Sodium Thiosulfate Titrations

Assign oxidation numbers

• Which is being oxidised? Reduced?• Which is the oxidising agent/ reducing agent?

I2(aq) + 2S2O3

2-(aq) 2I-

(aq) + S4O62- (aq)

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) + 2S2O3

2-(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

• 2MnO4- + 10I- +16H+ 2Mn+2 +5I2 +8H20

By doing the following reaction, an iodine solution of known concentration can be made..

Excess Potassium Iodide is used so that all of the potassium permanganate present will react to make iodine. It also increases the solubility of the iodine made

Potassium permanganate used is a standard solution (exp*)

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

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.

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

Potassium permanganate, KMnO4 solution can be standardised by

titration against a standard solution of ammonium iron(II) sulfate solution

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) + S4O6

2- (aq)

Sodium thiosulfate solution (unknown concentration)

Iodine solution of known concentration made by reacting a standard solution of potassium permanganate with acidified

potassium iodide in the conical flask

When iodine solution goes pale yellow...

Add starch indicator Blue/Back - colourless

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 Grams PER LITRE

0.125 x rmm = grams per litre0.125 x 248g = 31

There are 31 g of Na2S2O3.5H2O in one litre.

X RMM

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 Grams PER LITRE

0.1348 x rmm = grams per litre0.1348 x 248g = 33.4304

There are 33.4304 g of Na2S2O3.5H2O in one litre.

X RMM

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 water when it is being diluted.

Dilute sulfuric acid i is harmful to eyes and an irritant to skin.

Bleach i: 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) + S4O6

2- (aq)

Sodium thiosulfate solution (standard solution – last exp **** )

Iodine solution of unknown concentration - made by reaction of sodium hypochlorite

from bleach and potassium iodide under acidic conditions

When iodine solution goes pale yellow...

Add starch indicator Blue/Back - colourless

Calculations1. 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

V1 X M1 = V2 x M2

n1 n2 (20)X (M1) = (26.3) x (0.1) 1 2

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)

I2(aq) + 2S2O32-

(aq) 2I-(aq) + S4O6

2- (aq)

ii) Finding the conc of the sodium thiosulfate in the diluted bleach

• The eqt for the 1st reaction wasClO- + 2I- + 2H+ → Cl- + I2 + H2O

MOLARITY = 0.0658

SO MOLARITY = 0.0658

The concentration of NaOCl in the diluted bleach was 0.0658 moles per litre

• 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%

ii) Finding the w/v % of NaOCl in original bleach

Question 230

V1 X M1 = V2 x M2

n1 n2 (25)X (M1) = (19.56) x (0.25)

1 2

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)

I2(aq) + 2S2O32-

(aq) 2I-(aq) + S4O6

2- (aq)

ii) Finding the conc of the sodium thiosulfate in the diluted bleach

• The eqt for the 1st reaction wasClO- + 2I- + 2H+ → Cl- + I2 + H2O

MOLARITY = 0.0978

SO MOLARITY = 0.0978

The concentration of NaOCl in the diluted bleach was 0.0978 moles per litre

• 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%

ii) Finding the w/v % of NaOCl in original bleach

Question 231

V1 X M1 = V2 x M2

n1 n2 (25)X (M1) = (16.1) x (0.1) 1 2

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)

I2(aq) + 2S2O32-

(aq) 2I-(aq) + S4O6

2- (aq)

ii) Finding the conc of the sodium thiosulfate in the diluted bleach

• The eqt for the 1st reaction wasClO- + 2I- + 2H+ → Cl- + I2 + H2O

MOLARITY = 0.0322

SO MOLARITY = 0.0322

The concentration of NaOCl in the diluted bleach was 0.0322 moles per litre

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 =

e)f) Finding the w/v % of NaOCl 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)

Biochemical Oxygen Demand

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.

The Winkler method

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) + 2S2O3

2-(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 pointAdd 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 V2N1 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

Doing the calculations

1. Why is the reagent MnSO4 used?

2. Why is concentrated H2SO4 used?

3. Why must the bottles be shaken vigorously after adding the Manganese sulfate and alkaline potassium iodide?

4. Why are the bottles completely filled with water?5. 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?

6. State and explain what the letters B.O.D. mean. 7. Why are the bottles used during B.O.D.

measurements stored in the dark?

Questions relating to the experiment

Question 233g

V1 X M1 = V2 x M2

n1 n2 (50)X (M1) = (6) x (0.01) 1 2

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)

I2(aq) + 2S2O32-

(aq) 2I-(aq) + S4O6

2- (aq)

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 2I2

M1 = 0.0006

M1 = 0.0006 /2= .0003

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

e)f) Finding the concentration of oxygen in ppm

Question 235 – Sample 1

V1 X M1 = V2 x M2

n1 n2 (200)X (M1) = (27) x (0.01) 1 2

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)

I2(aq) + 2S2O32-

(aq) 2I-(aq) + S4O6

2- (aq)

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 2I2

M1 = 0.000675

M1 = 0.000675 /2= .0003375

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

e)f) Finding the concentration of oxygen in ppm

Question 235 – Sample 2

V1 X M1 = V2 x M2

n1 n2 (200)X (M1) = (4.8) x (0.01) 1 2

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)

I2(aq) + 2S2O32-

(aq) 2I-(aq) + S4O6

2- (aq)

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 2I2

M1 = 0.00012

M1 = 0.00012 /2= .00006

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

e)f) Finding the concentration of oxygen in ppm

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

n The disodium salt of edta is harmful. Contact with the skin and eyes should be avoided.

The buffer solution of pH 10 is corrosive. Avoid breathing vapours. Use in fume hood.

i Eriochrome Black T is an irritant and should not be allowed to come into contact with the skin.

1. Find the concentration of the Ca+2 ions in the water using results from the titration:

M1 X V1 = M2 X V2N1 N2

2. Find concentration of CaCO3 in the water in g/L and ppm

Doing the calculations

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 hardness2. 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.

It can be removed by:• using an ion exchanger

Removing permanent hardness from water

Deionisation –

Deionisation

In a deioniser any dissolved salts present ae replaced by water molecules.

• Deionised = no ions other than H+ and OH-• Distilled = No dissolved solids of any type

Relative purity ofdeionised and distilled water.

• 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