Sodium Thiosulfate Titrations
Assign oxidation numbers
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
• Which is being oxidised? Reduced?
• Which is the oxidising agent/ reducing agent?
An iodine/ sodium thiosulfate titration
• Since the ratio of the reaction is known, this
reaction can be used to find the concentration
of either solutions, once one is known
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
A suitable indicator
• The red/brown colour of iodine will fade to yellow and
then to colourless as it is used up during the reaction
• But this colour change is slow and it is difficult to
monitor the exact end point from it
• Add a starch indicator when the reaction mixture is
pale yellow..
• The iodine /starch complex will change suddenly from
Blue/Black to colourless at the end point of the
titration
Sodium thiosulfate
• Is not a primary standard as it is impure, so a
standard solution can not be made up directly.
• A standard solution of sodium thiosulfate can
only be prepared by titration against a
solution of known concentration
Man exp:
Preparing a standard solution of
Sodium thiosulfate
• Iodine is also not a primary standard as free Iodine (I2) is insoluble
in water and sublimes
By doing the following reaction, an iodine solution of known
concentration can be made..
• 2MnO4- + 10I- +16H+
2Mn+2 +5I2 +8H20
Excess Potassium Iodide is
Potassium
used so that all of the
permanganate
potassium permanganate
used is a
present will react to make
standard
iodine. It also increases the
solution (exp*)
solubility of the iodine made
Since Cl- will also react
with potassium
permanganate it is
important that none is
present in the reaction
mixture
Potassium manganate(VII) is not a
primary standard.
Potassium permanganate, KMnO4 solution can be standardised by
titration against a standard solution of ammonium iron(II) sulfate solution
Titrations involving Potassium permanganate, are always carried out under
acidic conditions. Acidic conditions are necessary, because in neutral or
alkaline conditions Mn+7 is reduced only as far as Mn+4 (a muddy brown
precipitate)
Acid used
• Dilute sulfuric acid. Sulfuric acid is a good source of H+, and the SO4-2 ions are
not reactive.
• Hydrochloric acid is a source of Cl- ions which would react with the KMnO4,
• Nitric acid is a source of NO3- ion is which would react with the KMnO4,
Both would cause more iodine to be released than would be calculated from
the balanced equation
Note - Manganese(IV) oxide, manganese(II) ions, light, heat, acids and bases all
catalyse the decomposition of potassium manganate(VII) solutions so they must
be standardised regularly
Using the iodine solution to standardise
sodium thiosulfate solution
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
Sodium thiosulfate solution
(unknown concentration)
When iodine solution
goes pale yellow...
Add starch indicator
Blue/Back - colourless
Iodine solution of known concentration
made by reacting a standard solution of
potassium permanganate with acidified
potassium iodide in the conical flask
Calculations
1. Use the known molarity of potassium
permanganate standard solution and the
given balanced equation to find the molarity
of the iodine solution made
2. Use to find the molarity of the sodium
thiosulfate solution
3. Once moles/ L is known for sodium
thiosulfate you can also work out grams/ L
Question 226 f
V1 X M1 = V2 x M2
n1
n2
(20)X (M1) = (25) x (0.05)
2
1
M1 = (25) x (0.05) x (2)
(1) x (20)
M1 = 0.125
The concentration of the sodium thiosulfate solution is 0.125M
(moles per litre)
(ii) What is the concentration in grams per litre?
• Moles PER LITRE
X RMM
Grams PER LITRE
0.125 x rmm = grams per litre
0.125 x 248g = 31
There are 31 g of Na2S2O3.5H2O in one litre.
Question 227 d
V1 X M1 = V2 x M2
n1
n2
(20)X (M1) = (15) x (0.1)
1
2
M1 = (15) x (0.1) x (1)
(2) x (20)
M1 = 0.0375
The concentration of the Iodine solution is 0.0375M (moles per litre)
Question 228
V1 X M1 = V2 x M2
n1
n2
(18.55)X (M1) = (25) x (0.05)
2
1
M1 = (25) x (0.05) x (2)
(1) x (18.55)
M1 = 0.1348
The concentration of the Sodium thiosulfate solution is 0.1348M
(moles per litre)
(ii) What is the concentration in grams per litre?
• Moles PER LITRE
X RMM
Grams PER LITRE
0.1348 x rmm = grams per litre
0.1348 x 248g = 33.4304
There are 33.4304 g of Na2S2O3.5H2O in one litre.
Sodium thiosulfate: Irritant to eyes.
Methanol
: Toxic by ingestion or inhalation. Much more poisonous than
ethanol. Highly flammable.
Potassium iodide may be harmful by ingestion. Eye irritant.
Starch: Starch powder is explosive when dry. Dust may irritate eyes and lungs.
