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Question 18N.3.hl.TZ0.3

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Date November 2018 Marks available [Maximum mark: 9] Reference code 18N.3.hl.TZ0.3
Level hl Paper 3 Time zone TZ0
Command term Calculate, Deduce, Determine, Outline, State Question number 3 Adapted from N/A
3.
[Maximum mark: 9]
18N.3.hl.TZ0.3

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

(a)

State the type of particle present in the plasma formed.

[1]

Markscheme

positive ions/cations/Pb2+

OR

free electrons ✔

 

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

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

(b.i)

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

[2]

Markscheme

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

[Pb2+] « = 0.50 × 10 − 6 g d m − 3 207.20 g mo l − 1 » = 2.4 × 10‒9 «mol dm‒3» ✔

 

Award [2] for correct final answer.

(b.ii)

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

Pb2+ (aq) + 2OH‒ (aq) ⇌ 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]

Markscheme

«Ksp = 1.43 × 10–20»

 

ALTERNATIVE 1:

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

 

Q > Ksp AND precipitate will form

OR

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

 

ALTERNATIVE 2:

critical [Pb2+] for hydroxide solution « = K s p [ O H − ] 2 = 1.43 × 10 − 20 ( 1.0 × 10 − 2 ) 2 » = 1.4 × 10–16 ✔

 

initial concentration > critical concentration AND precipitate will form

OR

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 > Ksp AND precipitate will form

OR

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

(c)

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]

Markscheme

«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 « = Q I = 9650 C 1.34 C s − 1 ≈ 7200 s  so  7200 s 60 × 60 s h − 1 » = 2.00 «hours» ✔

 

Award [2] for correct final answer.

Chelating agents can be used to treat heavy metal poisoning.

(d.i)

State one feature of a chelating agent.

[1]

Markscheme

Any one of:

two «or more» lone/non-bonding pairs on different atoms

OR

two «or more» atoms/centres that act as Lewis bases ✔

 

form «at least» two coordination/coordinate bonds

OR

«at least» two atoms can form coordination/coordinate bonds ✔

 

Reference to “on DIFFERENT atoms” required.

Accept “dative «covalent» bond” for “coordination/coordinate bond”.

(d.ii)

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]

Markscheme

increase in entropy

OR

ΔS > 0/ΔS positive ✔

 

Accept “ΔG < 0” but not “ΔH < 0”.

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