State the type of particle present in the plasma formed.

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

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.

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).

State one feature of a chelating agent.

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.

positive ions/cations/Pb²⁺

OR

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.

«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

OR

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

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 > K_sp AND precipitate will form

OR

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

«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» ✔

Award [2] for correct final answer.

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”.

increase in entropy

OR

ΔS > 0/ΔS positive ✔

Accept “ΔG < 0” but not “ΔH < 0”.