Mr Cole Chemistry

Redox Back Titrations Questions

  • Questions
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  1. A scientist wants to find out how much hydrated zinc hydroxide (Zn(H2O)4(OH)2) is present in a sample of absorbing agent. The scientist takes a 5.00g sample of absorbing agent and adds it to 250cm3 of 0.5mol dm-3 hydrochloric acid. Then, a 25.0 cm3 quotient of the remaining mixture was titrated against 0.400 mol dm-3 sodium hydroxide, requiring 18.6cm3 of solution to react. What percentage by mass of the absorbing agent is zinc hydroxide?
  2. A scientist investigated the percentage by mass of anhydrous chromium(II) sulfate in a sample of a metal coating agent. They dissolved a 0.400 g sample of the agent in 50.0 cm³ of excess 0.100 mol dm⁻³ acidified potassium permanganate(VII), ensuring all the chromium(II) ions were oxidised to chromium(III) ions. The excess potassium permanganate(VII) was then determined by titration.

    The resulting solution was then titrated against a 0.800 mol dm⁻³ solution of iron(II) sulfate. Iron(II) ions reduce permanganate(VII) ions in acidic conditions. The endpoint of the titration was determined using a suitable indicator. 30.0 cm³ of the iron(II) sulfate solution was required to completely react with the excess potassium permanganate(VII).
  3. A sample of group 2 metal ethanedioate (MC2O4) is analysed to determine the identity of the metal. A 5.25g sample of the metal ethanedioate is reacted against 100 cm3 of 0.2 moldm-3 potassium manganate. A 25cm3 quotient from the remaining solution is titrated against Periodic acid (HIO4) where the IO4– ions are reduced to IO3–, and the Manganese(II) ions are oxidised to manganate(VII) ions. In the titration, 22.2cm3 of 0.2moldm-3 HIO4 was required to react with the manganese ions. State the identity of the metal in the metal ethanedioate.
  4. The moles of fluorine gas in a 100 dm3 container is determined by a scientist. The scientist first takes a 1dm3 sample from container and reacts it with 32.0g of powdered iron. The mixture of iron and iron fluoride was mixed with water and filtered to remove the iron fluoride and isolate the remaining iron. The remaining iron was reacted with 250cm3 of sulfuric acid to form Fe2+ ions. The sulfuric acid was in excess. A 25.0 cm3 sample of the iron sulfate formed was titrated against potassium manganate, where it was found to require 24.9cm3 of 0.200 mol dm-3 of potassium manganate to fully react.
  5. A sample of anhydrous Iron (II) chloride was analysed by back titration. A 5.00g sample of iron(II) chloride was reacted with 200cm3 of 0.100 mol dm-3 potassium manganate(VII). Note that both the chloride and iron(II) ions are oxidised by the manganate. The excess potassium manganate was then portioned into 25cm3 aliquots, and these were found to fully react with 24.0cm3 of 0.1mol dm-3 potassium ethanedioate. What was the purity of the iron(II) chloride by percentage purity?
  1. A scientist wants to find out how much hydrated zinc hydroxide (Zn(H2O)4(OH)2) is present in a sample of absorbing agent. The scientist takes a 5.00g sample of absorbing agent and adds it to 250cm3 of 0.5mol dm-3 hydrochloric acid. Then, a 25.0 cm3 quotient of the remaining mixture was titrated against 0.400 mol dm-3 sodium hydroxide, requiring 18.6cm3 of solution to react. What percentage by mass of the absorbing agent is zinc hydroxide?
    Moles of NaOH: (18.6/1000) x 0.4 = 0.00744
    Moles if HCl in sample: 1:1 therefore 0.00744
    Moles of HCl in 250cm3: 0.0744
    Moles of HCl before the reaction: (250/1000) x 0.5 = 0.125
    Moles of HCl used up: 0.125 – 0.0744 = 0.0506
    Moles of Zn(H2O)4(OH)2: 2:1 therefore 0.0506 / 2 = 0.0253
    Mass of Zn(H2O)4(OH)2: 0.0253 x 135.4 = 3.43g
    Percentage: 3.43 / 5.00 = 68.5
  2. A scientist investigated the percentage by mass of anhydrous chromium(II) sulfate in a sample of a metal coating agent. They dissolved a 0.400 g sample of the agent in 50.0 cm³ of excess 0.100 mol dm⁻³ acidified potassium permanganate(VII), ensuring all the chromium(II) ions were oxidised to chromium(III) ions. The excess potassium permanganate(VII) was then determined by titration.

