Balance the following MnO4-  Mn2+ Fe2+  Fe3+.

Balance the following MnO4-  Mn2+ Fe2+  Fe3+

A potassium manganate(VII)/ammonium iron(II) sulfate titration
Potassium manganate (VII) (potassium permanganate, KMnO4) solution can be standardised by titration against a standard solution of ammonium iron(II) sulfate solution. MnO4- + 8H+ + 5Fe+2  Mn+2 + 5Fe+3 + 4H2O

Primary standard Standards are used in analytical chemistry. A primary standard is typically a reagent which can be weighed easily, and which is so pure that its weight is truly representative of the number of moles of substance contained.

Why is ammonium iron(II) sulfate suitable as a primary standard?
Because it is stable and available in a highly pure form.

Why do we add acid to the iron(II) sulfate solution?
Iron(II) is very susceptible to air oxidation, forming iron(III), under neutral or alkaline conditions but this oxidation is inhibited in the presence of acids.

Which acid is used for the reaction?
The normal source of acid used is dilute sulfuric acid. Sulfuric acid is a good source of H+, and the SO4-2 ions are not reactive. Hydrochloric acid is not suitable as the Cl- ions would be oxidised by the KMnO4 (instead of the Fe+2 ions being oxidised) Nitric acid is not suitable as the NO3- ion is reduced instead of the Mn+7.

Why do we add more acid before the reaction
The iron(II) solution is measured using a pipette and placed in the conical flask The titration is carried out under acidic conditions, so about 10 cm3 of dilute sulfuric acid is added to the Fe+2 solution before the titration

Why is acid needed in the reaction?
because in neutral or alkaline conditions Mn+7 is reduced only as far as Mn+4 which is brown and would make it difficult to determine when the end point of the titration happens. Acidic conditions ensure that Mn+7 is fully reduced to Mn+2

Read from the top of the meniscus
the potassium manganate(VII) solution is placed in a burette. Read from the top of meniscus because the very dark colour of the manganate(VII) solution makes the meniscus difficult to see.

No indicator is needed No indicator is needed, as the manganate(VII) ions are decolourised in the reaction until the end-point, when a pale pink colour persists.

Autocatalysis The reaction is catalysed by a product of the reaction - Mn2+ ions. The first droplet of MnO4- added decolourises slowly. As soon as some Mn2+ is made it acts as a catalyst and speeds up the next reaction – so the next drops of MnO4 decolourise quickly

A solution of potassium permanganate is standardised against a 0
A solution of potassium permanganate is standardised against a 0.11 M iron(II) sulfate solution. 25 cm3 of the iron(II) sulfate solution required 28.5 cm3 of the permanganate solution. Calculate the concentration of the potassium permanganate (KMnO4) solution in (a) moles per dm^3 (b) grams per dm^3 The equation for the reaction is: MnO4 ―+ 5Fe+2 + 8H Mn Fe+3 + 4H2O

(a) Find concentration of potassium permanganate solution in moles per dm^3
MnO4 ―+ 5Fe+2 + 8H Mn Fe+3 + 4H2O V1 X M1 = V2 x M2 n n2 Solution 1 – MnO4 - V1 = 28.5cm3 M1 = ? n1 = 1 Solution 2 – Fe+2 V2 = 25cm3 M2 = 0.11 n2 = 5

Question M1 = (25) x (0.11) x (1) (5) x (28.5) M1 = .0193
V1 X M1 = V2 x M2 n1 n2 (28.5)X (M1) = (25) x (0.11) M1 = (25) x (0.11) x (1) (5) x (28.5) M1 = .0193 The concentration of potassium permanganate solution is M (moles per dm^3)

(ii) What is the concentration in grams per dm^3?
X RMM Moles PER dm^ Grams PER dm^3 x rmm = grams per dm^3 x 158g = There are g of KMn04 in one dm^3.

