Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group whereas the melting points of the group 17 elements (F → I) increase down the group.

State the shape of the complex ion.

Deduce the charge on the complex ion and the oxidation state of cobalt.

Describe, in terms of acid-base theories, the type of reaction that takes place between the cobalt ion and water to form the complex ion.

Any three of:

Group 1:
atomic/ionic radius increases

smaller charge density

OR

force of attraction between metal ions and delocalised electrons decreases

Do not accept discussion of attraction between valence electrons and nucleus for M2.

Accept “weaker metallic bonds” for M2.

Group 17:
number of electrons/surface area/molar mass increase

London/dispersion/van der Waals’/vdw forces increase

Accept “atomic mass” for “molar mass”.

[Max 3 Marks]

«distorted» octahedral

Accept “square bipyramid”.

Charge on complex ion: 1+/+
Oxidation state of cobalt: +2

Lewis «acid-base reaction»

H2O: electron/e^– pair donor

OR

Co²⁺: electron/e^– pair acceptor

Describe the bonding in metals.

Titanium exists as several isotopes. The mass spectrum of a sample of titanium gave the following data:

Calculate the relative atomic mass of titanium to two decimal places.

State the number of protons, neutrons and electrons in the ${}_{\text{22}}^{\text{48}}\text{Ti}$ atom.

State the full electron configuration of the ${}_{\text{22}}^{\text{48}}\text{T}{\text{i}}^{2+}$ ion.

Suggest why the melting point of vanadium is higher than that of titanium.

Sketch a graph of the first six successive ionization energies of vanadium on the axes provided.

Explain why an aluminium-titanium alloy is harder than pure aluminium.

Describe, in terms of the electrons involved, how the bond between a ligand and a central metal ion is formed.

Outline why transition metals form coloured compounds.

State the type of bonding in potassium chloride which melts at 1043 K.

A chloride of titanium, $\text{TiC}{\text{l}}_{\text{4}}$, melts at 248 K. Suggest why the melting point is so much lower than that of KCl.

Formulate an equation for this reaction.

Suggest one disadvantage of using this smoke in an enclosed space.

electrostatic attraction

between «a lattice of» metal/positive ions/cations AND «a sea of» delocalized electrons

Accept “mobile electrons”.

Do not accept “metal atoms/nuclei”.

[2 marks]

$\frac{(46\times 7.98)\text{ + }(47\times 7.32)\text{ + }(48\times 73.99)\text{ + }(49\times 5.46)\text{ + }(50\times 5.25)}{100}=47.93$

Answer must have two decimal places with a value from 47.90 to 48.00.

Award [2] for correct final answer.

Award [0] for 47.87 (data booklet value).

[2 marks]

Protons: 22 AND Neutrons: 26 AND Electrons: 22

[1 mark]

$\text{1}{\text{s}}^{\text{2}}\text{2}{\text{s}}^{\text{2}}\text{2}{\text{p}}^{\text{6}}\text{3}{\text{s}}^{\text{2}}\text{3}{\text{p}}^{\text{6}}\text{3}{\text{d}}^{\text{2}}$

[1 mark]

vanadium has smaller ionic radius «leading to stronger metallic bonding»

Accept vanadium has «one» more valence electron«s» «leading to stronger metallic bonding».

Accept “atomic” for “ionic”.

[1 mark]

regular increase for first five AND sharp increase to the 6th

A log graph is acceptable.

Accept log plot on given axes (without amendment of y-axis).

Award mark if gradient of 5 to 6 is greater than “best fit line” of 1 to 5.

[1 mark]

titanium atoms/ions distort the regular arrangement of atoms/ions

OR

titanium atoms/ions are a different size to aluminium «atoms/ions»

prevent layers sliding over each other

Accept diagram showing different sizes of atoms/ions.

[2 marks]

pair of electrons provided by the ligand

Do not accept “dative” or “coordinate bonding” alone.

[1 mark]

partially filled d-orbitals

«ligands cause» d-orbitals «to» split

light is absorbed as electrons transit to a higher energy level «in d–d transitions»
OR
light is absorbed as electrons are promoted

energy gap corresponds to light in the visible region of the spectrum

colour observed is the complementary colour

[4 marks]

ionic

OR

«electrostatic» attraction between oppositely charged ions

[1 mark]

«simple» molecular structure

OR

weak«er» intermolecular bonds

OR

weak«er» bonds between molecules

Accept specific examples of weak bonds such as London/dispersion and van der Waals.

Do not accept “covalent”.

[1 mark]

$\text{TiC}{\text{l}}_{\text{4}}\text{(l)}+\text{2}{\text{H}}_{\text{2}}\text{O(l)}\to \text{Ti}{\text{O}}_{\text{2}}\text{(s)}+\text{4HCl(aq)}$ correct products
correct balancing

Accept ionic equation.

Award M2 if products are HCl and a compound of Ti and O.

[2 marks]

HCl causes breathing/respiratory problems

OR

HCl is an irritant

OR

HCl is toxic

OR

HCl has acidic vapour

OR

HCl is corrosive

Accept TiO₂ causes breathing

problems/is an irritant.

Accept “harmful” for both HCl and TiO₂.

Accept “smoke is asphyxiant”.

[1 mark]

Deduce the ratio of Fe²⁺:Fe³⁺ in Fe₃O₄.

State the type of spectroscopy that could be used to determine their relative abundances.

State the number of protons, neutrons and electrons in each species.

Iron has a relatively small specific heat capacity; the temperature of a 50 g sample rises by 44.4°C when it absorbs 1 kJ of heat energy.

Determine the specific heat capacity of iron, in J g⁻¹K⁻¹. Use section 1 of the data booklet.

A voltaic cell is set up between the Fe²⁺(aq) | Fe (s) and Fe³⁺ (aq) | Fe²⁺ (aq) half-cells.

Deduce the equation and the cell potential of the spontaneous reaction. Use section 24 of the data booklet.

The figure shows an apparatus that could be used to electroplate iron with zinc. Label the figure with the required substances.

Outline why, unlike typical transition metals, zinc compounds are not coloured.

Transition metals like iron can form complex ions. Discuss the bonding between transition metals and their ligands in terms of acid-base theory.

1:2 ✔

Accept 2 Fe³⁺: 1 Fe²⁺
Do not accept 2:1 only

mass «spectroscopy»/MS ✔

Award [1 max] for 4 correct values.

specific heat capacity « = $\frac{q}{m\times ∆T}/\frac{1000 \mathrm{J}}{50 \mathrm{g}\times 44 \mathrm{K}}$» = 0.45 «J g⁻¹K⁻¹» ✔

Equation:
2Fe³⁺(aq) + Fe(s) → 3Fe²⁺(aq) ✔

Cell potential:
«+0.77 V − (−0.45 V) = +»1.22 «V» ✔

Do not accept reverse reaction or equilibrium arrow.

Do not accept negative value for M2.

left electrode/anode labelled zinc/Zn AND right electrode/cathode labelled iron/Fe ✔

electrolyte labelled as «aqueous» zinc salt/Zn²⁺ ✔

Accept an inert conductor for the anode.

Accept specific zinc salts such as ZnSO₄.

« Zn²⁺» has a full d-shell
OR
does not form « ions with» an incomplete d-shell ✔

Do not accept “Zn is not a transition metal”.

Do not accept zinc atoms for zinc ions.

ligands donate pairs of electrons to metal ions
OR
forms coordinate covalent/dative bond✔

ligands are Lewis bases
AND
metal «ions» are Lewis acids ✔

State the electron configuration of the Cu⁺ ion.

Copper(II) chloride is used as a catalyst in the production of chlorine from hydrogen chloride.

4HCl (g) + O₂ (g) → 2Cl₂ (g) + 2H₂O (g)

Calculate the standard enthalpy change, ΔH^θ, in kJ, for this reaction, using section 12 of the data booklet.

The diagram shows the Maxwell–Boltzmann distribution and potential energy profile for the reaction without a catalyst.

Annotate both charts to show the activation energy for the catalysed reaction, using the label E_{a (cat)}.

Explain how the catalyst increases the rate of the reaction.

Solid copper(II) chloride absorbs moisture from the atmosphere to form a hydrate of formula CuCl₂•$x$H₂O.

A student heated a sample of hydrated copper(II) chloride, in order to determine the value of $x$. The following results were obtained:

Mass of crucible = 16.221 g
Initial mass of crucible and hydrated copper(II) chloride = 18.360 g
Final mass of crucible and anhydrous copper(II) chloride = 17.917 g

Determine the value of $x$.

State how current is conducted through the wires and through the electrolyte.

Wires: 

Electrolyte:

Write the half-equation for the formation of gas bubbles at electrode 1.

Bubbles of gas were also observed at another electrode. Identify the electrode and the gas.

Electrode number (on diagram):

Name of gas: 

Deduce the half-equation for the formation of the gas identified in (c)(iii).

Determine the enthalpy of solution of copper(II) chloride, using data from sections 18 and 20 of the data booklet.

