IB Chemistry on Titration Techniques and IA on Titrations

IB Chemistry on Titration Techniques and IA on Titrations

1. http://lawrencekok.blogspot.com
Prepared by
Lawrence Kok
IB Chemistry Titration Techniques, Acid/Base Titration and
IA on Titrations.

2. 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
2CO
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)

3. Titration
Redox Titration Acid Base Titration Complexometric titration
Neutralization
Indicator
pH sensor
Conductometric
Changes colour (end point) at equivalent point
Indicator pK
a
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- 10.0Colourless Pink
Video on pH sensor Video on indicator
Video on conductometric
measured using
measured using
Video on conductometric

4. 3
1
6
2
4
5
Titration Steps
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

5. 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
Equivalent point – 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 Na
2
CO
3

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
? ?
Stan
d
ardize N
aO
H

usin
g K
H
P
S
t
a
n
d
a
r
d
i
z
e

H
C
I





u
s
i
n
g

N
a
2C
O
3
10.6 g Na
2
CO
3

6. 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 volumetric flask
Step 3
Add water until 1L mark
Transfer to beaker, add water to dissolve
Step 4
Step 5
20.4 g
Video std solution preparation
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

7. 0.1M Na
2
CO
3
– 0.1 mole of Na
2
CO
3
in total vol of solution (1L)
Primary standard base
Mass of Na
2CO
3 0.1 mole Na
→
2CO
3 x M = 0.1 x 106 gStep 1
Step 2
Pour to 1L volumetric flask
Step 3
Add water until 1L mark
Transfer to beaker, add water to dissolve
Step 4
Step 5
10.6 g
Video std solution preparation
Molarity = 0. 1 mole
(0.1M) 1 L total vol (solute + solvent)
Primary standard
Preparing std solution – 0.1 M – 0.1 mole Na
2
CO
3
in 1 L

8. Pipette 25 ml NaOH in conical flask
Step 2
Fill burette with std 0.1M KHP sol
Step 3
Step 4
Step 5
2 drop phenolthalein – colourless to pink
Ini vol KHP recorded
Titrate until pink colour fades away
Final vol KHP recordedStep 6
Step 7 Repeat till consistent result agree within 0.1 cm
3
(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 burette with std 0.1M Na
2CO
3 sol
Ini vol Na
2CO
3 recorded
Pipette 25 ml HCI in conical flask
2 drop methyl orange to HCI – red
Titrate until red changes to pink
Final vol Na
2CO
3recorded
Repeat till consistent result agree within 0.1 cm
3
(triplicate)
Std 0.1M Na
2
CO
3

? Conc HCI
Step 1
Std 0.1M KHP

9. Standardization of (BASE) with std (ACID)
Vol KHP
Fin vol = 29.50 ± 0.05
Ini vol = 3.10 ± 0.05

Vol = 26.4 ± 0.1
Uncertainty in vol
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) cm
3
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 Processing Data collection
Vol KHP /cm
3
Titration trials
1 2 3
Final vol/(±0.05)cm
3
38.50 29.5045.90
Ini vol/ (±0.05)cm
3
12.10 3.10 19.50
Total vol/ (±0.1)cm
3
26.4 26.4 26.4
NaOH + KHP NaKP + H
→
2
O

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)

10. Standardization of (ACID) with std (BASE)
Vol Na
2CO
3
Fin vol = 29.50 ± 0.05
Ini vol = 3.10 ± 0.05

Vol = 26.4 ± 0.1
Uncertainty in vol
Add abs ∆ for final + ini
= (0.05 + 0.05) = ± 0.1
Ave Vol Na
2CO
3 ± uncertainty
= 26.4 + 26.4 + 26.4
3
= (26.4 ± 0.1) cm
3
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 Processing Data collection
Vol Na
2CO
3 /cm
3
Titration trials
1 2 3
Final vol/(±0.05)cm
3
38.50 29.5045.90
Ini vol/ (±0.05)cm
3
12.10 3.10 19.50
Total vol/ (±0.1)cm
3
26.4 26.4 26.4
Na
2
CO
3
+ 2HCI 2NaCI + H
→
2
O + CO
2


