A-LevelChemistryGroup 29701/42 · May/June 2020 · Q219 Marks

Describe and explain how the solubility of the Group 2 sulfates varies down the group.

(a)Describe and explain how the solubility of the Group 2 sulfates varies down the group.[4]
(b)The trend in the decomposition temperatures of Group 2 peroxides, MO₂, is similar to that of Group 2 carbonates. Suggest which of barium peroxide, BaO₂, and calcium peroxide, CaO₂, will decompose at the lower temperature. Explain your answer.[2]
(c)Magnesium iodate(V), Mg(IO₃)₂, decomposes when heated to form magnesium oxide, oxygen and iodine. Construct an equation for this reaction.[1]
(d)(i)Write an expression for the solubility product, Ksp, of Ca(IO₃)₂, and state its units. Ksp = units =[2]
(d)(ii)Calculate the numerical value for Ksp Ca(IO₃)₂ at 298 K. Ksp =[1]
(d)(iii)When a few cm³ of concentrated Ca(NO₃)₂(aq) is added to a saturated solution of Ca(IO₃)₂ a white precipitate forms. Identify the white precipitate and give an explanation for this observation.[2]
(e)Iodised salt is sodium chloride mixed with a small amount of sodium iodate(V), NaIO₃. • 50.00 g of iodised salt is dissolved in distilled water and the solution made up to 250 cm³ in a volumetric flask with distilled water. • 50.0 cm³ of this solution is pipetted into an excess of aqueous acidified potassium iodide. IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O • The iodine produced requires 12.40 cm³ of 0.00200 mol dm⁻³ aqueous sodium thiosulfate solution for complete reaction. I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻ Calculate the mass of sodium iodate(V) present in 50.00 g of iodised salt. mass of NaIO₃ = g [3][3]
(f)The half-equation for the reduction of iodate(V) ions is shown. IO₃⁻ + 6H⁺ + 5e⁻ → ½I₂ + 3H₂O E° = +1.19V Use data from the Data Booklet to predict whether a reaction is feasible when aqueous solutions of acidified iodate(V) ions and bromide ions are mixed. Explain your answer.[1]
(g)(i)Write the rate equation for this reaction, stating the units of the rate constant, k. rate = units of k = mol dm⁻³ s⁻¹ [2][2]
(g)(ii)The rate of reaction depends on the pH of the solution. Assume all other concentrations remain the same. Use the expression x = rate at pH 5.00 / rate at pH 4.00 to calculate the value of x. x =[1]

Mark scheme

Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.

[4] M1 solubility decreases down the group M2 ΔHlatt and ΔHhyd both become less exothermic / more endothermic M3 ΔHlatt changes less (than ΔHhyd as SO₄²⁻ being larger than M²⁺) M4 ΔHsol becomes less exothermic / less negative
[2] M1 CaO₂ and Ca²⁺ has a smaller ionic radii/ Ca²⁺ has a higher charge density M2 anion/O₂²⁻ becomes more polarised /distorted
[1] Mg(IO₃)₂(s) → MgO(s) + 2.5O₂(g) + I₂(s)
[2] M1 Ksp = [Ca²⁺][IO₃⁻]² M2 units = mol³ dm⁻⁹
[1] Ksp = 4 x (5.6 × 10⁻³ )³ Ksp = 7.03 x 10⁻⁷ 2sf min
[2] M1 Ca(IO₃)₂ AND as solubility of Ca(IO₃)₂ decreases M2 due to common ion effect
[3] M1 moles S₂O₃²⁻ = 0.002 x 12.4/1000 = 2.48 x 10⁻⁵ moles of I₂ = 1.24 x 10⁻⁵ M2 moles of IO₃⁻ = 4.13 x 10⁻⁶ in 50 cm³ moles of IO₃⁻ = 2.07 x 10⁻⁵ in 250 cm³ mass of NaIO₃ = 2.07 x 10⁻⁵ x 197.9 M3 mass of NaIO₃ = 0.0041
[1] It is feasible as the Ecell will be positive/+0.12 V
[2] M1 Rate = k[IO₃⁻][SO₃²⁻][H⁺] M2 units = mol⁻²dm⁶s⁻¹
[1] 0.10