The enthalpy of hydration of the copper(II) ion is −2161 kJ mol⁻¹.

Calculate the cell potential at 298 K for the disproportionation reaction, in V, using section 24 of the data booklet.

Comment on the spontaneity of the disproportionation reaction at 298 K.

Calculate the standard Gibbs free energy change, ΔG^θ, to two significant figures, for the disproportionation at 298 K. Use your answer from (e)(i) and sections 1 and 2 of the data booklet.

Suggest, giving a reason, whether the entropy of the system increases or decreases during the disproportionation.

Deduce, giving a reason, the sign of the standard enthalpy change, ΔH^θ, for the disproportionation reaction at 298 K.

Predict, giving a reason, the effect of increasing temperature on the stability of copper(I) chloride solution.

Describe how the blue colour is produced in the Cu(II) solution. Refer to section 17 of the data booklet.

Deduce why the Cu(I) solution is colourless.

When excess ammonia is added to copper(II) chloride solution, the dark blue complex ion, [Cu(NH₃)₄(H₂O)₂]²⁺, forms.

State the molecular geometry of this complex ion, and the bond angles within it.

Molecular geometry:

Bond angles: 

Examine the relationship between the Brønsted–Lowry and Lewis definitions of a base, referring to the ligands in the complex ion [CuCl₄]²⁻.

[Ar] 3d¹⁰
OR
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ ✔

ΔH^θ = ΣΔH^θ_f (products) − ΣΔH^θ_f (reactants) ✔
ΔH^θ = 2(−241.8 «kJ mol⁻¹») − 4(−92.3 «kJ mol⁻¹») = −114.4 «kJ» ✔

NOTE: Award [2] for correct final answer.

E_{a (cat)} to the left of E_a ✔                        

peak lower AND E_{a (cat)} smaller ✔

«catalyst provides an» alternative pathway ✔

«with» lower E_a
OR
higher proportion of/more particles with «kinetic» E ≥ E_a(cat) «than E_a» ✔

mass of H₂O = «18.360 g – 17.917 g =» 0.443 «g» AND mass of CuCl₂ = «17.917 g – 16.221 g =» 1.696 «g» ✔

moles of H₂O = «$\frac{0.443 \text{g}}{18.02 \text{g\hspace{0.05em}mo}{\text{l}}^{-1}}$=» 0.0246 «mol»
OR
moles of CuCl₂ =«$\frac{1.696 \text{g}}{134.45 \text{g\hspace{0.05em}mo}{\text{l}}^{-1}}$= » 0.0126 «mol» ✔

«water : copper(II) chloride = 1.95 : 1»

«$x$ =» 2 ✔

NOTE: Accept «$x$ =» 1.95.

NOTE: Award [3] for correct final answer.

Wires:
«delocalized» electrons «flow» ✔

Electrolyte:
«mobile» ions «flow» ✔

2Cl⁻ → Cl₂ (g) + 2e⁻
OR
Cl⁻ → $\frac{1}{2}$Cl₂ (g) + e⁻ ✔

NOTE: Accept e for e⁻.

«electrode» 3 AND oxygen/O₂ ✔

NOTE: Accept chlorine/Cl₂.

2H₂O (l) → 4H⁺ (aq) + O₂ (g) + 4e^– ✔

NOTE: Accept 2Cl^– (aq) → Cl₂ (g) + 2e^–.
Accept 4OH⁻ → 2H₂O + O₂ + 4e⁻

enthalpy of solution = lattice enthalpy + enthalpies of hydration «of Cu²⁺ and Cl⁻» ✔

«+2824 kJ mol^–1 − 2161 kJ mol^–1 − 2(359 kJ mol^–1) =» −55 «kJ mol^–1» ✔

NOTE: Accept enthalpy cycle.
Award [2] for correct final answer.

E^θ = «+0.52 – 0.15 = +» 0.37 «V» ✔

spontaneous AND E^θ positive ✔

ΔG^θ = «−nFE = −1 mol × 96 500 C Mol^–1 × 0.37 V=» −36 000 J/−36 kJ ✔

NOTE: Accept “−18 kJ mol^–1 «per mole of Cu⁺»”.

Do not accept values of n other than 1.

Apply SF in this question.

Accept J/kJ or J mol⁻¹/kJ mol⁻¹ for units.

2 mol (aq) → 1 mol (aq) AND decreases ✔

NOTE: Accept “solid formed from aqueous solution AND decreases”.
Do not accept 2 mol → 1 mol without (aq).

ΔG^θ < 0 AND ΔS^θ < 0 AND ΔH^θ < 0
OR
ΔG^θ + TΔS^θ < 0 AND ΔH^θ < 0 ✔

TΔS more negative «reducing spontaneity» AND stability increases ✔

NOTE: Accept calculation showing non-spontaneity at 433 K.

«ligands cause» d-orbitals «to» split ✔

light absorbed as electrons transit to higher energy level «in d–d transitions»
OR
light absorbed as electrons promoted ✔

energy gap corresponds to «orange» light in visible region of spectrum ✔

colour observed is complementary ✔

full «3»d sub-level/orbitals
OR
no d–d transition possible «and therefore no colour» ✔

octahedral AND 90° «180° for axial» ✔

NOTE: Accept square-based bi-pyramid.

Any two of:
ligand/chloride ion Lewis base AND donates e-pair ✔
not Brønsted–Lowry base AND does not accept proton/H⁺ ✔
Lewis definition extends/broader than Brønsted–Lowry definition ✔

Estimate the H−N−H bond angle in methanamine using VSEPR theory.

State the electron domain geometry around the nitrogen atom and its hybridization in methanamine.

Ammonia reacts reversibly with water.
$$\text{N}{\text{H}}_{\text{3}}\text{(g)}+{\text{H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{NH}}_{\text{4}}^{+}\text{(aq)}+\text{O}{\text{H}}^{-}\text{(aq)}$$
Explain the effect of adding ${\text{H}}^{+}\text{(aq)}$ ions on the position of the equilibrium.

Hydrazine reacts with water in a similar way to ammonia. (The association of a molecule of hydrazine with a second H⁺ is so small it can be neglected.)

$${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(aq)}+{\text{H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{N}}_{\text{2}}{\text{H}}_{\text{5}}^{+}\text{(aq)}+\text{O}{\text{H}}^{-}\text{(aq)}$$

$$\text{p}{K}_{\text{b}}\text{ (hydrazine)}=5.77$$

Calculate the pH of a $0.0100\text{ mol}\text{d}{\text{m}}^{-3}$ solution of hydrazine.

Suggest a suitable indicator for the titration of hydrazine solution with dilute sulfuric acid using section 22 of the data booklet.

Outline, using an ionic equation, what is observed when magnesium powder is added to a solution of ammonium chloride.

Determine the enthalpy change of reaction, $\mathrm{\Delta }H$, in kJ, when 1.00 mol of gaseous hydrazine decomposes to its elements. Use bond enthalpy values in section 11 of the data booklet.

$${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(g)}\to {\text{N}}_{\text{2}}\text{(g)}+\text{2}{\text{H}}_{\text{2}}\text{(g)}$$

The standard enthalpy of formation of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(l)}$ is $+50.6\text{ kJ}\text{mo}{\text{l}}^{-1}$. Calculate the enthalpy of vaporization, $\mathrm{\Delta }{H}_{\text{vap}}$, of hydrazine in $\text{kJ}\text{mo}{\text{l}}^{-1}$. $${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(l)}\to {\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(g)}$$ (If you did not get an answer to (f), use $-85\text{ kJ}$ but this is not the correct answer.)

Calculate, showing your working, the mass of hydrazine needed to remove all the dissolved oxygen from $\text{1000 d}{\text{m}}^{\text{3}}$ of the sample.

Calculate the volume, in $\text{d}{\text{m}}^{\text{3}}$, of nitrogen formed under SATP conditions. (The volume of 1 mol of gas = $\text{24.8 d}{\text{m}}^{\text{3}}$ at SATP.)

107°

Accept 100° to < 109.5°.

Literature value = 105.8°

[1 mark]

tetrahedral

sp³

No ECF allowed.

[2 marks]

removes/reacts with $\text{O}{\text{H}}^{-}$

moves to the right/products «to replace $\text{O}{\text{H}}^{-}$ ions»

Accept ionic equation for M1.

[2 marks]

K_b = 10^–5.77 / 1.698 x 10^–6
OR
$K_{\text{b}}=\frac{\left[{\text{N}}_{\text{2}}{\text{H}}_5^{+}\right]\times \left[\text{O}{\text{H}}^{-}\right]}{\left[{\text{N}}_{\text{2}}{\text{H}}_4\right]}$

 [OH^–]² «= 1.698 × 10^–6 × 0.0100» = 1.698 × 10^–8

OR

[OH^–] «$=\sqrt{1.698\times {10}^{-8}}$» = 1.303 × 10^–4 «mol dm^–3»

pH «$=-\text{lo}{\text{g}}_{10}\frac{1\times {10}^{-14}}{1.3\times {10}^{-4}}$» = 10.1

Award [3] for correct final answer.