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 (Na
2CO
3)
Acid (HCI)

11. NaOH
M = ?
V = 25.0ml
KHP
M = 0.100M
V = 26.4 ml
HCI
M = ?
V = 25.0ml
Sample Titration Calculation
Na
2
CO
3
M = 0.100M
V = 26.4 ml
Standardization of (BASE) with std (ACID) Standardization of (ACID) with std (BASE)
KHP + NaOH NaKP + H
→
2
O

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
==
´
´
=
Na
2
CO
3
+ 2HCI 2NaCI + H
→
2
O + CO
2


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

12. Sample Titration Calculation
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 26.5cm
3
of 1.0M H
2
SO
4

for neutralization.
Find its molarity of NaOH.
Find vol of 1.5M NH
3
required to neutralize
30 ml of 0.5M H
2
SO
4
H
2
SO
4
M = 1 M
V = 26.5 ml
NaOH
M = ?
V = 25 ml
2NaOH + H
2
SO
4
Na
→
2
SO
4
+ 2H
2
O
M = ? M = 1M
V = 25.0ml V = 26.5ml
H
2
SO
4
M = 0.5M
V = 30ml
NH
4
OH
M = 1.5M
V = ? ml
2NH
4
OH

+ H
2
SO
4
(NH
→
4
)
2
SO
4
+ 2H
2
O
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

13. Sample Titration Calculation
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 Na
2
CO
3
in 50ml water.
HCI
M = 2.0M
V = ? ml
Na
2
CO
3
M = 0.5M
V = 50ml
Na
2
CO
3
+ 2HCI 2NaCI
→

+ CO
2
+ H
2
O
M = 0.5M M = 2.0M
V = 50ml V = ? ml
Na
2
CO
3
M = 0.2M
V = 10.0ml
HCI
M = ?
V = 25 ml
10 cm
3
of 0.2 M Na
2
CO
3
need 25 cm
3
of HCI for
neutralization. Find molarity of HCI.
Na
2
CO
3
+ 2HCI 2NaCI + H
→
2
O + CO
2
M = 0.2M M = ?
V = 10 ml V = 25 ml

14. Titration for IA assessment
Acid Base Titration
Standardization HCI
with std Na
2
CO
3
Standardization NaOH
with std KHP
Titration bet NaOH
with std HCI
Titration bet HCI
with std NaOH
Determine water crystallization
in hydrated Na
2CO
3 with std HCI
Standardization KMnO
4
with std ammonium
iron(II) sulphate
Fe
2+
in iron pill
with std KMnO
4
Hypochlorite (OCI
-
) in bleach
with iodine/thiosulphate
Determine ethanoic acid
in vinegar
Cu
2+
in brass
with iodine/thiosulphate
Standardization KI/I
2
with std KIO
3
Determine acetylsalicylic
acid in aspirin
Vit C in fruits with
iodine/thiosulphate
Standardization Expt Acid/Base Expt
Standardization Expt
Redox Expt
Redox Titration
Standardization KI/I
2
with std
sodium thiosulphate
Iodine/thiosulphate (iodometric titration)

15. Equilibrium established when ethanoic acid and ethanol react together
in strong acid, using propanone as solvent. Eqn given.
CH
3
COOH + C
2
H
5
OH
↔
CH
3
COOC
2
H
5
+ H
2
O
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 cm
3
).
Liquid Vol/cm
3

Ethanoic acid 5.00 ± 0.05
Ethanol 5.00 ± 0.05
Hydrochloric acid1.00 ± 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
% uncertainty acid conc = % uncertainty in vol acid + % uncertainty in 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%
Uncertainty final – initial vol
(28.80 ±0.05 – 1.20 ±0.05 )
= (27.60 ± 0.1)
Add absolute uncertainty together