22.5 cm3 of a solution of 0.02 M KMnO4 solution reacted completely with 25 cm3 of a solution of ammonium iron(II) sulfate. Calculate the concentration of the ammonium iron(II) sulfate solution in (a) moles per dm^3 (b) grams of ammonium iron(II) sulfate, (NH4)2SO4.FeSO4.6H2O, per dm^3. The equation for the reaction is: MnO4 ―+ 5Fe+2 + 8H Mn Fe+3 + 4H2O

MnO4 ―+ 5Fe+2 + 8H Mn Fe+3 + 4H2O V1 X M1 = V2 x M2 n n2 Solution 1 – MnO4 - V1 = 22.5cm3 M1 = 0.02M n1 = 1 Solution 2 – Fe+2 V2 = 25cm3 M2 = ? n2 = 5

V1 X M1 = V2 x M2 n1 n2 (22.5)X (0.02) = (25) x (M2) (22.5)X (0.02)X (5) = M2 (1) x (25) 0.09 = M2 The concentration of the iron (II) sulfate solution was 0.09M (moles per dm^3)

0.09 moles x RMM = Grams per dm^3
(ii) What is the concentration in grams per dm^3? Know Moles PER dm^ Find Grams PER dm^3 0.09 moles x RMM = Grams per dm^3 0.09 mole x 392 = 35.28 There are g of (NH4)2SO4.FeSO4.6H2O in one dm^3.

(a) moles per dm^3 (b) grams per dm^3
A solution of ammonium iron(II) sulfate, (NH4)2SO4.FeSO4.6H2O was made up by dissolving 9.80 g of this crystalline solid in 250 cm3 of acidified solution. 25.0 cm3 of this solution completely reacted with cm3 of a potassium permanganate solution. Calculate the concentration of the potassium permanganate (KMnO4) solution in (a) moles per dm^3 (b) grams per dm^3 The equation for the reaction is: MnO4 ―+ 5Fe+2 + 8H Mn Fe+3 + 4H2O

MnO4 ―+ 5Fe+2 + 8H Mn Fe+3 + 4H2O V1 X M1 = V2 x M2 n n2 Solution 1 – MnO4 - V1 = 24.65cm3 M1 = ? n1 = 1 Solution 2 – Fe+2 V2 = 25cm3 M2 = ?? n2 = 5

Finding concentration of Ammonium Iron(II) sulfate solution
(i) Finding grams per dm^3 We know that 9.8g are in 250cm3 of solution In one dm^3 of solution there would be (9.8x4) = 39.2g (ii) Finding moles per dm^3 Grams per dm^ Moles per dm^3 39.2g / 392 = 0.1 Moles per dm^3 There are 0.1 moles of Ammonium Iron (II) sulfate in 1 dm^3. This is the concentration. / RMM

V1 X M1 = V2 x M2 n1 n2 (24.65)X (M1) = (25) x (0.1) M1 = (25) x (0.1) x (1) (5) x (24.65) M1 =0.0203 The concentration of potassium permanganate solution is M (moles per dm^3)

A standard solution of ammonium iron(II) sulfate, (NH4)2SO4. FeSO4
A standard solution of ammonium iron(II) sulfate, (NH4)2SO4.FeSO4.6H2O, was prepared by dissolving g of the crystals in dilute sulfuric acid and making up the solution to 250 cm3 with deionised water from a washbottle. (b) Calculate the concentration (moles per dm^3) of this solution. 25cm3 of this solution was taken and further acidified with dilute sulfuric acid, It required 33.3 cm3 of a potassium permanganate solution for complete reaction according to the equation. (e) Calculate the concentration of the potassium permanganate solution. MnO4 ―+ 5Fe+2 + 8H Mn Fe+3 + 4H2O

b. Finding concentration of Ammonium Iron(II) sulfate solution
(i) Finding grams per dm^3 We know that 11.76g are in 250cm3 of solution In one dm^3 of solution there would be (11.76x4) = 47.04g (ii) Finding moles per dm^3 Grams per dm^3/ RMM = moles per dm^3 47.04/ 392g = 0.12 There are 0.12 moles of Ammonium Iron (II) sulfate in 1 dm^3. This is the concentration.

