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Date	November 2018	Marks available	2	Reference code	18N.3.hl.TZ0.3
Level	HL	Paper	3	Time zone	TZ0
Command term	Determine	Question number	3	Adapted from	N/A

Question

The presence of very small amounts of lead in calcium-based antacids can be determined using inductively coupled plasma-mass spectroscopy (ICP-MS).

An unknown antacid sample has a lead ion concentration of 0.50 μg dm^‒3.

Chelating agents can be used to treat heavy metal poisoning.

State the type of particle present in the plasma formed.

[1]

Calculate the concentration of lead ions in the sample in mol dm^‒3.

[2]

b.i.

Lead ions are toxic and can be precipitated using hydroxide ions.

Pb²⁺ (aq) + 2OH^‒ (aq) $\rightleftharpoons$ Pb(OH)₂ (s)

Sufficient sodium hydroxide solid is added to the antacid sample to produce a 1.0 × 10^‒2 mol dm^‒3 hydroxide ion solution at 298 K.

Deduce if a precipitate will be formed, using section 32 of the data booklet.

If you did not calculate the concentration of lead ions in (b)(i), use the value of 2.4 × 10⁻⁴ mol dm^‒3, but this is not the correct value.

[2]

b.ii.

Electrolysis is used to obtain lead from Pb²⁺ (aq) solution.

Determine the time, in hours, required to produce 0.0500 mol lead using a current (I) of 1.34 A. Use section 2 of the data booklet and the equation, charge (Q) = current (I) × time (t, in seconds).

[2]

State one feature of a chelating agent.

[1]

d.i.

An aqueous lead(II) ion reacts with three ethane-1,2-diamine molecules to form an octahedral chelate ion.

Outline why the chelate ion is more stable than the reactants.

[1]

d.ii.

Markscheme

positive ions/cations/Pb²⁺

free electrons ✔

Accept “ions” OR “charged species/particle”.

[Pb²⁺] = 0.50 × 10^‒6/5.0 × 10^–7 «g dm^–3» ✔

[Pb²⁺] « $= \frac{{0.50 \times {{10}^{ - 6}}\,{\text{g}}\,{\text{d}}{{\text{m}}^{-3}}}}{{207.20\,{\text{g}}\,{\text{mo}}{{\text{l}}^{-1}}}}$ » = 2.4 × 10^‒9 «mol dm^‒3» ✔

Award [2] for correct final answer.

b.i.

«K_sp = 1.43 × 10^–20»

ALTERNATIVE 1:

«Q = [Pb²⁺] [OH^–]² = 2.4 × 10^–9 × (1.0 × 10^–2)²» = 2.4 × 10^–13 ✔

Q > K_sp AND precipitate will form

2.4 × 10^–13 > 1.43 × 10^–20 AND precipitate will form ✔

ALTERNATIVE 2:

critical [Pb²⁺] for hydroxide solution « $= \frac{{{K_{sp}}}}{{{{\left[ {{\text{O}}{{\text{H}}^ - }} \right]}^2}}} = \frac{{1.43 \times {{10}^{ - 20}}}}{{{{\left( {1.0 \times {{10}^{ - 2}}} \right)}^2}}}$ » = 1.4 × 10^–16 ✔

initial concentration > critical concentration AND precipitate will form

2.4 × 10^–9 > 1.4 × 10^–16 AND precipitate will form ✔

If value given is used:

ALTERNATIVE 3:

«Q = [Pb²⁺] [OH^–]² = 2.4 × 10^–4 × (1.0 × 10^–2)²» = 2.4 × 10^–8 ✔

Q > K_sp AND precipitate will form

2.4 × 10^–8 > 1.43 × 10^–20 AND precipitate will form ✔

b.ii.

«Faraday’s constant, F = 9.65 × 10⁴ C mol^‒1 and 1 A = 1 C s^–1»
Q «= 0.0500 mol × 2 × 96500 C mol^‒1» = 9650 «C» ✔

t « $= \frac{Q}{I} = \frac{{9650\,{\text{C}}}}{{1.34\,{\text{C}}\,{{\text{s}}^{-1}}}} \approx 7200\,{\text{s}}$ so $\frac{{7200\,{\text{s}}}}{{60 \times 60\,{\text{s}}\,{{\text{h}}^{-1}}}}$ » = 2.00 «hours» ✔