16. Two rxn kinetic investigated using iodine clock rxn.
Reaction A: H
2
O
2
+ 2I
−
+ 2H
+
I
→
2
+ 2H
2
O
Reaction B: I
2
+ 2S
2
O
3
2−
2I
→
−
+ S
4
O
6
2-
i. Find total uncertainty, in vol of rxn mixture
Mixture contained:
5.0 ± 0.1 cm
3
of 2M H
2
O
2
5.0 ± 0.1 cm
3
of 1 % starch
20.0 ± 0.1 cm
3
of 1M H
2
SO
4
20.0 ± 0.1 cm
3
of 0.01 M Na
2
S
2
O
3
50.0 ± 0.1 cm
3
of water with 0.0200 ± 0.0001 g
KI
i. Add all vol together: Add all absolute uncertainty together.
(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 = % uncertainty mass + % 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 cm
3
of 2M H
2
O
2
5.0 ± 0.1 cm
3
of 1 % starch
20.0 ± 0.1 cm
3
of 1M H
2
SO
4
20.0 ± 0.1 cm
3
of 0.01 M Na
2
S
2
O
3
50.0 ± 0.1 cm
3
of water with 0.0200 ± 0.0001 g KI
total vol
Only vol/mass/conc KI

17. 4.32 x 10
-5
x 176.14
= 7.61 x10
-3
g Vit C
KIO
3 + 5KI

+ 6H
+
3I
→
2 + 6K
+
+ 3H
2O

3C
6
H
8
O
6
+ 3I
2

3C
→
6
H
6
O
6
+ 6I
-
+ 6H
+
Iodometric titration on Vit C, (C
6
H
8
O
6
).
Vit C titrated with 0.002M KIO
3
, using excess KI and starch.
Redox Titration – Vit C quantification
KIO
3
M = 0.002M
Vit C
Amt = ?
Mole ratio (1 :3)
1 mol KIO
3
: 3 mol I
2
: 3 mol C
6
H
8
O
6
1 mol KIO
3
3 mol C
6
H
8
O
6
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 KIO
3
, required to prepare 0.250 dm
3
of 0.002M KIO
3
ii Titration results shown in table below
Find % uncertainty in mean vol of KIO
3
used.
Mean vol = (7.20 ± 0.10) cm
3
Find amt of KIO
3
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 KIO
3

→ 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

18. 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
CH
3
COOH
M = ?
V = 29.2ml
NaOH
M = 0.1M
V = 25.0ml
NaOH + CH
3
COOH CH
→
3
COONa + H
2
O

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
M
1
V
1
= M
2
V
2
M
1
= Ini molarity M
2
= Final molarity
V
1
= Ini vol V
2
= 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
=
´=´
=

19. Acid/Base Titration - Empirical formula Na
2
CO
3
. x H
2
O


HCI
M = 0.100 M
V = 48.8ml
Na
2
CO
3

M = ? M
V = 25 ml
2HCI + Na
2
CO
3
2NaCI
→

+ CO
2
+ H
2
O

M = 0.1M M = ?
V = 48.8ml V = 25.0ml
V = 1L
M = ?
25 ml
transfer
Mole ratio – 2: 1
Mass Na
2
CO
3
. x H
2
O = 27.82 g
Mass Na
2
CO
3
= 10.36 g
Mass of water = (27.82 – 10.36) g
= 17.46 g
Diuted to 1L
27.82g
Na
2CO
3. xH
2O
27.82 g of hydrated (Na
2
CO
3
. x H
2
O) 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 Na
2
CO
3
present in 1L of sol. Find x
Convert mol dm
-3
→ g dm
-3
Empirical formula
Na
2
CO
3
H
2
O
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
Na
2
CO
3
. 10 H
2
O
M
M
VM
VM
b
bb
aa
0976.0
1
2
0250.0
0488.01.0
1
2
==
´
´
=
0.0976 x 106 = 10.36g/dm
3
X RMM

