A-LevelChemistryGroup 29701/42 · Oct/Nov 2018 · Q522 Marks

Explain why the thermal stability of the Group 2 nitrates increases down the group.

(a)Explain why the thermal stability of the Group 2 nitrates increases down the group.[2]
(b)Sodium nitrite, NaNO2, is a decomposition product from heating sodium nitrate, NaNO3. A student analysed a sample of sodium nitrite by titration with aqueous cerium(IV) ions, Ce4+(aq). The equation for the titration reaction is shown. NO2-(aq) + 2Ce⁴+(aq) + H2O(l) → 2Ce³+(aq) + NO3¯(aq) + 2H+(aq) • 0.138g of impure sodium nitrite was dissolved in water and made up to 100 cm³ in a volumetric flask. • 25.0 cm³ of this solution required 21.80 cm³ of 0.0400 mol dm-³ Ce⁴+(aq) to reach the end-point. You should assume the impurity does not react with Ce⁴+(aq). Calculate the percentage purity of the sample of sodium nitrite. %[3]
(c)Acidified manganate(VII) ions, MnO4¯, can also be used to analyse solutions containing nitrite ions, NO2, by titration. In acidic solution, NO2 ions exist as HNO2.
(c)(i)Use the Data Booklet to construct an ionic equation for this reaction.[2]
(c)(ii)Use E° values to calculate the E°cell for this reaction. E°cell =[1]
(d)Nitrous acid, HNO2, is a weak acid with a Ka of 6.9 × 10-4 mol dm-³ at 298 K.
(d)(i)Explain the difference between a strong acid and a weak acid.[1]
(d)(ii)Write the expression for the acid dissociation constant, Ka, for HNO2. Ka =[1]
(d)(iii)Calculate the pH of 0.15 mol dm-³ HNO2. pH =[2]
(d)(iv)Calculate the percentage of HNO2 molecules that are ionised in 0.15 mol dm-³ HNO2. % ionisation =[1]
(e)Solutions containing a mixture of HNO2 and NaNO2 are buffer solutions.
(e)(i)Define what is meant by the term buffer solution.[2]
(e)(ii)Write two equations to show how a solution containing a mixture of HNO2 and NaNO2 acts as a buffer.[2]
(f)Nitrous acid is used in the preparation of diazonium salts. The –N2+ group in the diazonium ion can be replaced with Cl, Br or CN as shown. The reagent used is a copper(I) salt, CuX. [Figure 5.0] This reaction can be used in the synthesis of compound Z as shown. [Figure 5.1]
(f)(i)Suggest the reagent used in reaction 1.[1]
(f)(ii)Suggest structures of compounds Y and Z and draw them in the boxes above. [Figure 5.2][2]
(g)Compounds W and Z were analysed using carbon-13 NMR spectroscopy. Predict the number of peaks in the carbon-13 NMR spectra of W and Z. number of peaks W Z[2]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.

[1] ionic radius / ion size increases OR charge density decreases (down the group)
[1] less polarisation / distortion of anion / nitrate ion / NO3⁻ / nitrate group OR N-O / N=O bond is less weakened / distorted / polarised OR more energy to break N-O / N=O bond
[1] moles of Ce⁴⁺ = 0.0400 × 21.8 / 1000 = 8.72 × 10⁻⁴ (moles of Ce⁴⁺)
[1] moles of NO2⁻ = 8.72 × 10⁻⁴ / 2 = 4.36 × 10⁻⁴ in 25 cm³ (use of 2:1 ratio correctly)
[1] moles of NO2⁻ = 4.36 × 10⁻⁴ × 4 = 1.74(4) × 10⁻³ in 100 cm³ (use of 4:1 ratio correctly) mass NaNO2 = 1.74(4) × 10⁻³ × (23.0 + 14.0 + 32.0) = 0.120 g (use of Mr correctly) % purity = 0.120 /0.138 = 86.96% (use of 0.0138 correctly)
[1] all species correct
[1] balanced
[1] E°cell = 1.52 – 0.94 = 0.58 (V)
[1] weak acid is partly ionised and strong acid is completely ionised
[1] Ka = [H⁺][NO2⁻] / [HNO2]
[1] [H⁺] = √0.00069 × 0.15 = 1.02 × 10⁻²
[1] pH = -log[H⁺] = 2.0 (1.99) [minimum 2 significant figures]
[1] % ionisation = 100 × 1.02 × 10⁻² / 0.15 = 6.7–6.8%
[1] M1 A solution that resists changes in pH
[1] M2 when small amounts of acid or alkali are added to it
[1] M1 HNO2 + OH⁻ → NO2⁻ + H2O
[1] M2 NO2⁻ + H⁺ → HNO2
[1] CuCN / copper(I) cyanide
[1] × 2 [for (f)(ii) structures of Y and Z]
[1] × 2 [for (g) W=6, Z=3]