 x 100 % = 6% 0.200 Data ProcessingData collection Vol Na2CO3 /cm3 Titration trials 1 2 3 Final vol/(±0.05)cm3 38.50 29.50 45.90 Ini vol/ (±0.05)cm3 12.10 3.10 19.50 Total vol/ (±0.1)cm 3 26.4 26.4 26.4 Na2CO3 + 2HCI → 2NaCI + H2O + CO2 M = 0.1 M = ? V = 26.4 ml V = 25.0 ml M M VM VM a aa bb 2112.0 2 1 0250.0 0264.01.0 2 1     Error Analysis % ∆ - burette % ∆ - pipette Abs ∆ vol x 100% Ave vol = 0.1 x 100% 26.4 = 0.38% Abs ∆ vol x 100% vol = 0.03 x 100% 25.00 = 0.12 % Total % ∆ = % ∆ burette+ %∆ pipette = 0.38% + 0.12% = 0.5% Data Processing % uncertainty (Abs uncertainty) Base (Na2CO3) Acid (HCI)
  • 10. NaOH M = ? V = 25.0ml KHP M = 0.100M V = 26.4 ml HCI M = ? V = 25.0ml Sample TitrationCalculation Na2CO3 M = 0.100M V = 26.4 ml Standardizationof (BASE)with std (ACID) Standardizationof (ACID)with std (BASE) KHP + NaOH → NaKP + H2O M = 0.100 M = ? V = 26.40 ml V = 25.0 ml M M VM VM b bb aa 106.0 1 1 0250.0 0264.01.0 1 1     Na2CO3 + 2HCI → 2NaCI + H2O + CO2 M = 0.100 M = ? V = 26.4 ml V = 25.0 ml M M VM VM a aa bb 2112.0 2 1 0250.0 0264.01.0 2 1     Click here titration simulation Click here titration simulation Click here titration simulation Click here titration simulation Simulation on Titration
  • 11. Sample TitrationCalculation M M VM VM b aa bb 12.2 1 2 5.026.00.1 0250.0 1 2     3 20 1 2 305.0 5.1 1 2 cm V VM VM b aa bb     25 ml NaOH require 26.5cm3 of 1.0M H2SO4 for neutralization. Find its molarity of NaOH. Find vol of 1.5M NH3 required to neutralize 30 ml of 0.5M H2SO4 H2SO4 M = 1 M V = 26.5 ml NaOH M = ? V = 25 ml 2NaOH + H2SO4 → Na2SO4 + 2H2O M = ? M = 1M V = 25.0ml V = 26.5ml H2SO4 M = 0.5M V = 30ml NH4OH M = 1.5M V = ? ml 2NH4OH + H2SO4 → (NH4)2SO4 + 2H2O M = 1.5M M = 0.5M V = ? ml V = 30.0ml Simulation on Titration Click here titration simulation Click here titration simulation Click here titration simulation Click here titration simulation
  • 12. Sample TitrationCalculation 3 25 2 1 0.2 505.0 2 1 cm V VM VM a aa bb     M M VM VM a aa bb 16.0 2 1 25 102.0 2 1     Simulation on Titration Click here titration simulation Click here titration simulation Click here titration simulation Click here titration simulation Find vol of 2.0M HCI needed to neutralize 0.5M Na2CO3 in 50ml water. HCI M = 2.0M V = ? ml Na2CO3 M = 0.5M V = 50ml Na2CO3 + 2HCI → 2NaCI + CO2 + H2O M = 0.5M M = 2.0M V = 50ml V = ? ml Na2CO3 M = 0.2M V = 10.0ml HCI M = ? V = 25 ml 10 cm3 of 0.2 M Na2CO3 need 25 cm3 of HCI for neutralization. Find molarity of HCI. Na2CO3 + 2HCI → 2NaCI + H2O + CO2 M = 0.2M M = ? V = 10 ml V = 25 ml
  • 13. Titration for IA assessment Acid Base Titration StandardizationHCI with std Na2CO3 StandardizationNaOH with std KHP Titrationbet NaOH with std HCI Titrationbet HCI with std NaOH Determinewater crystallization in hydratedNa2CO3 with std HCI StandardizationKMnO4 with std ammonium iron(II)sulphate Fe 2+ in iron pill with std KMnO4 Hypochlorite(OCI-) in bleach with iodine/thiosulphate Determineethanoicacid in vinegar Cu2+ in brass with iodine/thiosulphate StandardizationKI/I2 with std KIO3 Determineacetylsalicylic acid in aspirin Vit C in fruitswith iodine/thiosulphate StandardizationExpt Acid/Base Expt StandardizationExpt Redox Expt Redox Titration StandardizationKI/I2 with std sodium thiosulphate Iodine/thiosulphate(iodometric titration)