Give appropriate credit for other methods containing errors that do not yield correct final answer.

[3 marks]

methyl red

OR

bromocresol green

OR

bromophenol blue

OR

methyl orange

[1 mark]

bubbles

OR

gas

OR

magnesium disappears

${\text{2NH}}_{\text{4}}^{+}\text{(aq)}+\text{Mg(s)}\to \text{M}{\text{g}}^{\text{2}+}\text{(aq)}+\text{2N}{\text{H}}_{\text{3}}\text{(aq)}+{\text{H}}_{\text{2}}\text{(g)}$

Do not accept “hydrogen” without reference to observed changes.

Accept "smell of ammonia".

Accept 2H⁺(aq) + Mg(s) $\to$ Mg²⁺(aq) + H₂(g)

Equation must be ionic.

[2 marks]

bonds broken:

E(N–N) + 4E(N–H)

OR

$158\ll \text{kJ}\text{mo}{\text{l}}^{-1}\gg +4\times 391\ll \text{kJ}\text{mo}{\text{l}}^{-1}\gg /1722\ll \text{kJ}\gg$

bonds formed:

E(N$\equiv$N) + 2E(H–H)

OR

$945\ll \text{kJ}\text{mo}{\text{l}}^{-1}\gg +2\times 436\ll \text{kJ}\text{mo}{\text{l}}^{-1}\gg /1817\ll \text{kJ}\gg$

$\ll \mathrm{\Delta }H=\text{bonds broken}-\text{bonds formed}=1722-1817=\gg -95\ll \text{kJ}\gg$

Award [3] for correct final answer.

Award [2 max] for +95 «kJ».

[3 marks]

OR

$\mathrm{\Delta }{H}_{\text{vap}}=-50.6\text{ kJ}\text{mo}{\text{l}}^{-1}-\text{(}-95\text{ kJ}\text{mo}{\text{l}}^{-1}\text{)}$

$\ll \mathrm{\Delta }{H}_{vap}=\gg +44\ll \text{kJ}\text{mo}{\text{l}}^{-1}\gg$

Award [2] for correct final answer. Award [1 max] for –44 «kJ mol^–1».

Award [2] for:

ΔH_vap = –50.6 kJ mol^–1 – (–85 J mol^–1) = ＋34 «kJ mol^–1».

Award [1 max] for –34 «kJ mol^–1».

[2 marks]

total mass of oxygen $\ll =8.0\times {10}^{-3}\text{ g}\text{d}{\text{m}}^{-3}\times 1000\text{ d}{\text{m}}^3\gg =8.0\ll \text{g}\gg$

$\text{n(}{\text{O}}_{\text{2}}\text{) }\ll =\frac{8.0\text{ g}}{32.00\text{ g}\text{mo}{\text{l}}^{-1}}=\gg 0.25\ll \text{mol}\gg$

OR

$\text{n(}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{)}=\text{n(}{\text{O}}_{\text{2}}\text{)}$

$\ll \text{mass of hydrazine}=0.25\text{ mol}\times 32.06\text{ g}\text{mo}{\text{l}}^{-1}=\gg 8.0\ll \text{g}\gg$

Award [3] for correct final answer.

[3 marks]

$\ll \text{n(}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{)}=\text{n(}{\text{O}}_{\text{2}}\text{)}=\frac{8.0\text{ g}}{32.00\text{ g}\text{mo}{\text{l}}^{-1}}=\gg 0.25\ll \text{mol}\gg$

$\ll \text{volume of nitrogen}=0.25\text{ mol}\times 24.8\text{ d}{\text{m}}^3\text{mo}{\text{l}}^{-1}\gg =6.2\ll \text{d}{\text{m}}^3\gg$

Award [1] for correct final answer.

[1 mark]

State the equation for the reaction of each substance with water.

Draw a diagram showing the delocalization of electrons in the conjugate base of butanoic acid.

Deduce the average oxidation state of carbon in butanoic acid.

A 0.250 mol dm⁻³ aqueous solution of butanoic acid has a concentration of hydrogen ions, [H⁺], of 0.00192 mol dm⁻³. Calculate the concentration of hydroxide ions, [OH⁻], in the solution at 298 K.

Determine the pH of a 0.250 mol dm⁻³ aqueous solution of ethylamine at 298 K, using section 21 of the data booklet.

Sketch the pH curve for the titration of 25.0 cm³ of ethylamine aqueous solution with 50.0 cm³ of butanoic acid aqueous solution of equal concentration. No calculations are required.

Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room temperature.

State a suitable reagent for the reduction of butanoic acid.

Deduce the product of the complete reduction reaction in (e)(i).

Butanoic acid:
CH₃CH₂CH₂COOH (aq) + H₂O (l) $\rightleftharpoons$ CH₃CH₂CH₂COO⁻ (aq) + H₃O⁺ (aq) ✔

Ethylamine:
CH₃CH₂NH₂ (aq) + H₂O (l) $\rightleftharpoons$ CH₃CH₂NH₃⁺ (aq) + OH⁻ (aq) ✔

Diagram showing:
dotted line along O–C–O AND negative charge

Accept correct diagrams with pi clouds.

–1 ✔

«$\frac{1.00\times {10}^{-14}\text{mo}{\text{l}}^2\text{d}{\text{m}}^{-6}}{0.00192\text{mol}\text{d}{\text{m}}^{-3}}$» = 5.21 × 10^–12 «mol dm^–3» ✔

«pK_b = 3.35, K_b = 10^–3.35 = 4.5 × 10^–4»

«C₂H₅NH₂ + H₂O $\rightleftharpoons$ C₂H₅NH₃⁺ + OH^–»

K_b = $\frac{\left[\text{O}{\text{H}}^{-}-\right]\left[\text{C}{\text{H}}_{\text{3}}\text{C}{\text{H}}_{\text{2}}\text{N}{{\text{H}}_{\text{3}}}^{\text{ + }}\right]}{\left[\text{C}{\text{H}}_{\text{3}}\text{C}{\text{H}}_{\text{2}}\text{N}{\text{H}}_{\text{2}}\right]}$

OR

«K_b =» 4.5 × 10^–4 = $\frac{\left[\text{O}{\text{H}}^{-}\right]\left[\text{C}{\text{H}}_{\text{3}}\text{C}{\text{H}}_{\text{2}}\text{N}{{\text{H}}_{\text{3}}}^{\text{ + }}\right]}{0.250}$

OR

«K_b =» 4.5 × 10^–4 = $\frac{x^2}{0.250}$ ✔

« x = [OH^–] =» 0.011 «mol dm^–3» ✔

«pH = –log$\frac{1.00\times {10}^{-14}}{0.011}=$» 12.04

OR

«pH = 14.00 – (–log 0.011)=» 12.04 ✔

Award [3] for correct final answer.

decreasing pH curve ✔

pH close to 7 (6–8) at volume of 25 cm³ butanoic acid ✔

weak acid/base shape with no flat «strong acid/base» parts on the curve ✔

Any two of:
butanoic acid forms more/stronger hydrogen bonds ✔
butanoic acid forms stronger London/dispersion forces ✔
butanoic acid forms stronger dipole–dipole interaction/force ✔

Accept “butanoic acid forms dimers”

Accept “butanoic acid has larger M_r/hydrocarbon chain/number of electrons” for M2.

Accept “butanoic acid has larger «permanent» dipole/more polar” for M3.

lithium aluminium hydride/LiAlH₄ ✔

butan-1-ol/1-butanol/CH₃CH₂CH₂CH₂OH ✔

The following curve was obtained using a pH probe.

State, giving a reason, the strength of the acid.

State a technique other than a pH titration that can be used to detect the equivalence point.

Deduce the pK_a for this acid.

The pK_a of an anthocyanin is 4.35. Determine the pH of a 1.60 × 10^–3 mol dm^–3 solution to two decimal places.

weak AND pH at equivalence greater than 7
OR
weak acid AND forms a buffer region

[1 mark]

calorimetry
OR
measurement of heat/temperature
OR
conductivity measurement

Accept “indicator” but not “universal indicator”.

[1 mark]

«pK_a = pH at half-equivalence =» 5.0

[1 mark]

K_a = ${10}^{-4.35}/4.46683\times {10}^{-5}$

[H₃O⁺] = $\sqrt{4.46683\times {10}^{-5}\times 1.60\times {10}^{-3}}/\sqrt{7.1469\times {10}^{-8}}/2.6734\times {10}^{-4}$ «mol dm^–3»

pH = «$-\log \sqrt{7.1469\times {10}^{-8}}=$» 3.57

Award [3] for correct final answer to two decimal places.

If quadratic equation used, then: [H₃O⁺] = 2.459 × 10^–4 «mol dm^–3» and pH = 3.61

[3 marks]

Write a balanced equation for the reaction.