V1 X M1 = V2 x M2 n1 n2 (33.3)X (M1) = (25) x () M1 = (25) x (0.12) x (1) (5) x (33.3) M1 = The concentration of potassium permanganate solution is M (moles per dm^3)

A solution of potassium manganate(VII) was prepared by a group of students. They then wished to standardise this solution against a 0.1 M standard solution of ammonium iron(II) sulfate,(NH4)2SO4.FeSO4.6H2O solution. The potassium manganate(VII) solution (KMnO4) was placed into a burette and titrated against 25 cm3 volumes of ammonium iron(II) sulfate. The titration results were 22.8 cm3, 22.4 cm3 and 22.5 cm3 Calculate the concentration of the potassium permanganate solutionand its concentration in grams per dm^3.

Find concentration of potassium permanganate solution in moles per dm^3
MnO4 ―+ 5Fe+2 + 8H Mn Fe+3 + 4H2O V1 X M1 = V2 x M2 n n2 Solution 1 – MnO4 - V1 = 22.45cm3 ( average) M1 = ? n1 = 1 Solution 2 – Fe+2 V2 = 25cm3 M2 = 0.1 n2 = 5

V1 X M1 = V2 x M2 n1 n2 (22.45)X (M1) = (25) x (0.1) M1 = (25) x (0.1) x (1) (5) x (22.45) M1 = The concentration of potassium permanganate solution is M (moles per dm^3)

Experiment Determination of the amount of iron in an iron tablet by titration against a standardised potassium permanganate solution

Past paper questions on this titration

Why are iron tablets sometimes prescribed?
To treat anaemia (iron deficiency)

Why must potassium permanganate solutions be standardised?
Potassium permanganate is unstable and pure and so is not a primary standard

Why should a potassium permanganate solution be standardised immediately before use?
It is unstable and will start to breakdown in the presence of light

What reagent must be used to standardise potassium permangante solution?
Ammonium Iron(II) sulfate

Why is it important to use the dilute sulfuric acid as well as the deionised water when making up the solution from the tablets? Sulfuric acid prevents Iron (II) being oxidised to Iron(III) by the air

Describe the procedure for making up the 250cm3 solution from the tablets
Weigh out on a clock glass Transfer to a beaker of dilute sulfuric acid Add rinsings to beaker Stir to dissolve Add to a volumetric flask with a funnel Add rinsings from rod and beaker Fill with deionised water near to the calibration mark Add deionised water dropwise until the bottom of the meniscus reaches the calibration mark Read at eye level/ on a vertical surface Invert 20 times to make a homogenous mixture

Explain why dilute sulfuric acid must be added to the titration flask before the starting?
Its stops Mn(7) being converted to Mn(4) which is brown and makes it difficult to see colour changes in the reaction. This may give an inaccurate titre result

How was the end point detected?
A permanent pink colour in the conical flask

Question Six iron tablets of total mass 1.47 g (consisting of iron(II) sulfate) were dissolved in dilute sulfuric acid and deionised water and the solution was made up to 250 cm3 in a volumetric flask. 20 cm3 portions of the resulting solution were titrated against M potassium permanganate. The average titration figure was 5.47 cm3. (b) Calculate: (i) the mass of anhydrous FeSO4 in each tablet, (ii) the mass of iron in each tablet and (iii) the percentage of FeSO4 in each tablet. The equation for the reaction is: MnO4 ―+ 5Fe+2 + 8H Mn Fe+3 + 4H2O

(a) Find concentration of iron sulfate solution in moles per dm^3
MnO4 ―+ 5Fe+2 + 8H Mn Fe+3 + 4H2O V1 X M1 = V2 x M2 n n2 Solution 1 – MnO4 - V1 = 5.47cm3 ( average) M1 = 0.015 n1 = 1 Solution 2 – Fe+2 V2 = 20cm3 M2 = ? n2 = 5

