potassium permanganate titrations. a potassium manganate(vii)/ammonium iron(ii) sulfate titration...

Potassium permanganate Titrations

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

Safety

Potassium manganate(VII) solution is an oxidising agent and can be a skin irritant. If it is washed off, it may leave a brown stain that will slowly disappear. Ammonium iron(II) sulfate is harmful if ingested in quantity, and is an eye irritant.

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

MnO4 ―+ Fe+2 + H+ Mn+2 + Fe+3 + H2O

x + 4(-2) = -1 x – 8 = -1 x = -1+ 8 x = 7

( + 7 ) ( - 2) ( +2) ( +1) ( +2) ( +1 ) ( - 2)

•Each Mn goes down 5 in number (reduction)•RIG – Each Mn is gaining 5 electrons.•It is the oxidising agent

•Each Fe goes down up 1 in number ( oxidation)•OIL – Each Fe is losing 1 electron•It is the reducing agent •Altogether ( 5 Fe) are losing 5 electrons

( +3)

5

Once the oxidation numbers are balanced, Make sure the overall equation still balances...

5 48

Why is ammonium iron(II) sulfate suitable as a primary standard?

Because it is stable and available in a highly pure form.

Adding 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.•

Adding 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

246. 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 litre (b) grams per litreThe equation for the reaction is:

MnO4 ―+ 5Fe+2 + 8H+ Mn+2 + 5 Fe+3 + 4H2O

(a) Find concentration of potassium permanganate solution in moles per litre

• V1 X M1 = V2 x M2

n1 n2

Solution 1 – MnO4 -

V1 = 28.5cm3

M1 = ?

n1 = 1

Solution 2 – Fe+2

V2 = 25cm3

M2 = 0.11

n2 = 5

MnO4 ―+ 5Fe+2 + 8H+ Mn+2 + 5 Fe+3 + 4H2O

Question 235(h)(i) V1 X M1 = V2 x M2

n1 n2

(28.5)X (M1) = (25) x (0.11) 1 5

M1 = (25) x (0.11) x (1)

(5) x (28.5)

M1 = .0193

The concentration of potassium permanganate solution is 0.0193 M (moles per litre)

(ii) What is the concentration in grams per litre?

• Moles PER LITRE Grams PER LITRE

0.0193 x rmm = grams per litre0.0193 x 158g = 3.0494

There are 3.0494g of KMn04 in one litre.

X RMM

Q247.22.5 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 litre(b) grams of ammonium iron(II) sulfate,

(NH4)2SO4.FeSO4.6H2O, per litre.The equation for the reaction is:MnO4 ―+ 5Fe+2 + 8H+ Mn+2 + 5 Fe+3 + 4H2O


• V1 X M1 = V2 x M2

n1 n2


V1 = 22.5cm3

M1 = 0.02M

n1 = 1

Solution 2 – Fe+2

V2 = 25cm3

M2 = ?

n2 = 5

MnO4 ―+ 5Fe+2 + 8H+ Mn+2 + 5 Fe+3 + 4H2O

V1 X M1 = V2 x M2

n1 n2

(22.5)X (0.02) = (25) x (M2) 1 5

0.09 = M2

The concentration of the iron (II) sulfate solution was 0.09M (moles per litre)

(22.5)X (0.02)X (5) = M2

(1) x (25)


• Know Moles PER LITRE Find Grams PER LITRE

0.09 moles x RMM = Grams per litre0.09 mole x 392 = 35.28

There are 35.28 g of (NH4)2SO4.FeSO4.6H2O in one litre.

248• 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 24.65 cm3 of a potassium permanganate solution.

• Calculate the concentration of the potassium permanganate (KMnO4) solution in

• (a) moles per litre (b) grams per litre• The equation for the reaction is:

MnO4 ―+ 5Fe+2 + 8H+ Mn+2 + 5 Fe+3 + 4H2O


• V1 X M1 = V2 x M2

n1 n2


V1 = 24.65cm3

M1 = ?

n1 = 1

Solution 2 – Fe+2

V2 = 25cm3

M2 = ??

n2 = 5

MnO4 ―+ 5Fe+2 + 8H+ Mn+2 + 5 Fe+3 + 4H2O

Finding molarity of Ammonium Iron(II) sulfate solution

(i) Finding grams per litre1. We know that 9.8g are in 250cm3 of solution 2. In one litre of solution there would be (9.8x4) = 39.2g

(ii) Finding moles per litre Grams per litre Moles per litre

39.2g / 392 = 0.1 Moles per litre

There are 0.1 moles of Ammonium Iron (II) sulfate in 1 litre. This is the molarity.