Identify the major species, other than water and potassium ions, at these points.

State a suitable indicator for this titration. Use section 22 of the data booklet

Suggest, giving a reason, which point on the curve is considered a buffer region.

State the $K_{\mathrm{a}}$ expression for ethanoic acid.

Calculate the $K_{\mathrm{b}}$ of the conjugate base of ethanoic acid using sections 2 and 21 of the data booklet.

In a titration, $25.00 {\mathrm{cm}}^3$ of vinegar required $20.75 {\mathrm{cm}}^3$ of $1.00 \mathrm{mol} {\mathrm{dm}}^{-3}$ potassium hydroxide to reach the end-point.

Calculate the concentration of ethanoic acid in the vinegar.

Potassium hydroxide solutions can react with carbon dioxide from the air. The solution was made one day prior to using it in the titration.

State the type of error that would result from the student’s approach.

Potassium hydroxide solutions can react with carbon dioxide from the air. The solution was made one day prior to using it in the titration.

Predict, giving a reason, the effect of this error on the calculated concentration of ethanoic acid in 5(e).

${\mathrm{CH}}_3\mathrm{COOH}\left(\mathrm{aq}\right)+\mathrm{KOH}\left(\mathrm{aq}\right)\to {\mathrm{CH}}_3\mathrm{COOK} \left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$ ✔

Accept the ionic equation.

B: ${\mathrm{CH}}_3\mathrm{COOH}$ AND ${\mathrm{CH}}_3{\mathrm{COO}}^{-}$ ✔

C: ${\mathrm{CH}}_3{\mathrm{COO}}^{-}$ ✔

Accept names.

Accept $C{H}_{\mathit{3}}COOK$ for $C{H}_{\mathit{3}}CO{O}^{\mathit{-}}$

phenolphthalein ✔

Accept “phenol red” or “bromothymol blue”.

B AND the region where small additions «of the base/$\mathrm{KOH}$ » result in little or no
change in $\mathrm{pH}$
OR
B AND the flattest region of the curve «at intermediate $\mathrm{pH}$/before equivalence
point »
OR
B AND half the volume needed to reach equivalence point
OR
B AND similar amounts of weak acid/${\mathrm{CH}}_3\mathrm{COOH}$/ethanoic acid AND conjugate base/${\mathrm{CH}}_3{\mathrm{COO}}^{-}$/ethanoate ✔

$K_{\mathrm{a}}=\frac{\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}$

Accept $H^{+}$ instead of $H_3{O}^{+}$.

$«K_{\mathrm{a}}={10}^{-4.76}=1.7\times {10}^{-5}»$
$«K_{\mathrm{w}}=K_{\mathrm{a}}·K_{\mathrm{b}}=1.0\times {10}^{-14}=1.7\times {10}^{-5}\times K_{\mathrm{b}}»$
$«{\mathrm{K}}_{\mathrm{b}}=»5.8\times {10}^{-10}$ ✔

Accept answers between $\mathit{5}\mathit{.}\mathit{7}\mathit{-}\mathit{5}\mathit{.}\mathit{9}\mathit{\times }{\mathit{10}}^{\mathit{-}\mathit{10}}$.

$«\mathrm{n}\left(\mathrm{KOH}\right)=0.02075 {\mathrm{dm}}^3\times 1.00 \mathrm{mol} {\mathrm{dm}}^{-3}=»0.0208 «\mathrm{mol}»$ ✔

$«\mathrm{n}\left(\mathrm{KOH}\right)=\mathrm{n}\left({\mathrm{CH}}_3\mathrm{COOH}\right)»$
$«\left[{\mathrm{CH}}_3\mathrm{COOH}\right]=\frac{0.0208 \mathrm{mol}}{0.02500 {\mathrm{dm}}^3}=»0.830 «\mathrm{mol} {\mathrm{dm}}^{-3}»$ ✔

Award [2] for correct final answer.

systematic «error» ✔

$\left[{\mathrm{CH}}_3\mathrm{COOH}\right]$ would be higher ✔

actual $\left[\mathrm{KOH}\right]$ is lower «than the value in calculation»
OR
larger volume of $\mathrm{KOH}$ «solution» needed to neutralize the acid ✔

Accept $KOH$ partially neutralised by $C{O}_2$ from air.

Most candidates could write a balanced neutralization equation.

Identifying species present at various points along a pH titration curve was one of the most poorly answered questions in the exam. Very few candidates realized there were two major species at point B even when they were able in general to realize that B was a buffer zone.

Almost all candidates could identify a suitable indicator to use in a titration of a weak acid with a strong base.

Most students could identify a buffer zone region in a titration but very few (50%) could coherently explain why.

Poorly answered with only 50% correctly writing a K_a expression. The major error was in candidates trying to calculate a K_a rather than write an expression for it.

Like with other calculations in this exam, the majority of candidates could correctly determine a concentration from titration data.

80% of candidates could identify the method used as a systematic error, with some stating human or random error.

Most candidates identified that the systematic error would result in the concentration of the alkali being lowered but then failed to propagate this through to the effect on the concentration of the acid.

The graph represents the titration of 25.00 cm³ of 0.100 mol dm⁻³ aqueous ethanoic acid with 0.100 mol dm⁻³ aqueous sodium hydroxide.

Deduce the major species, other than water and sodium ions, present at points A and B during the titration.

Calculate the pH of 0.100 mol dm⁻³ aqueous ethanoic acid.

K_a = 1.74 × 10⁻⁵

Outline, using an equation, why sodium ethanoate is basic.

Predict whether the pH of an aqueous solution of ammonium chloride will be greater than, equal to or less than 7 at 298 K.

Formulate the equation for the reaction of nitrogen dioxide, NO₂, with water to form two acids.

Formulate the equation for the reaction of one of the acids produced in (e)(i) with calcium carbonate.

A: CH₃COOH/ethanoic/acetic acid AND CH₃COO^–/ethanoate/acetate ions

B: CH₃COO^–/ethanoate/acetate ions

Penalize “sodium ethanoate/acetate” instead of “ethanoate/acetate ions” only once.

[2 marks]

$K_{\text{a}}=1.74\times {10}^{-5}=\frac{{\text{[}{\text{H}}^{+}\text{]}}^2}{0.10}$

OR

[H⁺] = 1.32 × 10^–3 «mol dm^–3»

«pH =» 2.88

Accept [2] for correct final answer.

[2 marks]

«forms weak acid and strong base, thus basic»

CH₃COO^–(aq) + H₂O(l) $\rightleftharpoons$ CH₃COOH(aq) + OH^–(aq)

Accept → for $\rightleftharpoons$.

[1 mark]

less than 7

[1 mark]

2NO₂(g) + H₂O(l) → HNO₂(aq) + HNO₃(aq)

[1 mark]

2HNO₂(aq) + CaCO₃(s) → Ca(NO₂)₂(aq) + CO₂(g) + H₂O(l)

OR

2HNO₃(aq) + CaCO₃(s) → Ca(NO₃)₂(aq) + CO₂(g) + H₂O(l)

[1 mark]

Calculate the pH of 0.0100 mol dm^–3 methanoic acid stating any assumption you make. K_a = 1.6 × 10^–4.

(i) Sketch a graph of pH against volume of a strong base added to a weak acid showing how you would determine pK_a for the weak acid.

(ii) Explain, using an equation, why the pH increases very little in the buffer region when a small amount of alkali is added.

Calculation:

ALTERNATIVE 1:
[H⁺] = (K_a × [HA])^1/2 / (1.6 × 10^–4 × 0.0100)^1/2 / 1.3 × 10^–3 «mol dm^–3»

pH = «–log₁₀[H⁺] ≈» 2.9

ALTERNATIVE 2:
pH = 0.5(pK_a - log₁₀[HA])
pH = 2.9

Award [2] for correct final answer

Assumption:
ionisation is << 0.0100 so 0.0100 - [A^–] ≈ 0.0100
OR
[HA]_eqm = [HA]_initial
OR
all H⁺ ions in the solution come from the acid «and not from the self-ionisation of water»
OR
[H⁺] = [HCOO^–]

Do not accept partial dissociation

i

correct shape of graph
pH at half neutralization/equivalence

M1: must show buffer region at pH < 7 and equivalence at pH > 7.
Accept graph starting from where two axes meet as pH scale is not specified.

ii

ALTERNATIVE 1:

HCOOH  HCOO^– + H⁺
H⁺ ions consumed in reaction with OH^– are produced again as equilibrium moves to the right «so [H⁺] remains almost unchanged»

ALTERNATIVE 2:
HCOOH + OH^–  HCOO^– + H₂O
added OH^- are neutralized by HCOOH
OR
strong base replaced by weak base 

Accept HA or any other weak acid in equations.
Equilibrium sign must be included in equation for M1

Calculate the initial pH before any sodium hydroxide was added, using section 21 of the data booklet.

The concentration of excess sodium hydroxide was 0.362 mol dm⁻³. Calculate the pH of the solution at the end of the experiment.