0.8205 = (20) x (M2) (5) M1 = 0.0205 (5.47)X (0.015) = (20) x (M2) 1 5
V1 X M1 = V2 x M2 n1 n2 (5.47)X (0.015) = (20) x (M2) = (20) x (M2) (5) M1 = The concentration of iron(II) sulfate solution is M (moles per dm^3)

Moles of Iron (II) sulfate in 250cm3?
moles of Iron(II) sulfate per dm^3 /4 = moles per 250cm3

There are 0.7752 g of Iron(II) Sulfate in 250cm3
(ii) What is the mass of FeSO4 in each tablet? X RMM Moles PER dm^ Grams PER dm^3 x rmm = grams per dm^3 x 152g = There are g of Iron(II) Sulfate in 250cm3 So in each tablet… There are / 6= g of Iron(II) Sulfate in each tablet

There are 0.2856 / 6 = 0.0476g of Iron in each tablet
(ii) What is the mass of iron in each tablet? X RMM Moles PER dm^ Grams PER dm^3 x rmm = grams per dm^3 x 56g = g There are g of Iron in 250cm3 So in each tablet… There are / 6 = g of Iron in each tablet

% of iron(II) Sulfate in each tablet?
Mass of iron(II) Sulfate per tablet x 100 Total mass per tablet 0.1292g x 100 0.245g %

In an analysis of iron tablets to determine the amount of iron sulfate per
tablet, five tablets whose total mass was 1.20 g were dissolved in dilute sulfuric acid and the solution made up to 250 cm3 in a volumetric flask. 25 cm3 portions of the resulting solution were titrated against M potassium permanganate. The average titration figure was 5.75 cm3. (a) What would be observed if the student forgot to add the dilute sulfuric acid? (b) Calculate: (i) the mass of anhydrous FeSO4 in each tablet, (ii) the mass of iron in each tablet and (iii) the percentage of FeSO4 in each tablet. The equation for the reaction is: MnO4 ―+ 5Fe+2 + 8H Mn Fe+3 + 4H2O

MnO4 ―+ 5Fe+2 + 8H Mn Fe+3 + 4H2O V1 X M1 = V2 x M2 n n2 Solution 1 – MnO4 - V1 = 5.75cm3 ( average) M1 = 0.015 n1 = 1 Solution 2 – Fe+2 V2 = 25cm3 M2 = ? n2 = 5

0.862 = (25) x (M2) (5) M1 = 0.01725 (5.75)X (0.015) = (25) x (M2) 1 5
V1 X M1 = V2 x M2 n1 n2 (5.75)X (0.015) = (25) x (M2) = (25) x (M2) (5) M1 = The concentration of iron solution is M (moles per dm^3)

Moles of Iron (II) sulfate in 250cm3
moles of Iron(II) sulfate per dm^3 0.0173/4 = moles per 250cm3

There are 0.6536 g of Iron(II) Sulfate in 250cm3
(ii) What is the mass of FeSO4 in each tablet? X RMM Moles Grams x rmm = grams x 152g = There are g of Iron(II) Sulfate in 250cm3 So in each tablet… There are / 5= g of Iron(II) Sulfate in each tablet

There are 0.2408 / 5 = 0.0482g of Iron in each tablet

% of iron(II) Sulfate in each tablet?
Mass of iron(II) Sulfate per tablet x 100 Total mass per tablet 0.0482g x 100 0.24g %

Iron tablets may be used in the treatment of anaemia
Iron tablets may be used in the treatment of anaemia. To analyse the iron(II) content of commercially available iron tablets a student used four tablets, each of mass g, to make up 250 cm3 of solution in a volumetric flask using dilute sulfuric acid and deionised water. About 15 cm3 of dilute sulfuric acid was added to 25 cm3 portions of this iron(II) solution and the mixture then titrated with a M solution of potassium manganate(VII), KMnO4.The titration reaction is described by the equation MnO4 ― + 5Fe2+ + 8H+  Mn2+ + 5Fe3+ + 4H2O (e) In the titrations the 25 cm3 portions of the iron(II) solution made from the tablets required 13.9 cm3 of the M KMnO4 solution. Calculate (i) the concentration of the iron(II) solution in moles per dm^3 (ii) the mass of iron(II) in one tablet (iii) the percentage by mass of iron(II) in each tablet.