/ RMM


• V1 X M1 = V2 x M2

n1 n2


V1 = 24.65cm3

M1 = ?

n1 = 1

Solution 2 – Fe+2

V2 = 25cm3

M2 = 0.1

n2 = 5

MnO4 ―+ 5Fe+2 + 8H+ Mn+2 + 5 Fe+3 + 4H2O

V1 X M1 = V2 x M2

n1 n2

(24.65)X (M1) = (25) x (0.1) 1 5

M1 = (25) x (0.1) x (1)

(5) x (24.65)

M1 =0.0203

The concentration of potassium permanganate solution is 0.0203 M (moles per litre)




There are 3.2074g of KMn04 in one litre.

X RMM

249 A standard solution of ammonium iron(II) sulfate, (NH4)2SO4.FeSO4.6H2O, wasprepared by dissolving 11.76 g of the crystals in dilute sulfuric acid and making upthe solution to 250 cm3 with deionised water from a washbottle.(b) Calculate the molarity (moles per litre) of this solution.25cm3 of this solution was taken and further acidified with dilute sulfuric acid, Itrequired 33.3 cm3 of a potassium permanganate solution for complete reactionaccording to the equation.(e) Calculate the molarity of the potassium permanganate solution.

MnO4 ―+ 5Fe+2 + 8H+ Mn+2 + 5 Fe+3 + 4H2O

b. Finding molarity of Ammonium Iron(II) sulfate solution

(i) Finding grams per litre1. We know that 11.76g are in 250cm3 of solution 2. In one litre of solution there would be (11.76x4) = 47.04g

(ii) Finding moles per litre Grams per litre/ RMM = moles per litre

47.04/ 392g = 0.12

There are 0.12 moles of Ammonium Iron (II) sulfate in 1 litre. This is the molarity.

MnO4 ―+ Fe+2 + H+ Mn+2 + Fe+3 + H2O

x + 4(-2) = -1 x – 8 = -1 x = -1+ 8 x = 7

( - 7 ) ( - 2) ( +2) ( +1) ( +2) ( +1 ) ( - 2)

•Each Mn goes down 5 in number (reduction)•RIG – Each Mn is gaining 5 electrons.•It is the oxidising agent

•Each Fe goes down up 1 in number ( oxidation)•OIL – Each Fe is losing 1 electron•It is the reducing agent •Altogether ( 5 Fe) are losing 5 electrons

( +3)

5

Once the oxidation numbers are balanced, Make sure the overall equation still balances...

5 48


• V1 X M1 = V2 x M2

n1 n2


V1 = 33.3cm3

M1 = ?

n1 = 1

Solution 2 – Fe+2

V2 = 25cm3

M2 = 0.12

n2 = 5

MnO4 ―+ 5Fe+2 + 8H+ Mn+2 + 5 Fe+3 + 4H2O

V1 X M1 = V2 x M2

n1 n2

(33.3)X (M1) = (25) x () 1 5

M1 = (25) x (0.12) x (1)

(5) x (33.3)

M1 = 0.0180

The concentration of potassium permanganate solution is 0.0180M (moles per litre)

• 250. 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 molarity of the potassium permanganate solutionand its concentration in grams per litre.

(f) Find concentration of potassium permanganate solution in moles per litre

• V1 X M1 = V2 x M2

n1 n2


V1 = 22.45cm3 ( average)

M1 = ?

n1 = 1

Solution 2 – Fe+2

V2 = 25cm3

M2 = 0.1

n2 = 5

MnO4 ―+ 5Fe+2 + 8H+ Mn+2 + 5 Fe+3 + 4H2O

Question 250 f V1 X M1 = V2 x M2

n1 n2

(22.45)X (M1) = (25) x (0.1) 1 5

M1 = (25) x (0.1) x (1)

(5) x (22.45)

M1 = 0.0223

The concentration of potassium permanganate solution is 0.0223M (moles per litre)




There are 3.6814 g of KMn04 in one litre.

X RMM

Mandatory experiment

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

LC 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

252. Six iron tablets of total mass 1.47 g (consisting of iron(II) sulfate)

weredissolved 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 0.015 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+2 + 5 Fe+3 + 4H2O

(a) Find concentration of iron sulfate solution in moles per litre

• V1 X M1 = V2 x M2

n1 n2



M1 = 0.015

n1 = 1

Solution 2 – Fe+2

V2 = 20cm3

M2 = ?

n2 = 5

MnO4 ―+ 5Fe+2 + 8H+ Mn+2 + 5 Fe+3 + 4H2O

V1 X M1 = V2 x M2

n1 n2

(5.47)X (0.015) = (20) x (M2)

1 5

0.8205 = (20) x (M2) (5)

M1 = 0.0205

The concentration of iron(II) sulfate solution is 0.0205 M (moles per litre)

Moles of Iron (II) sulfate in 250cm3?