20. Redox Titration - % Fe in iron tablet
Iron tablet contain FeSO
4
.7H
2
O. 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 H
2
SO
4
and titrated with 0.002M
KMnO
4
. Ave 24.5ml need to reach end point. Cal % Fe in iron tablet.
10ml transfer
20ml acid added
1.863 g
250ml
KMnO
4
M = 0.002M
V = 24.5 ml
Fe
2+
M = ?
V = 30ml
MnO
4
-
+ 5Fe
2+
+ 8H
+
Mn
→
2+
+ 5Fe
2+
+ 4H
2
O
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 FeSO
4
10ml sol contain - 2.45 x 10
-4
Fe
2+
250ml sol contain - 250 x 2.45 x 10
-4
Fe
2+
10
= 6.125 x 10
-3
mole Fe
2+
42
2
1045.2.
5
1
.
0245.0002.0
5
1
-+
+
´=
=
´
=
FeMole
FeMole
VM
VM
bb
aa
Convert mole

→ Mass

X RMM

21. Mole bef dil = Mole aft dil
M
1
V
1
= M
2
V
2
M
1
x 10 = 1.78 x 10
-2
x 250
M
1
= 1.78 x 10
-2
x 250
10
M
1
= 0.445M
2CIO
-
+ 2I
-
+ 2H
+
I
→
2
+ 2CI
-
+ H
2
O

I
2 + 2S
2O
3
2-
→ S
4O
6
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 Na
2
S
2
O
3
.Using starch indicator, end point was 17.3ml.
Cal molarity of CIO in bleach.
Redox Titration – CIO
-
in Bleach
Na
2
S
2
O
3
M = 0.0206M
V = 17.3ml
I
2
M = ?
Mole ratio ( 1 : 1)
2 mole CIO
-
: 1 mole I
2
: 2 mole S
2
O
3
2-
2 mole CIO
-
2 mole S
2
O
3
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

22. KIO
3
+ 5KI

+ 6H
+
3I
→
2
+ 6K
+
+ 3H
2
O

3C
6
H
8
O
6
+ 3I
2

3C
→
6
H
6
O
6
+ 6I
-
+ 6H
+
Iodometric titration on Vit C, (C
6
H
8
O
6
). 25 ml Vit C titrated with 0.002M KIO
3
from burette,
using excess KI and starch. Ave vol KIO
3
is 25.5ml. Cal molarity of Vit C.
Redox Titration – Vit C quantification
KIO
3
M = 0.002M
V = 25.5ml
Vit C
M = ?
V = 25ml
Mole ratio (1 :3)
1 mol KIO
3
: 3 mol I
2
: 3 mol C
6
H
8
O
6
1 mol KIO
3
3 mol C
6
H
8
O
6
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

23. 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%
2Cu
2+
+ 4I
-
I
→
2
+ 2CuI

I
2
+ 2S
2
O
3
2-
S
→
4
O
6
2-
+ 2I
-
2.5g brass react with 10ml HNO
3
producing Cu
2+
ion. Sol made up to 250ml using water. Pipette 25ml of sol to
flask. Na
2
CO
3
add to neutralize excess acid. 1g KI (excess) and starch added. Titrate with 0.1M S
2
O
3
2-
and end
point, is 28.2 ml. Find molarity Cu
2+
and % Cu found in brass.
Redox Titration - % Cu in Brass

Na
2
S
2
O
3
M = 0.1M
V = 28.2ml
I
2
M = ?
Mole ratio (1 : 1)
2 mol Cu
2+
: 1 mol I
2
: 2 mol S
2
O
3
2-
2 mol Cu
2+
2 mol S
2
O
3
2-
Pour into
Volumetric flask
V = 250ml
M = ?
25ml transfer
1g KI excess/ starch

