Potassium permanganate Titrations. A potassium manganate(VII)/ammonium iron(II) sulfate titration...

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Transcript of Potassium permanganate Titrations. A potassium manganate(VII)/ammonium iron(II) sulfate titration...

  • Slide 1
  • Potassium permanganate Titrations
  • Slide 2
  • A potassium manganate(VII)/ammonium iron(II) sulfate titration Potassium manganate (VII) (potassium permanganate, KMnO 4 ) solution can be standardised by titration against a standard solution of ammonium iron(II) sulfate solution. MnO 4 - + 8H + + 5Fe +2 Mn +2 + 5Fe +3 + 4H 2 O
  • Slide 3
  • Safety
  • Slide 4
  • MnO 4 + Fe +2 + H + Mn +2 + Fe +3 + H 2 O 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 4 8
  • Slide 5
  • Why is ammonium iron(II) sulfate suitable as a primary standard? Because it is stable and available in a highly pure form.
  • Slide 6
  • 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.
  • Slide 7
  • 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 SO 4 -2 ions are not reactive. Hydrochloric acid is not suitable as the Cl- ions would be oxidised by the KMnO 4 (instead of the Fe +2 ions being oxidised) Nitric acid is not suitable as the NO 3 - ion is reduced instead of the Mn +7.
  • Slide 8
  • 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 cm 3 of dilute sulfuric acid is added to the Fe +2 solution before the titration
  • Slide 9
  • 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
  • Slide 10
  • 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.
  • Slide 11
  • 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.
  • Slide 12
  • Autocatalysis The reaction is catalysed by a product of the reaction - Mn 2+ ions. The first droplet of MnO 4 - added decolourises slowly. As soon as some Mn 2+ is made it acts as a catalyst and speeds up the next reaction so the next drops of MnO 4 decolourise quickly
  • Slide 13
  • 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: MnO 4 + 5Fe +2 + 8H + Mn +2 + 5 Fe +3 + 4H 2 O
  • Slide 14
  • ( a) Find concentration of potassium permanganate solution in moles per litre V 1 X M 1 = V 2 x M 2 n 1 n 2 Solution 1 MnO 4 - V 1 = 28.5cm 3 M 1 = ? n 1 = 1 Solution 2 Fe +2 V 2 = 25cm 3 M 2 = 0.11 n 2 = 5 MnO 4 + 5Fe +2 + 8H + Mn +2 + 5 Fe +3 + 4H 2 O
  • Slide 15
  • Question 235(h)(i) V 1 X M 1 = V 2 x M 2 n 1 n 2 (28.5)X (M 1 ) = (25) x ( 0.11 ) 1 5 M 1 = (25) x (0.11) x (1) (5) x (28.5) M 1 =.0193 The concentration of potassium permanganate solution is 0.0193 M (moles per litre)
  • Slide 16
  • (ii) What is the concentration in grams per litre? Moles PER LITRE Grams PER LITRE 0.0193 x rmm = grams per litre 0.0193 x 158g = 3.0494 There are 3.0494g of KMn0 4 in one litre. X RMM
  • Slide 17
  • 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: MnO 4 + 5Fe +2 + 8H + Mn +2 + 5 Fe +3 + 4H 2 O
  • Slide 18
  • ( a) Find concentration of potassium permanganate solution in moles per litre V 1 X M 1 = V 2 x M 2 n 1 n 2 Solution 1 MnO 4 - V 1 = 22.5cm 3 M 1 = 0.02M n 1 = 1 Solution 2 Fe +2 V 2 = 25cm 3 M 2 = ? n 2 = 5 MnO 4 + 5Fe +2 + 8H + Mn +2 + 5 Fe +3 + 4H 2 O
  • Slide 19
  • V 1 X M 1 = V 2 x M 2 n 1 n 2 (22.5)X (0.02) = (25) x ( M 2 ) 1 5 0.09 = M 2 The concentration of the iron (II) sulfate solution was 0.09M (moles per litre) (22.5)X (0.02)X (5) = M 2 (1) x (25)
  • Slide 20
  • (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 35.28 g of (NH 4 ) 2 SO 4.FeSO 4.6H 2 O in one litre.
  • Slide 21
  • 248 A solution of ammonium iron(II) sulfate, (NH 4 ) 2 SO 4.FeSO 4.6H 2 O 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: MnO 4 + 5Fe +2 + 8H + Mn +2 + 5 Fe +3 + 4H 2 O
  • Slide 22
  • ( a) Find concentration of potassium permanganate solution in moles per litre V 1 X M 1 = V 2 x M 2 n 1 n 2 Solution 1 MnO 4 - V 1 = 24.65cm 3 M 1 = ? n 1 = 1 Solution 2 Fe +2 V 2 = 25cm 3 M 2 = ?? n 2 = 5 MnO 4 + 5Fe +2 + 8H + Mn +2 + 5 Fe +3 + 4H 2 O
  • Slide 23
  • Finding molarity of Ammonium Iron(II) sulfate solution (i) Finding grams per litre 1.We know that 9.8g are in 250cm 3 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
  • Slide 24
  • ( a) Find concentration of potassium permanganate solution in moles per litre V 1 X M 1 = V 2 x M 2 n 1 n 2 Solution 1 MnO 4 - V 1 = 24.65cm 3 M 1 = ? n 1 = 1 Solution 2 Fe +2 V 2 = 25cm 3 M 2 = 0.1 n 2 = 5 MnO 4 + 5Fe +2 + 8H + Mn +2 + 5 Fe +3 + 4H 2 O
  • Slide 25
  • V 1 X M 1 = V 2 x M 2 n 1 n 2 (24.65)X (M 1 ) = (25) x (0.1) 1 5 M 1 = (25) x (0.1) x (1) (5) x (24.65) M 1 =0.0203 The concentration of potassium permanganate solution is 0.0203 M (moles per litre)
  • Slide 26
  • (ii) What is the concentration in grams per litre? Moles PER LITRE Grams PER LITRE 0.0203 x rmm = grams per litre 0.0203 x 158g = 3.2074 There are 3.2074g of KMn0 4 in one litre. X RMM
  • Slide 27
  • 249 A standard solution of ammonium iron(II) sulfate, (NH4)2SO4.FeSO4.6H2O, was prepared by dissolving 11.76 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. MnO 4 + 5Fe +2 + 8H + Mn +2 + 5 Fe +3 + 4H 2 O
  • Slide 28
  • b. Finding molarity of Ammonium Iron(II) sulfate solution (i) Finding grams per litre 1.We know that 11.76g are in 250cm 3 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.
  • Slide 29
  • MnO 4 + Fe +2 + H + Mn +2 + Fe +3 + H 2 O 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 4 8
  • Slide 30
  • ( a) Find concentration of potassium permanganate solution in moles per litre V 1 X M 1 = V 2 x M 2 n 1 n 2 Solution 1 MnO 4 - V 1 = 33.3cm 3 M 1 = ? n 1 = 1 Solution 2 Fe +2 V 2 = 25cm 3 M 2 = 0.12 n 2 = 5 MnO 4 + 5Fe +2 + 8H + Mn +2 + 5 Fe +3 + 4H 2 O
  • Slide 31
  • V 1 X M 1 = V 2 x M 2 n 1 n 2 (33.3)X (M 1 ) = (25) x () 1 5 M 1 = (25) x (0.12) x (1) (5) x (33.3) M 1 = 0.0180 The concentration of potassium permanganate solution is 0.0180M (moles per litre)
  • Slide 32
  • 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 solu