• 0.0205 moles of Iron(II) sulfate per litre• 0.0205 /4 = .0051 moles per 250cm3

(ii) What is the mass of FeSO4 in each tablet?



X RMM

So in each tablet…

There are 0.7752 / 6= 0.1292g of Iron(II) Sulfate in each tablet

There are 0.7752 g of Iron(II) Sulfate in 250cm3

(ii) What is the mass of iron in each tablet?


0.0051 x rmm = grams per litre0.0051 x 56g = 0.2856g

X RMM

There are 0.2856g of Iron in 250cm3


There are 0.2856 / 6 = 0.0476g of Iron in each tablet

% of iron(II) Sulfate in each tablet?

Mass of iron(II) Sulfate per tablet x 100Total mass per tablet 1

0.1292g x 100 0.245g 1

52.7347%

253In an analysis of iron tablets to determine the amount of iron sulfate

pertablet, five tablets whose total mass was 1.20 g were dissolved in dilutesulfuric acid and the solution made up to 250 cm3 in a volumetric flask.25 cm3 portions of the resulting solution were titrated against 0.015 Mpotassium permanganate. The average titration figure was 5.75 cm3.(a) What would be observed if the student forgot to add the dilutesulfuric 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+2 + 5 Fe+3 + 4H2O


• V1 X M1 = V2 x M2

n1 n2



M1 = 0.015

n1 = 1

Solution 2 – Fe+2

V2 = 25cm3

M2 = ?

n2 = 5

MnO4 ―+ 5Fe+2 + 8H+ Mn+2 + 5 Fe+3 + 4H2O

V1 X M1 = V2 x M2

n1 n2

(5.75)X (0.015) = (25) x (M2)

1 5

0.862 = (25) x (M2) (5)

M1 = 0.01725

The concentration of iron solution is 0.0173 M (moles per litre)

Moles of Iron (II) sulfate in 250cm3

• 0.0173 moles of Iron(II) sulfate per litre• 0.0173/4 = .0043 moles per 250cm3

(ii) What is the mass of FeSO4 in each tablet?

• Moles Grams

0.0043 x rmm = grams0.0043 x 152g = 0.6536

X RMM


There are 0.6536 / 5= 0.1307g of Iron(II) Sulfate in each tablet

There are 0.6536 g of Iron(II) Sulfate in 250cm3




X RMM



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 100Total mass per tablet 1

0.0482g x 100 0.24g 1

20.0833%

Iron tablets may be used in the treatment of anaemia.To analyse the iron(II) content of commercially available iron tablets a studentused four tablets, each of mass 0.360 g, to make up 250 cm3 of solution in avolumetric 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 0.010 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 thetablets required 13.9 cm3 of the 0.010 M KMnO4 solution.Calculate(i) the concentration of the iron(II) solution in moles per litre(ii) the mass of iron(II) in one tablet(iii) the percentage by mass of iron(II) in each tablet.


• V1 X M1 = V2 x M2

n1 n2


V1 = 13.9 cm3 ( average)

M1 = 0.01

n1 = 1

Solution 2 – Fe+2

V2 = 25cm3

M2 = ?

n2 = 5

MnO4 ―+ 5Fe+2 + 8H+ Mn+2 + 5 Fe+3 + 4H2O

V1 X M1 = V2 x M2

n1 n2

(13.9)X (0.01) = (25) x (M2) 1 5

0.139 = (25) x (M2) (5)

M1 = 0.0278

The concentration of iron(II) sulfate solution is 0.0278M (moles per litre)

Moles of Iron in 250cm3?• 0.0278 moles of Iron(II) sulfate per litre0.0278/4 = .0070 moles per 250cm3




X RMM



There are 0.392/ 4 = 0.098g of Iron in each tablet

% of iron in each tablet?

Mass of iron per tablet x 100Total mass per tablet 1

0.098g x 100 0.360g 1

27.2222%

• 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 molarity of this solution was found by titrating it in 25 cm3• volumes against a 0.005 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 themass of iron in each tablet.


• V1 X M1 = V2 x M2

n1 n2


V1 = 17.5 cm3 ( average)

M1 = 0.005

n1 = 1

Solution 2 – Fe+2

V2 = 25cm3

M2 = ?

n2 = 5

MnO4 ―+ 5Fe+2 + 8H+ Mn+2 + 5 Fe+3 + 4H2O

V1 X M1 = V2 x M2

n1 n2

(17.5)X (0.005) = (25) x (M2)

1 5

0.0875 = (25) x (M2)

(5)

M1 = 0.0175

The concentration of iron(II) sulfate solution is 0.0175 M (moles per litre)

Moles of Iron in 250cm3?• 0.0175 moles of Iron per litre0.0175/4 = .0044 moles per 250cm3




X RMM



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

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