  • 14. Equilibrium established when ethanoic acid and ethanol react together in strong acid, using propanone as solvent. Eqn given. CH3COOH + C2H5OH ↔ CH3COOC2H5 + H2O Density ethanoic acid is 1.05 g cm–3. i. Find amt, mol, of acid present ii. Conc acid is 1.748 mol dm–3. Find % uncertainty of conc. Titration performed on acid using a base. Result shown below Find absolute uncertainty of titre for Titration 1 (27.60 cm3). Liquid Vol/cm3 Ethanoic acid 5.00 ± 0.05 Ethanol 5.00 ± 0.05 Hydrochloric acid 1.00 ± 0.02 Propanone 39.0 ± 0.5 gmass voldenstiymass 25.500.505.1   vol mass Density  molMol RMM mass Mol 0874.0 60 25.5   RMM acid = 60 % uncertaintyacid conc = % uncertaintyin vol acid + % uncertaintyin total vol vol mol acidConc . (0.05/5.00) x 100 % = 1 % (0.62/50) x 100% = 1.24 % Total % uncertainty = (1 + 1.24) % = 2.24% Uncertaintyfinal – initial vol (28.80 ±0.05 – 1.20 ±0.05 ) = (27.60 ± 0.1) Add absolute uncertaintytogether
  • 15. Two rxn kinetic investigated using iodine clock rxn. Reaction A: H2O2 + 2I− + 2H+ → I2 + 2H2 O Reaction B: I2 + 2S2O3 2− → 2I− + S4O6 2- i. Find total uncertainty, in vol of rxn mixture Mixture contained: 5.0 ± 0.1 cm3 of 2M H2O2 5.0 ± 0.1 cm3 of 1 % starch 20.0 ± 0.1 cm3 of 1M H2SO4 20.0 ± 0.1 cm3 of 0.01 M Na2S2O3 50.0 ± 0.1 cm3 of water with 0.0200 ± 0.0001 g KI i. Add all vol together:Add all absolute uncertaintytogether. (5.0 ± 0.1)+ (5.0 ± 0.1) + (20.0 ± 0.1)+ (20.0 ± 0.1 ) + (50.0 ± 0.1) = (100 ± 0.5) cm 3 ii. Conc KI =Mass/ vol % uncertainty conc KI = % uncertaintymass + % uncertainty vol KI % ∆ mass = (0.0001/0.02) x 100% = 0.5 % % ∆ vol = (0.1/50) x 100% = 0.2 % % conc KI = (0.5 + 0.2)% = 0.7 % iii. Final Conc KI = Conc KI in total mixture ii. Find % uncertainty for KI conc in final rxn sol. iii. Find % uncertainty for KI conc in overall rxn mixture % ∆ final conc KI = % ∆ conc KI + % ∆ total vol KI % ∆ conc KI = 0.7 % % ∆ total vol = (0.5/100) x 100 % = 0.5% % conc KI = (0.5 + 0.7) = 1.2 % Mixture contained: 5.0 ± 0.1 cm3 of 2M H2O2 5.0 ± 0.1 cm3 of 1 % starch 20.0 ± 0.1 cm3 of 1M H2SO4 20.0 ± 0.1 cm3 of 0.01 M Na2S2O3 50.0 ± 0.1 cm3 of water with 0.0200 ± 0.0001 g KI total vol Only vol/mass/conc KI
  • 16. 4.32 x 10-5 x 176.14 = 7.61 x10-3 g Vit C KIO3 + 5KI + 6H+ → 3I2 + 6K+ + 3H2O 3C6H8O6 + 3I2 → 3C6H6O6 + 6I- + 6H+ Iodometric titration on Vit C, (C6H8O6). Vit C titrated with 0.002M KIO3 , using excess KI and starch. Redox Titration – Vit C quantification KIO3 M = 0.002M Vit C Amt = ? Mole ratio (1 :3) 1 mol KIO3 : 3 mol I2 : 3 mol C6H8O6 1 mol KIO3 3 mol C6H8O6 Amt = ? transfer 1g KI excess + starch titrated Vit C 5 3 1032.4)..( 3 1 )..( 0072.0002.0 3 1 ).( )(      CVitMole CVitMole CVitMV KIOMV 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 4 1000.5 250.0002.0 250.0 002.0 .      mol mol mol vol mol acidConc 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
  • 17. 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 diluted vinegar added. Find its molarity. mole ratio Moles bef dilution = Moles aft dilution M1 V1 = M2V2 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 0856.0 1 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   
  • 18. 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 M M VM VM b bb aa 0976.0 1 2 0250.0 0488.01.0 1 2     0.0976 x 106 = 10.36g/dm3 X RMM
  • 19. 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