Sketch the neutralisation curve obtained and label the equivalence point.

«Ka = 10^–2.87 = 1.35 × 10^–3 »

«1.35 × 10^–3 = $\frac{\left[\mathrm{chloroethanoate}\right]\times \left[{\mathrm{H}}^{+}\right]}{0.50 \mathrm{mol} {\mathrm{dm}}^{-3}}=\frac{{\mathrm{x}}^2}{0.50 \mathrm{mol} {\mathrm{dm}}^{-3}}$ »

«x = [H⁺] =$\sqrt{1.4\times {10}^{-3}\times 0.50}$=» 2.6 × 10^–2 «mol dm^–3» ✔

«pH = –log[H⁺] = –log(2.6 × 10^–2) =» 1.59 ✔

Accept final answer in range 1.58–1.60.

Award [2] for correct final answer.

«pOH = –log(0.362) = 0.441»

«pH = 14.000 – 0.441 =» 13.559 ✔

OR

starts at 1.6 AND finishes at 13.6 ✔

approximately vertical at the correct volume of alkali added ✔

equivalence point labelled AND above pH 7 ✔

Accept any range from 1.1-1.9 AND 13.1-13.9 for M1 or ECF from 11c(i) and 11c(ii).

Award M2 for vertical climb at 28 cm³ OR 15 cm³.

Equivalence point must be labelled for M3.

Calculate the pH of 0.00100 mol dm^–3 propanoic acid solution. Use section 21 of the data booklet.

Sketch the general shape of the variation of pH when 50 cm³ of 0.001 mol dm^–3 NaOH (aq) is gradually added to 25 cm³ of 0.001 mol dm^–3 CH₃CH₂COOH (aq).

K_a = 10^−4.87 / 1.35 × 10⁻⁵ ✔

[H⁺] = «$\sqrt{1.35\times {10}^{-5}\times 0.001}=\sqrt{1.35\times {10}^{-8}}=$» 1.16 × 10⁻⁴ «mol dm⁻³» ✔

pH = 3.94 ✔

Accept alternative methods of calculation.

Award [3] for correct final answer.

Award [3] for 3.96 {answer if solved by quadratic}.

Any three of:

correct “S” shape ✔

equivalence point at 25 cm³ ✔

final pH tends to 11 ✔

pH at equivalence point >7 ✔

starting pH between 3.8 - 4 ✔

pH at half equivalence approx. 5 ✔

Do not penalize for incorrect points.
Award any 3 correct.

State why NH₃ is a Lewis base.

Calculate the pH of a 1.00 × 10⁻² mol dm⁻³ aqueous solution of ammonia.

pK_b = 4.75 at 298 K.

Justify whether a 1.0 dm³ solution made from 0.10 mol NH³ and 0.20 mol HCl will form a buffer solution.

Sketch the shape of one sigma ($\text{σ}$) and one pi ($\mathrm{\pi }$) bond.

Identify the number of sigma and pi bonds in HCN.

State the hybridization of the carbon atom in HCN.

Suggest why hydrogen chloride, HCl, has a lower boiling point than hydrogen cyanide, HCN.

Explain why transition metal cyanide complexes are coloured.

donates «lone/non-bonding» pair of electrons ✔

Kb = 10^-4.75 /1.78 x 10^-5
OR
Kb = $\frac{{\left[{\mathrm{OH}}^{-}\right]}^2}{\left[{\mathrm{NH}}_3\right]}$ ✔

[OH^–] = « $\sqrt{\left(1.00\times {10}^{-2}\times {10}^{-4.75}\right)}$ =» 4.22 × 10^–4 «(mol dm^–3)» ✔

pOH« = –log₁₀ (4.22 × 10^–4)» = 3.37
AND
pH = «14 – 3.37» = 10.6

OR

[H⁺]« =$\frac{1.00\times {10}^{-14}}{4.22\times {10}^{-4}}$» = 2.37 × 10^–11
AND
pH« = –log₁₀ 2.37 × 10^–11» = 10.6 ✔

Award [3] for correct final answer.

no AND is not a weak acid conjugate base system

OR

no AND weak base «totally» neutralized/ weak base is not in excess

OR

no AND will not neutralize small amount of acid ✔

Accept “no AND contains 0.10 mol NH₄Cl + 0.10 mol HCl”.

Sigma ($\sigma$):

Pi ($\pi$):

Accept overlapping p-orbital(s) with both lobes of equal size/shape.

Shaded areas are not required in either diagram.

Sigma ($\text{σ}$): 2 AND Pi ($\mathrm{\pi }$): 2 ✔

sp ✔

HCN has stronger dipole–dipole attraction ✔

Do not accept reference to H-bonds.

Any three from:

partially filled d-orbitals ✔

«CN- causes» d-orbitals «to» split ✔

light is absorbed as electrons transit to a higher energy level «in d–d transitions»
OR
light is absorbed as electrons are promoted ✔

energy gap corresponds to light in the visible region of the spectrum ✔

Do not accept “colour observed is the complementary colour” for M4.

The main error was the omission of lone electron "pair", though there was also a worrying amount of very confused answers for a very basic chemistry concept where 40% provided very incorrect answers.

Rather surprisingly, many students got full marks for this multi-step calculation; others went straight to the pH/pKa acid/base equation so lost at least one of the marks: students often seem less prepared for base calculations, as opposed to acid calculations.

Poorly answered revealing little understanding of buffering mechanisms, which is admittedly a difficult topic.

This proved to be the most challenging question (10%). It was a good question, where candidates had to explain a huge difference in boiling point of two covalent compounds, requiring solid understanding of change of state where breaking bonds cannot be involved). Yet most considered the triple bonds in HCN as the cause, suggesting covalent bonds break when substance boil, which is very worrying. Others considered H-bonds which at least is an intermolecular force, but shows they are not too familiar with the conditions necessary for H-bonding.

This question appears frequently in exams but with slightly different approaches. In general candidates ignore the specific question and give the same answers, failing in this case to describe why complexes are coloured rather than what colour is seen. These answers appear to reveal that many candidates don't really understand this phenomenon, but learn the answer by heart and make mistakes when repeating it, for example, stating that the ‘d-orbitals of the ligands were split’- an obvious misconception. The average mark was 1.6/3, with a MS providing 4 ideas that would merit a mark

Write a balanced equation for the reaction that occurs.

Identify a metal, in the same period as magnesium, that does not form a basic oxide.

Calculate the amount of magnesium, in mol, that was used.

Determine the percentage uncertainty of the mass of product after heating.

Assume the reaction in (a)(i) is the only one occurring and it goes to completion, but some product has been lost from the crucible. Deduce the percentage yield of magnesium oxide in the crucible.

Evaluate whether this, rather than the loss of product, could explain the yield found in (b)(iii).

Suggest an explanation, other than product being lost from the crucible or reacting with nitrogen, that could explain the yield found in (b)(iii).

Calculate coefficients that balance the equation for the following reaction.

Ammonia is added to water that contains a few drops of an indicator. Identify an indicator that would change colour. Use sections 21 and 22 of the data booklet.

Determine the oxidation state of nitrogen in Mg₃N₂ and in NH₃.

Deduce, giving reasons, whether the reaction of magnesium nitride with water is an acid–base reaction, a redox reaction, neither or both.

State the number of subatomic particles in this ion.

Some nitride ions are ¹⁵N^3–. State the term that describes the relationship between ¹⁴N^3– and ¹⁵N^3–.

The nitride ion and the magnesium ion are isoelectronic (they have the same electron configuration). Determine, giving a reason, which has the greater ionic radius.

Suggest, giving a reason, whether magnesium or nitrogen would have the greater sixth ionization energy.

Suggest two reasons why atoms are no longer regarded as the indivisible units of matter.

State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.

2 Mg(s) + O₂(g) → 2 MgO(s) ✔

Do not accept equilibrium arrows. Ignore state symbols

aluminium/Al ✔

$⟨⟨\frac{53.726 \mathrm{g}-47.372 \mathrm{g}}{244.31 \mathrm{g} {\mathrm{mol}}^{-1}}=\frac{6.354 \mathrm{g}}{24.31 \mathrm{g} {\mathrm{mol}}^{-1}}⟩⟩=0.2614 «\mathrm{mol}»✔$

mass of product $«=56.941 \mathrm{g}-47.372 \mathrm{g}»=9.569 «\mathrm{g}»$ ✔

$\text{⟨⟨100 × }\frac{2\times 0.001 \mathrm{g}}{9.569 \mathrm{g}}\text{=0.0209⟩⟩ = 0.02 «\%»}$ ✔

Award [2] for correct final answer

Accept 0.021%

$⟨⟨ 0.2614 \mathrm{mol} \times (24.31 \mathrm{g} {\mathrm{mol}}^{-1}+16.00 \mathrm{g} {\mathrm{mol}}^{-1})=0.2614 \mathrm{mol}\times 40.31 \mathrm{g} {\mathrm{mol}}^{-1}⟩⟩=10.536 «\mathrm{g}»$ ✔

$⟨⟨100\times \frac{9.569 \mathrm{g}}{10.536 \mathrm{g}}= 90.822⟩⟩=91 «\%»$ ✔

Award «0.2614 mol x 40.31 g mol^–1»

Accept alternative methods to arrive at the correct answer.

