‘Lawn sand’ is spread over the grass in gardens to reduce the growth of moss. Lawn sand is a mixture of sand and iron(II) sulfate crystals, FeSO₄.7H₂O. Lawn sand usually contains 6–10% FeSO₄.7H₂O by mass. To determine the exact percentage by mass of FeSO₄.7H₂O present in a sample of lawn sand, a student devises the following experiment. step 1 Use a known mass of lawn sand to prepare 250.0 cm³ of solution A containing Fe²⁺(aq) ions. Solution A must have dilute sulfuric acid, H₂SO₄(aq), added to it before it is made up to 250 cm³. step 2 To determine the concentration of Fe²⁺(aq) in solution A, titrate a 25.00 cm³ sample of solution A against 0.0200 mol dm⁻³ aqueous potassium manganate(VII), KMnO₄(aq). The reaction which takes place during the titration is shown. MnO₄⁻(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 4H₂O(l) + 5Fe³⁺(aq)
- (a)(i)The end-point of the titration is 25.00 cm³ of 0.0200 mol dm⁻³ KMnO₄(aq). Determine the concentration of Fe²⁺(aq) that was present in the 25.00 cm³ sample of solution A at the start of the titration. concentration of Fe²⁺(aq) = ........................ mol dm⁻³[1]
- (a)(ii)If you were unable to calculate the concentration in (i), assume for (ii) and (iii) that the concentration of Fe²⁺(aq) is 0.300 mol dm⁻³. This is not the correct answer. Determine the mass of lawn sand needed to prepare the 250.0 cm³ of solution A at the concentration calculated in (i). Assume that lawn sand contains 8% FeSO₄.7H₂O by mass. [Ar: Fe, 55.8; S, 32.1; O, 16.0; H, 1.0] mass of lawn sand = ........................ g[3]
- (a)(iii)Solution A must contain enough H⁺(aq) ions for the reaction to take place during the titration. MnO₄⁻(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 4H₂O(l) + 5Fe³⁺(aq) Use the concentration of Fe²⁺(aq) from (i) to determine the minimum volume of 2.00 mol dm⁻³ H₂SO₄(aq) which must be added to prepare the 250.0 cm³ of solution A. volume = ........................ cm³[2]
- (b)Describe a method to prepare 250.0 cm³ of solution A starting with a glass beaker which contains the known mass of lawn sand determined in (a)(ii) as shown. [Figure 2.1: Diagram of glass beaker with known mass of lawn sand] Assume that common laboratory apparatus is available. You may find it helpful to write your answer as a series of smaller steps.[5]
- (c)State the colour change in the conical flask at the end-point of the titration. from ................................ to ................................[1]
- (d)Aqueous potassium manganate(VII) is a powerful oxidising agent. Suggest the effect, if any, on the end-point volume if the student acidified the mixture with dilute hydrochloric acid, HCl(aq), instead of dilute sulfuric acid, H₂SO₄(aq). Explain your answer. effect, if any, on the end-point volume .......................................................................................... explanation ....................................................................................................................................................[2]
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Mark scheme
Marking points from the official mark scheme (Feb/Mar 2018). The mark or mark type is shown in brackets.
- [1] 0.0200 × 5 = 0.1(00) mol dm⁻³
- [3] M1 = 277.9 (seen anywhere) M2 = M1 × 0.1(00) × 250 / 1000 = 6.9475 M3 = M2 × 100/8 = 86.84 g
- [2] M1 = total moles of H⁺ = (0.100 × 8/5 × 250 / 1000) = 0.04(00) mol M2 = volume of 2 mol dm⁻³ sulfuric acid = (M1/2) × (1000 / 2) = 10 cm³
- [5] M1 Dissolve the iron(II) sulfate crystals (in the beaker) using (distilled) water M2 Filter M3 Rinse the residue If no M2 (filtration), M3 can be applied to mixture in M1 as part of transfer in M5. M4 Add H2SO4 M5 Transfer / add to a 250 cm³ volumetric flask and make up to mark with (distilled) water
- [1] Colourless to pink / pale purple
- [2] M1 Higher M2 Some of the MnO₄⁻ (aq) would be used oxidising Cl⁻(aq) ions
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