    The resulting solution was then titrated against a 0.800 mol dm⁻³ solution of iron(II) sulfate. Iron(II) ions reduce permanganate(VII) ions in acidic conditions. The endpoint of the titration was determined using a suitable indicator. 30.0 cm³ of the iron(II) sulfate solution was required to completely react with the excess potassium permanganate(VII).
    Equation 1: MnO₄⁻ + 8H⁺ + 5Cr2+ → Mn²⁺ + 4H₂O + 5Cr3+
    Equation 2: MnO₄⁻ + 8H⁺ + 5Fe2+ → Mn²⁺ + 4H₂O + 5Fe3+
    Moles of Fe2+: (30.0/1000) x 0.800 = 0.024
    Moles of excess MnO4–: 0.024 / 5 = 0.0048
    Moles of MnO4– before reacting: (50/1000) x 0.2 = 0.01
    Moles of MnO4– used up: 0.01 – 0.0048 = 0.0052
    Moles of Cr2+: 0.0052 x 5 = 0.026Mass of CrSO4: 0.026 x 148.1 = 3.85
    Percentage mass of CrSO4: 96.3%
  3. A sample of group 2 metal ethanedioate (MC2O4) is analysed to determine the identity of the metal. A 5.25g sample of the metal ethanedioate is reacted against 100 cm3 of 0.2 moldm-3 potassium manganate. A 25cm3 quotient from the remaining solution is titrated against Periodic acid (HIO4) where the IO4– ions are reduced to IO3–, and the Manganese(II) ions are oxidised to manganate(VII) ions. In the titration, 22.2cm3 of 0.2moldm-3 HIO4 was required to react with the manganese ions. State the identity of the metal in the metal ethanedioate.
    Equation 1: 2MnO4– + 16H+ 5C2O42- –> 2Mn2+ + 8H2O + 10CO2
    Equation 2: 2Mn2+ + 3H2O + 5IO4– –> 2MnO4– + 6H+ + 5IO3–
    Moles of IO4–: (22.2/1000) x 0.3 = 0.00666
    Moles of Mn2+in quotient: 5:2 ratio, therefore 0.00266 moles
    Moles of Mn2+ after reaction 1: 0.00266 x 4 = 0.0107 moles
    Moles of Mn2+ before reaction 1: (100/1000) x 0.2 = 0.02
    Moles of Mn2+ used up during reaction 1: =0.02 – 0.0107 = 0.00934
    Moles of C2O42-: 2:5 ratio, therefore 0.00934 x 2.5 = 0.0234
    Mr of MC2O4: 5.25 / 0.0234 = 224.74
    Mr of M: 224.74 – 88 = 136.74
    Identity of M: Ba
  4. The moles of fluorine gas in a 100 dm3 container is determined by a scientist. The scientist first takes a 1dm3 sample from container and reacts it with 32.0g of powdered iron. The mixture of iron and iron fluoride was mixed with water and filtered to remove the iron fluoride and isolate the remaining iron. The remaining iron was reacted with 250cm3 of sulfuric acid to form Fe2+ ions. The sulfuric acid was in excess. A 25.0 cm3 sample of the iron sulfate formed was titrated against potassium manganate, where it was found to require 24.9cm3 of 0.200 mol dm-3 of potassium manganate to fully react.
    Equation 1: Fe + 1.5F2 –> FeF3
    Equation 2: MnO₄⁻ + 8H⁺ + 5Fe2+–> Mn²⁺ + 4H₂O + 5Fe3+
    Moles of MnO4–: (24.9/.1000) x 0.2 = 0.00498
    Moles of Fe2+ in 25cm3: 0.00498 x 5 = 0.0249
    Moles of Fe2+ in 250cm3: 0.249
    Moles of Fe2+ initial: 32 / 55.8 = 0.573
    Moles Fe2+ used up: = 0.573 – 0.249 = 0.324
    Moles of F2 in 1dm3: 0.324 x 1.5 = 0.487
    Moles of F2 in 100dm3: = 48.7 moles
  5. A sample of anhydrous Iron (II) chloride was analysed by back titration. A 5.00g sample of iron(II) chloride was reacted with 200cm3 of 0.100 mol dm-3 potassium manganate(VII). Note that both the chloride and iron(II) ions are oxidised by the manganate. The excess potassium manganate was then portioned into 25cm3 aliquots, and these were found to fully react with 24.0cm3 of 0.1mol dm-3 potassium ethanedioate. What was the purity of the iron(II) chloride by percentage purity?
    Equation 1: 5FeCl2 + 3MnO4– + 24H+ –> 5Fe3+ + 5Cl2 + 3Mn2+ + 12H2O
    Equation 2: 2MnO4– + 16H+ 5C2O42- –> 2Mn2+ + 8H2O + 10CO2
    Moles of C2O42-: (24.0/1000) x 0.1 = 0.0024
    Moles of MnO4– in 25cm3: 5:2 ratio, therefore 0.0024 x 0.4 = 0.000960
    Moles of MnO4– in 200cm3: 0.00096 x 8 = 0.00768
    Moles of MnO4– before reacting: (200/1000) x 0.1 = 0.02
    Moles of MnO4– used up in reaction 1: 0.02 – 0.00768 = 0.0123
    Moles of FeCl2: 3:5 Ratio therefore 0.01232 x (5/3) = 0.0205
    Mass of FeCl2: 0.0205 x 126.75 = 2.6026
    Percentage Purity of FeCl2: 52.1%
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