Accept final answers in the range 90.5-91.5%

[2] for correct final answer.

yes
AND
«each Mg combines with $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ N, so» mass increase would be 14x$\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ which is less than expected increase of 16x
OR
3 mol Mg would form 101g of Mg₃N₂ but would form 3 x MgO = 121 g of MgO
OR
0.2614 mol forms 10.536 g of MgO, but would form 8.796 g of Mg₃N₂ ✔

Accept Yes AND “the mass of N/N₂ that combines with each g/mole of Mg is lower than that of O/O₂”

Accept YES AND “molar mass of nitrogen less than of oxygen”.

incomplete reaction
OR
Mg was partially oxidised already
OR
impurity present that evaporated/did not react ✔

Accept “crucible weighed before fully cooled”.

Accept answers relating to a higher atomic mass impurity consuming less O/O₂.

Accept “non-stoichiometric compounds formed”.

Do not accept "human error", "wrongly calibrated balance" or other non-chemical reasons.

If answer to (b)(iii) is >100%, accept appropriate reasons, such as product absorbed moisture before being weighed.

«1» Mg₃N₂ (s) + 6 H₂O (l) → 3 Mg(OH)₂ (s) + 2 NH₃ (aq) ✔

phenol red ✔

Accept bromothymol blue or phenolphthalein.

Mg₃N₂: -3
AND
NH₃: -3 ✔

Do not accept 3 or 3-

Acid–base:
yes AND N^3- accepts H⁺/donates electron pair«s»
OR
yes AND H₂O loses H⁺ «to form OH^-»/accepts electron pair«s» ✔

Redox:
no AND no oxidation states change ✔

Accept “yes AND proton transfer takes place”

Accept reference to the oxidation state of specific elements not changing.

Accept “not redox as no electrons gained/lost”.

Award [1 max] for Acid–base: yes AND Redox: no without correct reasons, if no other mark has been awarded

Protons: 7 AND Neutrons: 7 AND Electrons: 10 ✔

isotope«s» ✔

nitride AND smaller nuclear charge/number of protons/atomic number ✔

nitrogen AND electron lost from first «energy» level/s sub-level/s-orbital AND magnesium from p sub-level/p-orbital/second «energy» level
OR
nitrogen AND electron lost from lower level «than magnesium» ✔

Accept “nitrogen AND electron lost closer to the nucleus «than magnesium»”.

Any two of:

subatomic particles «discovered»
OR
particles smaller/with masses less than atoms «discovered»
OR
«existence of» isotopes «same number of protons, different number of neutrons» ✔

charged particles obtained from «neutral» atoms
OR
atoms can gain or lose electrons «and become charged» ✔

atom «discovered» to have structure ✔

fission
OR
atoms can be split ✔

Accept atoms can undergo fusion «to produce heavier atoms»

Accept specific examples of particles.

Award [2] for “atom shown to have a nucleus with electrons around it” as both M1 and M3.

Award [1] for all bonding types correct.

Award [1] for each correct description.

Apply ECF for M2 only once.

Done very well. However, it was disappointing to see the formula of oxygen molecule as O and the oxide as Mg₂O and MgO₂ at HL level.

Average performance; the question asked to identify a metal; however, answers included S, Si, P and even noble gases besides Be and Na. The only choice of aluminium; however, since its oxide is amphoteric, it could not be the answer in the minds of some.

Very good performance; some calculated the mass of oxygen instead of magnesium for the calculation of the amount, in mol, of magnesium. Others calculated the mass, but not the amount in mol as required.

Mediocre performance; instead of calculating percentage uncertainty, some calculated percentage difference.

Satisfactory performance; however, a good number could not answer the question correctly on determining the percentage yield.

Poorly done. The question asked to evaluate and explain but instead many answers simply agreed with the information provided instead of assessing its strength and limitation.

Mediocre performance; explaining the yield found was often a challenge by not recognizing that incomplete reaction or Mg partially oxidized or impurities present that evaporated or did not react would explain the yield.

Calculating coefficients that balance the given equation was done very well.

Well done; some chose bromocresol green or methyl red as the indicator that would change colour, instead of phenol red, bromothymol blue or phenolphthalein.

Good performance; however, surprising number of candidates could not determine one or both oxidation states correctly or wrote it as 3 or 3−, instead of −3.

Average performance; choosing the given reaction as an acid-base or redox reaction was not done well. Often answers were contradictory and the reasoning incorrect.

Stating the number of subatomic particles in a ¹⁴N^3- was done very well. However, some answers showed a lack of understanding of how to calculate the number of relevant subatomic particles given formula of an ion with charge and mass number.

Exceptionally well done; A few candidates referred to isomers, rather than isotopes.

There was reference to nitrogen and magnesium, rather than nitride and magnesium ions. Also, instead identifying smaller nuclear charge in nitride ion, some referred to core electrons, Z_eff, increased electron-electron repulsion or shielding.

Common error in suggesting nitrogen would have the greater sixth ionization energy was that for nitrogen, electron is lost from first energy level without making reference to magnesium losing it from second energy level.

Good performance; some teachers were concerned about the expected answers. However, generally, students were able to suggest two reasons why matter is divisible.

One teacher commented that not asking to describe bonding in terms of electrostatic attractions as in earlier papers would have been confusing and some did answer in terms of electrostatic forces of attractions involved. However, the question was clear in its expectation that the answer had to be in terms of how the valence electrons produce the three types of bonds and the overall performance was good. Some had difficulty identifying the bond type for Mg, O₂ and MgO.

Distinguish between a weak and strong acid.

Weak acid: 

Strong acid: 

The hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.

State the formula of its conjugate base.

When a bottle of carbonated water is opened, these equilibria are disturbed.

State, giving a reason, how a decrease in pressure affects the position of Equilibrium (1).

At 298 K the concentration of aqueous carbon dioxide in carbonated water is 0.200 mol dm⁻³ and the pK_a for Equilibrium (2) is 6.36.

Calculate the pH of carbonated water.

Identify the type of bonding in sodium hydrogencarbonate.

Between sodium and hydrogencarbonate:

Between hydrogen and oxygen in hydrogencarbonate:

Predict, referring to Equilibrium (2), how the added sodium hydrogencarbonate affects the pH.(Assume pressure and temperature remain constant.)

100.0cm³ of soda water contains 3.0 × 10⁻²g NaHCO₃.

Calculate the concentration of NaHCO₃ in mol dm⁻³.

The uncertainty of the 100.0cm³ volumetric flask used to make the solution was ±0.6cm³.

Calculate the maximum percentage uncertainty in the mass of NaHCO₃ so that the concentration of the solution is correct to ±1.0 %.

The reaction of the hydroxide ion with carbon dioxide and with the hydrogencarbonate ion can be represented by Equations 3 and 4.

Equation (3)     OH⁻ (aq) + CO₂ (g) → HCO₃⁻ (aq)
Equation (4)     OH⁻ (aq) + HCO₃⁻ (aq) → H₂O (l) + CO₃²⁻ (aq)

Discuss how these equations show the difference between a Lewis base and a Brønsted–Lowry base.

Equation (3):

Equation (4):

Aqueous sodium hydrogencarbonate has a pH of approximately 7 at 298 K.

Sketch a graph of pH against volume when 25.0cm³ of 0.100 mol dm⁻³ NaOH (aq) is gradually added to 10.0cm³ of 0.0500 mol dm⁻³ NaHCO₃ (aq).

Weak acid: partially dissociated/ionized «in aqueous solution/water»
AND
Strong acid: «assumed to be almost» completely/100 % dissociated/ionized «in aqueous solution/water»    [✔]

CO₃^2-    [✔]

shifts to left/reactants AND to increase amount/number of moles/molecules of gas/CO₂ (g)    [✔]

Note: Accept “shifts to left/reactants AND to increase pressure”.

«K_a =» 10^–6.36/4.37 × 10^–7 = $\frac{{[{\text{H}}^{+}]}^2}{[\text{C}{\text{O}}_2]}$
OR
«K_a =» 10^–6.36/4.37 × 10^–7 = $\frac{{[{\text{H}}^{+}]}^2}{0.200}$  [✔]

[H⁺] « $\sqrt{0.200\times 4.37\times {10}^{-7}}$  » = 2.95 × 10^–4 «mol dm^–3»     [✔]
«pH =» 3.53     [✔]

Note: Award [3] for correct final answer.

Between sodium and hydrogencarbonate:
ionic    [✔]

Between hydrogen and oxygen in hydrogencarbonate:
«polar» covalent     [✔]

«additional HCO₃^-» shifts position of equilibrium to left   [✔]

pH increases   [✔]

Note: Do not award M2 without any justification in terms of equilibrium shift in M1.