MnO4 ―+ 5Fe+2 + 8H Mn Fe+3 + 4H2O V1 X M1 = V2 x M2 n n2 Solution 1 – MnO4 - V1 = 13.9 cm3 ( average) M1 = 0.01 n1 = 1 Solution 2 – Fe+2 V2 = 25cm3 M2 = ? n2 = 5

0.139 = (25) x (M2) (5) M1 = 0.0278 (13.9)X (0.01) = (25) x (M2) 1 5

Moles of Iron in 250cm3? 0.0278 moles of Iron(II) sulfate per dm^3
0.0278/4 = moles per 250cm3

There are 0.392/ 4 = 0.098g of Iron in each tablet
(ii) What is the mass of iron in each tablet? X RMM Moles PER dm^ Grams PER dm^3 x rmm = grams per dm^3 x 56g = 0.392g There are 0.392g of Iron in 250cm3 So in each tablet… There are 0.392/ 4 = 0.098g of Iron in each tablet

% of iron in each tablet? Mass of iron per tablet x 100
Total mass per tablet 0.098g x 100 0.360g %

iron tablet. A 250 cm3 volume of solution containing five tablets
The following experiment was carried out to find the mass of iron in an iron tablet. A 250 cm3 volume of solution containing five tablets dissolved in dilute sulfuric acid was carefully made up in a volumetric flask. The concentration of this solution was found by titrating it in 25 cm3 volumes against a M solution of potassium manganate(VII) which had been previously standardised. The potassium manganate(VII) solution was put in the burette and a number of titrations were carried out. The average titration figure was 17.5 cm3. d) Assuming that the tablets are exactly equal in mass, calculate the mass of iron in each tablet.

MnO4 ―+ 5Fe+2 + 8H Mn Fe+3 + 4H2O V1 X M1 = V2 x M2 n n2 Solution 1 – MnO4 - V1 = 17.5 cm3 ( average) M1 = 0.005 n1 = 1 Solution 2 – Fe+2 V2 = 25cm3 M2 = ? n2 = 5

0.0875 = (25) x (M2) (5) M1 = 0.0175 (17.5)X (0.005) = (25) x (M2) 1 5

Moles of Iron in 250cm3? 0.0175 moles of Iron per dm^3
0.0175/4 = moles per 250cm3

There are 0.2464/ 4 = 0.0493g of Iron in each tablet

Sodium Thiosulphate (Na2S2O3)
Sodium thiosulphate is an important reducing agent used in chemistry Used in developing photographs and in analytical chemistry

Most commonly encountered as a colourless
crystalline solid of the form: Na2S2O3.5H2O (pentahydrate)

Reaction of Iodine and Thiosulphate Ions
Sodium thiosulphate is a reducing agent and reacts with iodine molecules (I2), to convert them to iodide (I- ions): I2(aq) S2 O32-(aq) S4 O62-(aq) I-(aq) Thiosulphate Ion Tetrathionate Ion Iodide Ion Iodine +2 +2.5 -1 Oxidising Agent Reducing Agent

Ratio I2 : S2O32- 1 :

Thiosulphate Ion S2O32- Tetrathionate Ion S4O62-

Standardising Sodium Thiosulphate using Iodine
Sodium Thiosulphate is NOT usually a primary standard because it is difficult to obtain in the anhydrous form (no water) and the crystals lose water of crystallisation to the atmosphere and so are unstable. However, it is possible to determine the concentration of a sodium thiosulphate solution by titrating it against a standard iodine solution: Standardising Sodium Thiosulphate using Iodine

Iodine is NOT usually a primary standard. A standard
solution of iodine cannot be made up by direct weighing because: - it is too volatile (vaporises slightly at room temp.) - it does NOT dissolve in water A STANDARD SOLUTION of IODINE (0.06M) is made up by reacting a standard solution of acidified potassium permanganate (0.02M) with excess potassium iodide.