  • 20. 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 + 2S2O3 2- → 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. Cal molarity of CIO in bleach. Redox Titration – CIO- in Bleach Na2S2O3 M = 0.0206M V = 17.3ml I2 M = ? Mole ratio ( 1 : 1) 2 mole CIO- : 1 mole I2 : 2 mole S2O3 2- 2 mole CIO- 2 mole S2O3 2- 10.0ml CIO- transfer V = 250ml M = 1.78 x 10-2 M 20ml transfer 1g KI excess added 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
  • 21. 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. Redox Titration – Vit C quantification KIO3 M = 0.002M V = 25.5ml Vit C M = ? V = 25ml Mole ratio (1 :3) 1 mol KIO3 : 3 mol I2 : 3 mol C6H8O6 1 mol KIO3 3 mol C6H8O6 V = 25ml M = ? 25ml transfer 1g KI excess + starch titrated Vit C 4 3 1053.1)..( 3 1 )..( 0255.0002.0 3 1 ).( )(      CVitMole CVitMole CVitMV KIOMV M x V = Mol 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
  • 22. 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- → S4O6 2- + 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. Redox Titration - % Cu in Brass Na2S2O3 M = 0.1M V = 28.2ml I2 M = ? Mole ratio (1 : 1) 2 mol Cu2+ : 1 mol I2 : 2 mol S2O3 2- 2 mol Cu2+ 2 mol S2O3 2- Pour into Volumetric flask V = 250ml M = ? 25ml transfer 1g KI excess/ starch 10 ml HNO3 titrated Water added 250ml 2.5g brass 32 2 2 32 2 1082.2).( 2 2 0282.01.0 ).( 2 2 )( )(         CuMole CuMole OSMV CuMV M x V = Mol Cu 2+ M x V = 2.82 x 10-3 M x 0.025 = 2.82 x 10-3 M = 2.82 x 10-3 0025 M = 1.13 x 10-3 M Convert mole Cu→ Mass Cu X RMM X 10
  • 23. 2Cu2+ + 4I- → I2 + 2CuI I2 + 2S2O3 2- → S4O6 2- + 2I- 0.456 g brass react with 25ml HNO3 producing Cu2+ ions. Sol was titrate with 0.1M S2O3 2- and end point, reached when 28.5ml added. Cal mol, mass, molarity of Cu 2+ and % Cu in brass. Redox Titration - % Cu in Brass Na2S2O3 M = 0.1M V = 28.5ml I2 M = ? Mole ratio (1 : 1) 2 mol Cu2+ : 1 mol I2 : 2 mol S2O3 2- 2 mol Cu2+ 2 mol S2O3 2- transfer 1g KI excess starch 25ml HNO3 titrated 0.456g brass 32 2 2 32 2 1085.2).( 2 2 0285.01.0 ).( 2 2 )( )(         CuMole CuMole OSMV CuMV M x V = Mol Cu 2+ 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 %
  • 24. % 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 2HNO3 + CaCO3 → (CaNO3)2 + H2O + 2CO2 Mole Mole 0.483 ? Mole ratio (2 : 1) 2 mol HNO3 - 1 mol CaCO3 0.483 mol HNO3 - o.242 molCaCO3 2 107.1.. 1 1 ).( 017.000.1 1 1      acidMole acidMole VM VM aa bb 25 g impure CaCO3 in egg shell Convert mole CaCO3 → Mass /g X RMM 0.242 x 100 = 24.2 g CaCO3 % CaCO3 = mass CaCO3 x 100% mass egg = 24.2 x 100% 25.0 = 96.8 %
  • 25. % Calcium carbonate in egg shell - Back Titration Amt of HCI added Amt of base (egg) Amt of HCI left Titrate NaOH M = 0.10 V = 23.8 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 = 0.1 M mol = ? V = 23.8 ml Amt HCI add = M x V = 0.2 x 0.272 = 0.0544 mol Amt HCI react = Amt HCI add – Amt HCI left = 0.0544 – 2.38 x 10-3 = 0.00306 mol 2HCI + CaCO3 → CaCI3 + H2O + CO2 Mole Mole 0.00306 ? Mole ratio (2 : 1) 2 mol HCI - 1 mol CaCO3 0.00306 mol HCI - o.00153 molCaCO3 3 1038.2.. 1 1 ).( 238.01.0 1 1      acidMole acidMole VM VM aa bb Convert mole CaCO3 → Mass /g X RMM 0.00153 x 100 = 0.153 g CaCO3 % 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
  • 26. 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 molCa(OH)2 3 10727.3.. 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
  • 27. 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 3 108.8. 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