Concentrated sulfuric acid
is very corrosive to eyes and skin. Due to its very
considerable heat of reaction with water, it is essential that the acid be added to wate
when it is being diluted.
Dilute sulfuric acid
Bleach
i:
i
is harmful to eyes and an irritant to skin.
Hypochlorite solutions are corrosive, harmful and irritant.
Finding the concentration of sodium
hypochlorite in bleach
• Many commercial bleaches are simply solutions
of hypochlorite salts such as sodium hypochlorite
(NaOCl)
• Hypochlorite ion reacts with excess iodide ion in
the presence of acid to generate an iodine
solution:
• ClO- + 2I- + 2H+ → Cl- + I2 + H2O
Using the iodine solution to standardise
sodium thiosulfate solution
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
Sodium thiosulfate solution
(standard solution – last
exp **** )
When iodine solution
goes pale yellow...
Add starch indicator
Blue/Back - colourless
Iodine solution of unknown concentration
- made by reaction of sodium hypochlorite
from bleach and potassium iodide under acidic
conditions
Calculations
1. Use the known molarity of the standard solution and
the given balanced equation to find the molarity of
the iodine solution from the titration
2. Use the balanced equation to see what the molarity
of the diluted sodium hypochlorite solution is.
3. Work out the molarity of sodium hypochlorite in the
original bleach solution.
4. Once moles/ L is known for sodium hypochlorite you
can also work out grams/ L and v/w%
Question
229 f
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
(20)X (M1) = (26.3) x (0.1)
1
2
V1 X M1 = V2 x M2
n1
n2
M1 = (26.3) x (0.1) x (1)
(2) x (20)
M1 = 0.0658
The concentration of the iodine solution is 0.0658M (moles per litre)
ii) Finding the conc of the sodium
thiosulfate in the diluted bleach
• The eqt for the 1st reaction was
ClO- + 2I- + 2H+ → Cl- + I2 + H2O
SO MOLARITY = 0.0658
MOLARITY = 0.0658
The concentration of NaOCl in the diluted bleach was 0.0658 moles
per litre
ii) Finding the w/v % of NaOCl in
original bleach
• Molarity of NaOCl in diluted bleach = 0.0658m/L
• Molarity of NaOCl in original bleach =
0.0658 x 10 = 0.658m/L
• Concentration of NaOCl in g/L in original bleach =
0.658 x 74.5 = 49.02 g/L
• Concentation of NaOCl in w/v % ( grams per 100cm3) in
original bleach =
49.02/10 = 4.9%
Question
230
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
V1 X M1 = V2 x M2
(25)X (M1) = (19.56) x (0.25)
1
2
n1
n2
M1 = (19.56) x (0.25) x (1)
(2) x (25)
M1 = 0.0978
The concentration of the iodine solution is 0.0978M (moles per litre)
ii) Finding the conc of the sodium
thiosulfate in the diluted bleach
• The eqt for the 1st reaction was
ClO- + 2I- + 2H+ → Cl- + I2 + H2O
SO MOLARITY = 0.0978
MOLARITY = 0.0978
The concentration of NaOCl in the diluted bleach was 0.0978 moles
per litre
ii) Finding the w/v % of NaOCl in
original bleach
• Molarity of NaOCl in diluted bleach = 0.0978m/L
• Molarity of NaOCl in original bleach =
0.0978 x 10 = 0.489m/L
• Concentration of NaOCl in g/L in original bleach =
0.489 x 74.5 = 34.4305g/L
• Concentation of NaOCl in w/v % ( grams per 100cm3) in
original bleach =
34.4305/10 = 3.6405%
Question
231
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
V1 X M1 = V2 x M2
(25)X (M1) = (16.1) x (0.1)
1
2
n1
n2
M1 = (16.1) x (0.1) x (1)
(2) x (25)
M1 = 0.0322
The concentration of the iodine solution is 0.0322M (moles per litre)
ii) Finding the conc of the sodium
thiosulfate in the diluted bleach
• The eqt for the 1st reaction was
ClO- + 2I- + 2H+ → Cl- + I2 + H2O
SO MOLARITY = 0.0322
MOLARITY = 0.0322
The concentration of NaOCl in the diluted bleach was 0.0322 moles
per litre
e)f) Finding the w/v % of NaOCl in
original bleach
e)i)Molarity of NaOCl in diluted bleach = 0.0322m/L
ii)Molarity of NaOCl in original bleach =
0.0322 x 20 = 0.644m/L
f) i)Concentration of NaOCl in g/L in original bleach
=
ii) Concentation of NaOCl in w/v % ( grams per
100cm3) in original bleach =


Organic mattere.g. sewage,
industrial waste, silage, milk.
discharged into a water acts as a
food source for the bacteria
present there.