10 ml HNO
3
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

24. 2Cu
2+
+ 4I
-
I
→
2
+ 2CuI

I
2
+ 2S
2
O
3
2-
S
→
4
O
6
2-
+ 2I
-
0.456 g brass react with 25ml HNO
3
producing Cu
2+
ions. Sol was titrate with 0.1M S
2
O
3
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
Na
2
S
2
O
3
M = 0.1M
V = 28.5ml
I
2
M = ?
Mole ratio (1 : 1)
2 mol Cu
2+
: 1 mol I
2
: 2 mol S
2
O
3
2-
2 mol Cu
2+
2 mol S
2
O
3
2-
transfer
1g KI excess starch
25ml HNO
3
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 %

25. % Calcium carbonate in egg shell - Back Titration
250ml,
2M HNO
3
Amt of HNO
3
added
Amt of base (egg)
Amt of
HNO
3 left
Titrate NaOH
M = 1.0
V = 17.0ml
Amt HNO
3
react = Amt HNO
3
– Amt HNO
3

add left
HNO
3
left

Transfer
to flask
Left overnight in acid
added
25 g of egg shell (CaCO
3
) dissolved in 250 ml, 2 M HNO
3
. Sol titrated with NaOH.
17 ml, 1M NaOH needed to neutralize excess acid. Cal % CaCO
3
by mass in egg shell.
NaOH + HNO
3
NaNO
→
3

+ H
2
O

M = 1.00M mol = ?
V = 17 ml
Amt HNO
3
add = M x V
= 2.0 x 0.250
= 0.50 mol
Amt HNO
3
react = Amt HNO
3
add – Amt HNO
3
left
= 0.50 – 1.7 x 10
-2

= 0.483 mol
2HNO
3 + CaCO
3 (CaNO
→
3)
2 + H
2O + 2CO
2

Mole Mole
0.483 ?
Mole ratio (2 : 1)
2 mol HNO
3
- 1 mol CaCO
3
0.483 mol HNO
3
- o.242 mol

CaCO
3
2
107.1..
1
1
).(
017.000.1
1
1
-
´=
=
´
=
acidMole
acidMole
VM
VM
aa
bb
25 g impure
CaCO
3
in egg shell
Convert mole CaCO
3

→ Mass /g
X RMM
0.242 x 100 = 24.2 g CaCO
3
% CaCO
3
= mass CaCO
3
x 100%
mass egg
= 24.2 x 100%
25.0
= 96.8 %

26. % 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
→

+ H
2
O
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 + CaCO
3 CaCI
→
3 + H
2O + CO
2

Mole Mole
0.00306 ?
Mole ratio (2 : 1)
2 mol HCI

- 1 mol CaCO
3
0.00306 mol HCI - o.00153 mol

CaCO
3
3
1038.2..
1
1
).(
238.01.0
1
1
-
´=
=
´
=
acidMole
acidMole
VM
VM
aa
bb
Convert mole CaCO
3

→ Mass /g
X RMM
0.00153 x 100 = 0.153 g CaCO
3
% CaCO
3
= mass CaCO
3
x 100%
mass egg
= 0.153 x 100%
0.188
= 81.4 %
0.188g of egg shell (CaCO
3
) 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 CaCO
3
by mass in egg shell.
0.188g impure
CaCO
3
in egg shell
27.20ml, 0.2M
HCI

27. 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
→

+ H
2
O

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
CaCI
→
3
+ H
2
O
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
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

28. 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
→

+ H
2
O
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 + H
2A Na
→
3 A+ 2H
2O

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 H
2
A
20 ml, 2M NaOH
Molar Mass Acid
0.0156 mol acid - 2.04 g
1 mol acid - 2.04
0.0156
= 131

29. Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenses
http://creativecommons.org/licenses/

Prepared by Lawrence Kok
Check out more video tutorials from my site and hope you enjoy this tutorial
http://lawrencekok.blogspot.com