«molar mass of NaHCO₃ =» 84.01 «g mol^-1»    [✔]

«concentration = $\frac{3.0\times {10}^{-2}\text{g}}{84.01\text{ g mo}{\text{l}}^{-1}}\times \frac{1}{0.100\text{ d}{\text{m}}^3}$ =» 3.6 × 10^–3 «mol dm^-3»     [✔]

Note: Award [2] for correct final answer.

«1.0 – 0.6 = ± » 0.4 «%»    [✔]

Equation (3):
OH^- donates an electron pair AND acts as a Lewis base     [✔]

Equation (4):
OH^- accepts a proton/H⁺/hydrogen ion AND acts as a Brønsted–Lowry base     [✔]

S-shaped curve from ~7 to between 12 and 14     [✔]

equivalence point at 5 cm³     [✔]

Note: Accept starting point >6~7.

As expected, many candidates were able to distinguish between strong and weak acids; some candidates referred to “dissolve” rather than dissociate.

More than half the candidates were able to deduce that carbonate was the conjugate base but a significant proportion of those that did, wrote the carbonate ion with an incorrect charge.

Many students gave generic responses referring to a correct shift without conveying the idea of compensation or restoration of pressure or moles of gas. This generic reply reflects the difficulty in applying a theoretical concept to the practical situation described in the question.

Most candidates calculated the pH of the aqueous CO₂. Some candidates attempted to use the Henderson-Hasselback equation and others used the quadratic expression to calculate [H⁺] (these two options were very common in the Spanish scripts) getting incorrect solutions. These answers usually ended in pH of approx. 1 which candidates should realize cannot be correct for soda water.

This was an easy question, especially the identification of the type of bond between H and O, yet some candidates interpreted that the question referred to intermolecular bonding.

A significant number of candidates omitted the “equilibrium” involved in the dissolution of a weak base.

This is another stoichiometry question that most candidates were able to solve well, with occasional errors when calculating M_r of hydrogen carbonate.

Mixed responses, more attention should be given to this simple calculation which is straightforward and should be easy as required for IA reports.

This was a good way to test this topic because answers showed that, while candidates usually knew the topic in theory, they could not apply this to identify the Lewis and Bronsted-Lowry bases in the context of a reaction that was given to them. In some cases, they failed to specify the base, OH^- or also lost marks referring just to electrons, an electron or H instead of hydrogen ions or H⁺ for example.

Most students that got 1mark for this titration curve was for the general shape, because few realized they had the data to calculate the equivalence point. There were also some difficulties in establishing the starting point even if it was specified in the stem.

Outline why ethanoic acid is classified as a weak acid.

A solution of bleach can be made by reacting chlorine gas with a sodium hydroxide solution.

Cl₂ (g) + 2NaOH (aq) ⇌ NaOCl (aq) + NaCl (aq) + H₂O (l)

Suggest, with reference to Le Châtelier’s principle, why it is dangerous to mix vinegar and bleach together as cleaners.

Draw a Lewis (electron dot) structure of chloramine.

State the hybridization of the nitrogen atom in chloramine.

Deduce the molecular geometry of chloramine and estimate its H–N–H bond angle.

Molecular geometry:

H–N–H bond angle:

State the type of bond formed when chloramine is protonated.

Sketch a graph of pH against volume of hydrochloric acid added to ammonia solution, showing how you would determine the pK_a of the ammonium ion.

Suggest a suitable indicator for the titration, using section 22 of the data booklet.

Explain, using two equations, how an equimolar solution of ammonia and ammonium ions acts as a buffer solution when small amounts of acid or base are added.

partial dissociation «in aqueous solution»    [✔]

ethanoic acid/vinegar reacts with NaOH    [✔]

moves equilibrium to left/reactant side    [✔]

releases Cl₂ (g)/chlorine gas
OR
Cl₂ (g)/chlorine gas is toxic    [✔]

Note: Accept “ethanoic acid produces H⁺ ions”

Accept “ethanoic acid/vinegar reacts with NaOCl”.

Do not accept “2CH₃COOH + NaOCl + NaCl → 2CH₃COONa + Cl₂ + H₂O” as it does not refer to equilibrium.

Accept suitable molecular or ionic equations for M1 and M3.

    [✔]

Note: Accept any combination of dots/crosses or lines to represent electron pairs.

sp³    [✔]

Molecular geometry:
«trigonal» pyramidal   [✔]

H–N–H bond angle:
107°    [✔]

Note: Accept angles in the range of 100–109.

covalent/dative/coordinate    [✔]

correct shape of graph AND vertical drop at Vn    [✔]

pK_a = pH at $\frac{\text{Vn}}{2}$/half neutralization/half equivalence    [✔]

Note: M1: must show buffer region at pH > 7 and equivalence point at pH < 7. Graph must start below pH = 14.

methyl orange
OR
bromophenol blue
OR
bromocresol green
OR
methyl red    [✔]

NH₃ (aq) + H⁺ (aq) → NH₄ + (aq)    [✔]

NH₄ + (aq) + OH⁻ (aq) → NH₃ (aq) + H₂O(l)    [✔]

Note: Accept reaction arrows or equilibrium signs in both equations.

Award [1 max], based on two correct reverse equations but not clearly showing reacting with acid or base but rather dissociation.

Majority of candidates understood weak acids do not fully dissociate.

The average score was 1 out 3. Many could not suggest why it is dangerous to mix chlorine with vinegar. Most students gained at least one mark for stating that “chlorine gas will be produced” but couldn’t link it to equilibrium ideas.

Most candidates correctly drew the Lewis structure of chloramine. Some left off lone pair electrons.

Mostly correct with a surprising number stating sp or sp² hybridization.

Generally well done with some candidates misinterpreting the bond angle from the stated geometry.

“Ionic bond”, “hydrogen bond” and “intermolecular forces” were some common answers.

Quite poorly done with many candidates not indicating a vertical drop but rather a weak acid/weak base curve. Some did not have the correct location for the equivalence point.

Generally well done although a number of candidates chose bromothymol blue as a suitable indicator for weak base with a strong acid.

Nearly 30 % of candidates did not attempt to answer this question about buffer equations. It was also poorly answered because equations were not used to explain buffer action or the dissociation equations for the base and acid were given rather than their reactions with H⁺ or OH^- .

A sample of ethanoic acid was titrated with sodium hydroxide solution, and the following pH curve obtained.

Annotate the graph to show the buffer region and the volume of sodium hydroxide at the equivalence point.

Identify the most suitable indicator for the titration using section 22 of the data booklet.

Describe, using a suitable equation, how the buffer solution formed during the titration resists pH changes when a small amount of acid is added.

buffer region on graph ✔
equivalence point/V_eq on graph ✔

NOTE: Construction lines not required.

phenolphthalein ✔

NOTE: Accept phenol red.

ALTERNATIVE 1:
H⁺ (aq) + CH₃COO^– (aq) → CH₃COOH (aq) ✔

added acid neutralised by ethanoate ions
OR
«weak» CH₃COOH (aq)/ethanoic acid replaces H⁺ (aq)
OR
CH₃COOH/CH₃COO^– ratio virtually/mostly unchanged ✔

ALTERNATIVE 2:
CH₃COOH (aq) $\rightleftharpoons$ H⁺ (aq) + CH₃COO^– (aq) ✔

equilibrium shifts to the ethanoic acid side
OR
CH₃COOH/CH₃COO⁻ ratio virtually/mostly unchanged ✔

State the relationship between NH₄⁺ and NH₃ in terms of the Brønsted–Lowry theory.

Determine the concentration, in mol dm^–3, of the solution formed when 900.0 dm³ of NH₃(g) at 300.0 K and 100.0 kPa, is dissolved in water to form 2.00 dm³ of solution. Use sections 1 and 2 of the data booklet.

Calculate the concentration of hydroxide ions in an ammonia solution with pH = 9.3. Use sections 1 and 2 of the data booklet.

Calculate the concentration, in mol dm^–3, of ammonia molecules in the solution with pH = 9.3. Use section 21 of the data booklet.

An aqueous solution containing high concentrations of both NH₃ and NH₄⁺ acts as an acid-base buffer solution as a result of the equilibrium:

NH₃ (aq) + H⁺ (aq) $\rightleftharpoons$ NH₄⁺ (aq)

Referring to this equilibrium, outline why adding a small volume of strong acid would leave the pH of the buffer solution almost unchanged.

Magnesium salts form slightly acidic solutions owing to equilibria such as:

Mg²⁺(aq) + H₂O (l) $\rightleftharpoons$ Mg(OH)⁺(aq) + H⁺(aq)

Comment on the role of Mg²⁺ in forming the Mg(OH)⁺ ion, in acid-base terms.

