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**Potassium permanganate Titrations**

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** 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

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Safety

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**MnO4 ―+ Fe+2 + H+ Mn+2 + Fe+3 + H2O**

Once the oxidation numbers are balanced, Make sure the overall equation still balances... Each Mn goes down 5 in number (reduction) RIG – Each Mn is gaining 5 electrons. It is the oxidising agent MnO4 ―+ Fe+2 + H+ Mn Fe H2O 5 8 5 4 ( + 7 ) ( - 2) ( +2) ( +1) ( +2) ( +3) ( +1 ) ( - 2) x + 4(-2) = -1 x – 8 = -1 x = -1+ 8 x = 7 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

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

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

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**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.

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**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.

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**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

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**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

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**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.

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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.

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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

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**246. A solution of potassium permanganate is standardised against a 0**

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 litre The equation for the reaction is: MnO4 ―+ 5Fe+2 + 8H Mn Fe+3 + 4H2O

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

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

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**Question 235(h)(i) 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 litre)

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**0.0193 x rmm = grams per litre 0.0193 x 158g = 3.0494**

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

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Q 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 Fe+3 + 4H2O

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

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

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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 litre)

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**0.09 moles x RMM = Grams per litre**

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

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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 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 Fe+3 + 4H2O

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

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

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**Finding molarity of Ammonium Iron(II) sulfate solution**

(i) Finding grams per litre We know that 9.8g are in 250cm3 of solution 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

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

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 = 0.1 n2 = 5

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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 litre)

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**0.0203 x rmm = grams per litre 0.0203 x 158g = 3.2074**

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

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249 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 molarity (moles per litre) 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 molarity of the potassium permanganate solution. MnO4 ―+ 5Fe+2 + 8H Mn Fe+3 + 4H2O

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

(i) Finding grams per litre We know that 11.76g are in 250cm3 of solution 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.

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**MnO4 ―+ Fe+2 + H+ Mn+2 + Fe+3 + H2O**

Each Mn goes down 5 in number (reduction) RIG – Each Mn is gaining 5 electrons. It is the oxidising agent Once the oxidation numbers are balanced, Make sure the overall equation still balances... MnO4 ―+ Fe+2 + H+ Mn Fe H2O 5 8 5 4 ( - 7 ) ( - 2) ( +2) ( +1) ( +2) ( +3) ( +1 ) ( - 2) x + 4(-2) = -1 x – 8 = -1 x = -1+ 8 x = 7 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

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

MnO4 ―+ 5Fe+2 + 8H Mn Fe+3 + 4H2O V1 X M1 = V2 x M2 n n2 Solution 1 – MnO4 - V1 = 33.3cm3 M1 = ? n1 = 1 Solution 2 – Fe+2 V2 = 25cm3 M2 = 0.12 n2 = 5

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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 litre)

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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.

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

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

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**Question 250 f M1 = (25) x (0.1) x (1) (5) x (22.45) M1 = 0.0223**

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 litre)

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**0.0223 x rmm = grams per litre 0.0223 x 158g = 3.6814**

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

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

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**LC Past paper questions on this titration**

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**Why are iron tablets sometimes prescribed?**

To treat anaemia (iron deficiency)

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**Why must potassium permanganate solutions be standardised?**

Potassium permanganate is unstable and pure and so is not a primary standard

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**Why should a potassium permanganate solution be standardised immediately before use?**

It is unstable and will start to breakdown in the presence of light

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**What reagent must be used to standardise potassium permangante solution?**

Ammonium Iron(II) sulfate

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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

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**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

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**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

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**How was the end point detected?**

A permanent pink colour in the conical flask

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252. 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

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

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

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**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 litre)

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**Moles of Iron (II) sulfate in 250cm3?**

moles of Iron(II) sulfate per litre /4 = moles per 250cm3

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**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 LITRE Grams PER LITRE x rmm = grams per litre 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

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**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 LITRE Grams PER LITRE x rmm = grams per litre x 56g = g There are g of Iron in 250cm3 So in each tablet… There are / 6 = g of Iron in each tablet

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**% 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 %

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253 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

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

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

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**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 litre)

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**Moles of Iron (II) sulfate in 250cm3**

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

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**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

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

(ii) What is the mass of iron in each tablet? X RMM Moles PER LITRE Grams PER LITRE x rmm = grams per litre x 56g = g There are g of Iron in 250cm3 So in each tablet… There are / 5 = g of Iron in each tablet

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**% 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 %

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**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 litre (ii) the mass of iron(II) in one tablet (iii) the percentage by mass of iron(II) in each tablet.

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

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

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

V1 X M1 = V2 x M2 n1 n2 (13.9)X (0.01) = (25) x (M2) = (25) x (M2) (5) M1 = The concentration of iron(II) sulfate solution is M (moles per litre)

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**Moles of Iron in 250cm3? 0.0278 moles of Iron(II) sulfate per litre**

0.0278/4 = moles per 250cm3

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**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 LITRE Grams PER LITRE x rmm = grams per litre 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

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**% of iron in each tablet? Mass of iron per tablet x 100**

Total mass per tablet 0.098g x 100 0.360g %

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**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 molarity 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.

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

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

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

V1 X M1 = V2 x M2 n1 n2 (17.5)X (0.005) = (25) x (M2) = (25) x (M2) (5) M1 = The concentration of iron(II) sulfate solution is M (moles per litre)

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**Moles of Iron in 250cm3? 0.0175 moles of Iron per litre**

0.0175/4 = moles per 250cm3

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

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

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