The bacteria will multiply and use
up the available dissolved
oxygen
This may cause
 fish kills.
 bacteria will produce hydrogen
sulphide and ammonia(
anaerobic conditions)

The level of dissolved oxygen in a water sample is an
indicator of the quality of the sample.
Biochemical oxygen demand test
 The amount of dissolved oxygen used up by
biochemical action when a sample of water is
kept in the dark at 20oC for 5 days.
Note –
Compare before and after readings!
Sample is kept in the dark to prevent new oxygen being
produced by photosynthesis



It is not possible to directly measure the
amount of dissolved oxygen in a water
sample directly.
The dissolved oxygen does not directly react
with another suitable reagent, an indirect
procedure was developed by Winkler.
An iodine/thiosulfate titration can be used to
measure the dissolved oxygen present in a
water sample.
Manganese sulfate in alkaline conditions
1. Mn2+(aq) + 2OH-(aq)  Mn(OH)2(s) ( white precipitate)
Mix with sample under water - This reacts with the dissolved
oxygen to produce a brown precipitate.
2. 4Mn(OH)2(s) + H20 + O2
(aq)
 4Mn(OH)3(s)
Adding concentrated H2SO4 - enables the Mn(IV) compound to
release free iodine from KI.
3. 2Mn(OH)3(s) + 6H+(aq) + 2I-
(aq) 
2Mn2+(aq) + I2(aq) + 6H2O(l)
Titration : The free iodine is then titrated with standard sodium
thiosulfate
4. I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-
(aq)

Standard solution = Sodium thiosulfate
Iodine solution = unknown concentration
Colour change =
Red/ Brown – Pale yellow but hard to see the end point
Add starch indicator near the end point to see a clear colour change.
Blue /Black ( iodine present) – Colourless ( iodine absent
1.
Find the concentration of the iodine
solution using results from the titration:
M1 X V1 = M2 X V2
N1
N2
2. Using the known ratios of how the dissolved
oxygen reacted to make the iodine solution
you can work out the molarity of the oxygen
in the solution
3. Use the molarity of the oxygen in the
solution to find g/L and ppm
1.
2.
3.
4.
5.
6.
7.
Why is the reagent MnSO4 used?
Why is concentrated H2SO4 used?
Why must the bottles be shaken vigorously after
adding the Manganese sulfate and alkaline
potassium iodide?
Why are the bottles completely filled with water?
If the white precipitate remains on addition of
manganese(II) sulfate solution and alkaline
potassium iodide solution, what does this
indicate about the water sample?
State and explain what the letters B.O.D. mean.
Why are the bottles used during B.O.D.
measurements stored in the dark?
Question
233g
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
V1 X M1 = V2 x M2
(50)X (M1) = (6) x (0.01)
1
2
n1
n2
M1 = (6) x (0.01) x (1)
(2) x (50)
M1 = 0.0006
The concentration of the iodine solution is 0.006M (moles per litre)
Ratios of reactions making I2
1. Mn2+(aq) + 2OH-(aq)  Mn(OH)2(s)
( white precipitate)
2. 4Mn(OH)2(s) + H20 + O2 (aq)  4Mn(OH)3(s)
.
3. 2Mn(OH)3(s) + 6H+(aq) + 2I- (aq)  2Mn2+(aq) + I2(aq) + 6H2O(l)
Overall ratio is 1O2
M1 = 0.0006 /2=
.0003
2I2
M1 = 0.0006
e)f) Finding the concentration of
oxygen in ppm
Ppm = parts per million ( mg in 1000cm3)
f) i)Concentration of O2 in g/L in water
= 0.0003 X 32 = 0.0096g/L
ii) Concentation of of O2 in mg/L in water (ppm)
.0096 x 1000 = 9.6ppm
Question
235 –
Sample 1
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
(200)X (M1) = (27) x (0.01)
1
2
V1 X M1 = V2 x M2
n1
n2
M1 = (27) x (0.01) x (1)
(2) x (200)
M1 = 0.000675
The concentration of the iodine solution is 0.000675M (moles per
litre)
Ratios of reactions making I2
1. Mn2+(aq) + 2OH-(aq)  Mn(OH)2(s)
( white precipitate)
2. 4Mn(OH)2(s) + H20 + O2 (aq)  4Mn(OH)3(s)
.