Reaction equations for formation of standard
solution of iodine MnO H e Mn H2O (Each permanganate ion needs 5e-) 2I I2 + 2e- (Each iodide loses 1e- (2I- = 2e- and forms iodine as a result)

Upon inspection of equations A & B it is clear to see that every MnO4- ion must receive 5e- in order for the reaction to go to completion. But the 2I- only results in the loss of 2e-, so in order to balance the e- gain and loss we must: (A) x2: 2MnO H e Mn H2O (B) x5: 10I I e- Adding these two half equations: 2MnO I- + 16H Mn I H2O

In the previous equation the I- ions (from potassium iodide) are kept in excess. As a result the amount of iodine formed in this reaction depends on the amount of permanganate present. So, by using a standard solution of KMnO4 we can determine the amount of iodine formed. This known amount of iodine formed can then be used to standardise a sodium thiosulphate solution

**Note: Iodine is non-polar and water is a polar solvent. For this reason iodine will NOT dissolve in water but will dissolve in a solution of potassium iodide (KI). Adding I2 to KI: - I2 molecule reacts with I- to form I3- (tri-iodide ion) I2 + I I3- state of equilibrium - I3- is identical in its chemical behaviour to I2 and because of its charge it dissolves easily in water - As a result a solution of iodine in potassium iodide may be treated as if it were a solution of iodine in water

Determining the end point of an iodine/thiosulphate titration
As thiosulphate is added to the iodine the reddish/brown colour gradually turns yellow which gradually turns paler and paler until it turns colourless – very difficult to detect end point. Therefore need to use an indicator – starch indicator I2(aq) S2 O32-(aq) S4O62-(aq) I-(aq) (Colourless) Reddish Brown Colourless

Starch forms a blue-black colour with iodine
therefore when ALL of the iodine has reacted with the thiosulphate the blue-black colour suddenly goes colourless. Note: starch is only added when most of iodine has disappeared i.e. when reddish/brown colour of iodine has gone pale(straw) yellow. Why??? If too much added or added too early starch participates in rxn and removes to much iodine When pale colour is observed it indicates end point is near

Colour Change Reddish/brown Straw Yellow Blue/Black Colourless
End Point Colour Change Add Starch

Points to Note: A standard solution of sodium thiosulphate can be used to standardise an iodine solution In an iodine-thiosulphate titration: - iodine in conical flask - sodium thiosulphate in burette Starch indicator must be prepared fresh as it deteriorates quickly on standing. A 0.02M standard solution of potassium iodate may be used in place of potassium permanganate in preparing the iodine solution.

To prepare a solution of sodium thiosulphate and standardise it by titration against a solution of iodine Two reactions: Formation of iodine in conical flask: 2. Main rxn – Reaction of Iodine with Sodium Thiosulphate So, 2MnO4- : 5I2 : 10S2O32- Ratio : 2MnO I- + 16H Mn I H2O I2(aq) S2O32-(aq) S4O62-(aq) I-(aq)

Applications of Iodine –Thiosulphate Titrations
Most household bleaches contain chlorine in the form of sodium hypochlorite, NaClO. NaClO contains the hypochlorite ion (chlorate(I) ion) – ClO- Hypochlorite ion is an oxidising agent and converts iodide ions to iodine: ClO- + 2I- + 2H Cl- + I2 + H2O I2(aq) + 2S2O32-(aq) S4O62-(aq) + 2I-(aq)

Ratio: ClO- : I2 : 2S2 O32- 1 : : Where, ClO- is the oxidising agent and 2S2O32- is the reducing agent

Balance the following MnO4-  Mn2+ Fe2+  Fe3+.

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Balance the following MnO4-  Mn2+ Fe2+  Fe3+.

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