A-LevelChemistryAnalytical techniques9701/42 · Feb/Mar 2023 · Q69 Marks

A student uses thin-layer chromatography (TLC) to analyse a mixture containing different metal cations. The student repeats the experiment using different solvents. Fig. 6.1 shows the chromatogram obtained by the student using water as a solvent. [Figure 6.1]

(a)(i)Suggest a compound that could be used as the stationary phase in this experiment.[1]
(a)(ii)Table 6.1 shows the Rf values for different metal cations when separated by TLC using water as a solvent. Table 6.1 cation Rf value (water) Cd²⁺(aq) 0.40 Co²⁺(aq) 0.77 Cu²⁺(aq) 0.32 Fe³⁺(aq) 0.12 Hg²⁺(aq) 0.23 Ni²⁺(aq) 0.75 Suggest the identity of the cation that causes the spot at M in Fig. 6.1. Explain your answer.[1]
(b)The student repeats the experiment using butan-1-ol as a solvent. The metal cations do not travel as far up the TLC plate in this experiment. Suggest why the metal cations do not move as far up the TLC plate with butan-1-ol as a solvent.[1]
(c)The student sprays the TLC plate in Fig. 6.1 with KSCN(aq). The colour of some of the spots changes, as some of the metal cations undergo a ligand exchange reaction. Identify the ligands involved in the ligand exchange reaction.[1]
(d)(i)Complete the equation for the reaction of C8H5O4K(aq) with KOH(aq). C8H5O4K + ..............................................[1]
(d)(ii)Calculate the pH of the buffer solution. Show all your working.[4]

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Mark scheme

Marking points from the official mark scheme (Feb/Mar 2023). The mark or mark type is shown in brackets.

[1] SiO2 OR Al2O3 OR silica OR alumina OWTTE
[1] Cd2⁺ AND Rf of M (= 2/5) = 0.40 / same Rf as in the Table 6.1
[1] metal cations are less soluble in butan-1-ol (than in water) OR metal cations weaker ion-dipole forces with butan-1-ol
[1] H2O AND SCN⁻
[1] C8H5O4K + KOH → C8H4O4K2 + H2O OR C8H5O4K + OH⁻ → C8H4O4K⁻ + H2O
[1] M1 initially: moles of KOH = 0.150 × 20.0 ÷ 1000 = 3.0×10⁻³ AND moles of acid HA = 0.100 × 50.0÷1000 = 5.0×10⁻³
[1] M2 at equilibrium: moles of salt KA = 3.0×10⁻³ ecf AND moles of acid HA = 5.0×10⁻³ – moles of NaOH = 2.0×10⁻³ ecf
[1] M3 Ka = 10⁻⁵·⁴⁰ = 3.98×10⁻⁶ 3.98×10⁻⁶ = [H⁺] [3.0×10⁻³] / [2.0×10⁻³] [H⁺] = 2.65 ×10⁻⁶ ecf
[1] M4 pH = -log(2.65 ×10⁻⁶) = 5.58 / 5.6 ecf on a calculated and identified [H⁺] ALLOW alternative approach using Henderson-Hasselbalch equation for M3 and M4

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