3. 2Mn(OH)3(s) + 6H+(aq) + 2I- (aq)  2Mn2+(aq) + I2(aq) + 6H2O(l)
Overall ratio is 1O2
M1 = 0.000675 /2=
.0003375
2I2
M1 =
0.000675
e)f) Finding the concentration of
oxygen in ppm
Ppm = parts per million ( mg in 1000cm3)
f) i)Concentration of O2 in g/L in water tested
= 0.0003375 X 32 = 0.0108g/L
ii) Concentation of of O2 in mg/L in water tested(ppm)
.0108 x 1000 = 10.8ppm
Water tested was diluted down ten times so original
sample had 10.8 x 10 = 108ppm
Question
235 –
Sample 2
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62- (aq)
(200)X (M1) = (4.8) x (0.01)
1
2
V1 X M1 = V2 x M2
n1
n2
M1 = (4.8) x (0.01) x (1)
(2) x (200)
M1 = 0.00012
The concentration of the iodine solution is 0.00012M (moles per litre)
Ratios of reactions making I2
1. Mn2+(aq) + 2OH-(aq)  Mn(OH)2(s)
( white precipitate)
2. 4Mn(OH)2(s) + H20 + O2 (aq)  4Mn(OH)3(s)
.
3. 2Mn(OH)3(s) + 6H+(aq) + 2I- (aq)  2Mn2+(aq) + I2(aq) + 6H2O(l)
Overall ratio is 1O2
M1 = 0.00012 /2=
.00006
2I2
M1 = 0.00012
e)f) Finding the concentration of
oxygen in ppm
Ppm = parts per million ( mg in 1000cm3)
f) i)Concentration of O2 in g/L in water tested
= 0.00006 X 32 = 0.00192g/L
ii) Concentation of of O2 in mg/L in water tested(ppm)
.00192x 1000 = 1.92ppm
Water tested was diluted down ten times so original
sample had 1.92 x 10 = 19.2 ppms
BOD of the water
• 1st reading – 2nd reading
• 108ppm – 19.2ppm = 88.8ppm
• Describe what hardness in water is
• Describe how to estimate of the total hardness of a
water sample
Hardness in water
• Def: Water is said to be hard when it is difficult to
form a lather with soap.
• It is caused by the presence of Ca+2 and Mg+2 ions
being dissolved in water
Experiment – estimation of the total
hardness of a water sample
The disodium salt of edta is harmful. Contact with the skin and eyes should be
avoided.
n
The buffer solution of pH 10 is corrosive. Avoid breathing vapours. Use in fume
hood.
Eriochrome Black T is an irritant and should not be allowed to come into contact
with the skin.
i
1.
Find the concentration of the Ca+2 ions in
the water using results from the titration:
M1 X V1 = M2 X V2
N1
N2
2. Find concentration of CaCO3 in the water in
g/L and ppm
Questions relating to the experiment
1.
Why is it important that the reaction between the edta and the
metal ions in solution (i) is rapid and (ii) goes to completion?
2.
The water sample could contain metal ions other than Ca2+ and
Mg2+. How would the reliability of the result be affected if this were the
case? Suggest two other metal ions that could be present in the water.
3.
This reagent cannot distinguish between temporary and permanent
hardness. List the compounds of calcium and magnesium that cause
hardness, and indicate those which cause temporary hardness.
4.
Suggest a method of establishing the amount of permanent
hardness in a water sample.
5.
What is the function of the buffer solution?
Q243
• Describe what hardness in water is
• Explain causes of temporary and permanent
hardness
• Do tests on scale deposits in a kettle
• Know that removal of hardness is by boiling or ion
exchange
• Describe what deionisation is and how it works
Hardness in water
• Def: Water is said to be hard when it is difficult to
form a lather with soap.
• It is caused by the presence of Ca+2 and Mg+2 ions
being dissolved in water
• There are two different types of hardness in water:
1. Temporary hardness
2. Permanent hardness
1. Temporary hardness
• Def: Temporary hardness is hardness in water
that can be removed by boiling.
Found in parts of the
country where
limestone (calcium
carbonate) is found in
the earth.
What causes temporary hardness
• Water (rain), carbon dioxide( in the air) and
Calcium carbonate (limestone) all react
together:
•Calcium hydrogen carbonate
Ca(HCO3)2 is a made and
dissolves in the water.
Removing temporary hardness from water
• 1. It can be removed by boiling!
This reaction occurs:
• Ca(HCO3)2 + heat
CaCO3 + H2O + CO2
Limescale inside a kettle
Removing temporary hardness from water
• 2. It can be removed using an ion exchanger the hardness causing calcium and magnesium ions are
swapped with sodium ions, which do not cause hardness.
Permanent hardness
• Permanent hardness is hardness in water that
cannot be removed by boiling.
• It is caused by the presence of calcium sulfate
CaSO4 and magnesium sulfate MgSO4 in water.
Removing permanent hardness from
water
It can be removed by:
• using an ion exchanger
Deionisation –
Deionisation
In a deioniser any dissolved salts present ae
replaced by water molecules.
Relative purity of
deionised and distilled water.
• Deionised = no ions other than H+ and OH• Distilled = No dissolved solids of any type
• Describe what hardness in water is
• Explain causes of temporary and permanent
hardness
• Know that removal of hardness is by boiling or
ion exchange
• Describe what deionisation is and how it
works