Mg(OH)⁺ is a complex ion, but Mg is not regarded as a transition metal. Contrast Mg with manganese, Mn, in terms of one characteristic chemical property of transition metals, other than complex ion formation.

conjugate «acid and base» ✔

amount of ammonia $⟨⟨=\frac{P.V}{R.T}=\frac{100.0 kPa\times 900.0 d{m}^3}{8.31 J K^{-1}mo{l}^{-1}\times 300.0 K}⟩⟩ = 36.1 «\mathrm{mol}»$ ✔

concentration $⟨⟨=\frac{n}{V}=\frac{36.1}{2.00}⟩⟩=18.1 «\mathrm{mol} {\mathrm{dm}}^{-3}»$ ✔

Award [2] for correct final answer.

$\left[{\mathrm{OH}}^{-}\right] ⟨⟨=\frac{{\mathrm{K}}_{\mathrm{W}}}{\left[{\mathrm{H}}^{+}\right]}=\frac{{10}^{-14}}{{10}^{-9.3}}={10}^{-4.7}⟩⟩=2.0 \times {10}^{-5}⟨⟨\mathrm{mol} {\mathrm{dm}}^{-3}⟩⟩$  ✔

$K_b=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_3\right]}/\frac{{10}^{-4.7}\times {10}^{-4.7}}{\left[N{H}_3\right]}⟨⟨={10}^{-4.75}⟩⟩$ ✔

$\left[{\mathrm{NH}}_3\right]=⟨⟨=\frac{{10}^{-9.4}}{{10}^{-4.75}}={10}^{-4.65}⟩⟩=2.24\times {10}^{-5}«\mathrm{mol} {\mathrm{dm}}^{-3}»$ ✔

Accept other methods of carrying out the calculation.

Award [2] for correct answer.

equilibrium shifts to right/H⁺ reacts with NH₃ ✔

«as large excess» ratio [NH₃]:[NH₄⁺] «and hence pH» almost unchanged ✔

Accept “strong acid/H⁺ converted to a weak acid/NH₄⁺ «and hence pH almost unchanged».

Lewis acid ✔

accepts «a lone» electron pair «from the hydroxide ion» ✔

Do not accept electron acceptor without mention of electron pair.

ALTERNATIVE 1

Property: variable oxidation state ✔

Comparison: Mn compounds can exist in different valencies/oxidation states AND Mg has a valency/oxidation state of +2 in all its compounds ✔

Accept valency.
Accept for second statement “Mg «always» has the same oxidation state”.

ALTERNATIVE 2

Property: coloured ions/compounds/complexes ✔

Comparison: Mn ions/compounds/complexes coloured AND Mg ions/compounds white/«as solids»/colourless «in aqueous solution» ✔

Accept Mn forms coloured ions/compounds/complexes and Mg does not.

ALTERNATIVE 3

Property: catalytic activity ✔

Comparison: «many» Mn compounds act as catalysts AND Mg compounds do not «generally» catalyse reactions ✔


For any property accept a correct specific example, for example manganate(VII) is purple.
Do not accept differences in atomic structure, such as partially filled d sub-levels, but award ECF for a correct discussion.

Well done; However, instead of identifying the conjugate acid-base relationship, some simply identified these as Brønsted–Lowry base and acid.

Good performance. Some teachers suggested the question had an error in units, but this was not the case. The question had to be solved, first by using the data provided for application of gas law to determine the number of moles of gas. Next, given volume of solution, V = 2.00 dm³, determine its concentration.

Concentration of [OH^˗] was asked for but some calculated [H₃O⁺] instead. On the whole, question was done well.

Mediocre performance. Since a mark was given for the K_b expression, that mark could also be scored for the Henderson Hasselbalch (HH) equation, provided it is specific to the equilibrium reaction. Unfortunately, there was poor understanding of the application of the equation in most cases. Students should be strongly encouraged to use the HH equation only when a buffer is involved. Appropriate K_a or K_b expressions should be used when buffer solutions are not involved.

Mediocre performance. One mark was scored for suggesting equilibrium shifts to right or H⁺ reacts with NH₃. However, some made reference to ammonia being a strong base or no reference to the strong acid, H⁺ being converted to a weak acid, NH₄⁺.

Mediocre performance; although some Mg²⁺ was identified as a Lewis acid, the reasoning given was that it accepts an electron, rather than an electron pair or references were made to Bronsted-Lowry theory.

Write the chemical equation for the complete combustion of ethanol.

Deduce the change in enthalpy, ΔH, in kJ, when 56.00 g of ethanol is burned. Use section 13 in the data booklet.

Oxidation of ethanol with potassium dichromate, K₂Cr₂O₇, can form two different organic products. Determine the names of the organic products and the methods used to isolate them.

Write the equation and name the organic product when ethanol reacts with methanoic acid.

Sketch the titration curve of methanoic acid with sodium hydroxide, showing how you would determine methanoic acid pK_a.

Identify an indicator that could be used for the titration in 5(d)(i), using section 22 of the data booklet.

Determine the concentration of methanoic acid in a solution of pH = 4.12. Use section 21 of the data booklet.

Identify if aqueous solutions of the following salts are acidic, basic, or neutral.

CH₃CH₂OH (l) + 3O₂(g) → 2CO₂(g) + 3H₂O (g) ✓

«n = $\frac{56.00 \mathrm{g}}{46.08 \mathrm{g} {\mathrm{mol}}^{-1}}$ =» 1.215 «mol» ✓

«1.215mol × (−1367 kJ mol⁻¹) =» −1661 «kJ» ✓

Award [2] for correct final answer.

Award [1 max] for “«+»1661 «kJ»”.

ethanal AND distillation ✓

ethanoic acid AND reflux «followed by distillation» ✓

Award [1] for both products OR both methods.

Equation:
CH₃CH₂OH + HCOOH $\rightleftharpoons$ HCOOCH₂CH₃ + H₂O ✓

Product name:
ethyl methanoate ✓

Accept equation without equilibrium arrows.

Accept equation with molecular formulas (C₂H₆O + CH₂O₂ $\rightleftharpoons$ C₃H₆O₂ + H₂O) only if product name is correct.

increasing S-shape pH curve ✓

pK_a: pH at half neutralization/equivalence ✓

M1: Titration curve must show buffer region at pH <7 and equivalence at pH >7.

Ignore other parts of the curve, i.e., before buffer region, etc.

Accept curve starting from where two axes meet as pH scale is not specified.

phenolphthalein
OR
phenol red ✓

Alternative 1:
K_a = $\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{HCOO}}^{-}\right]}{\left[\mathrm{HCOOH}\right]}$
OR
[HCOOH] = $\frac{{\left({10}^{-4.12}\right)}^2}{{10}^{-3.75}}$✓

«[HCOOH] =» 3.24 × 10⁻⁵ «mol dm⁻³» ✓

Alternative 2:
«pH = pK_a + log $\frac{\left[{\mathrm{HCOO}}^{-}\right]}{\left[\mathrm{HCOOH}\right]}$»
4.12 = 3.75 + log$\frac{{10}^{-4.12}}{\left[\mathrm{HCOOH}\right]}$ ✓

«[HCOOH] =» 3.24 × 10⁻⁵ «mol dm⁻³» ✓

Award [2] for correct final answer.

Sodium methanoate: basic

Ammonium chloride: acidic

Sodium nitrate: neutral ✓ ✓

Award [2] for three correct.

Award [1] for two correct.

State the equilibrium constant expression, K_c , for this reaction.

The following equilibrium concentrations in mol dm^–3 were obtained at 761 K.

Calculate the value of the equilibrium constant at 761 K.

Determine the value of ΔG^θ, in kJ, for the above reaction at 761 K using section 1 of the data booklet.

Calculate [H₃O⁺] in the solution and the dissociation constant, K_a , of the acid at 25 °C.

Calculate K_b for HCO₃^– acting as a base.

K_c = $\frac{{\text{[HI]}}^{\text{2}}}{\text{[}{\text{H}}_{\text{2}}\text{][}{\text{I}}_{\text{2}}\text{]}}$

45.6

ΔG^θ = «– RT ln K = – (0.00831 kJ K⁻¹ mol⁻¹ x 761 K x ln 45.6) =» – 24.2 «kJ»

[H₃O⁺] = 6.76 x 10^–5 «mol dm^–3»

K_a = $\frac{{\left(6.76\times {10}^{-5}\right)}^2}{\left(0.010-6.76\times {10}^{-5}\right)}/\frac{{\left(6.76\times {10}^{-5}\right)}^2}{0.010}$

4.6 x 10^–7

Accept 4.57 x 10^–7

Award [3] for correct final answer.

«$\frac{1.00\times {10}^{-14}}{4.6\times {10}^{-7}}$ =» 2.17 x 10^–8

OR

«$\frac{1.00\times {10}^{-14}}{4.57\times {10}^{-7}}$ =» 2.19 x 10^–8

Predict, giving a reason, a difference between the reactions of the same concentrations of hydrochloric acid and ethanoic acid with samples of calcium carbonate.

Dissolved carbon dioxide causes unpolluted rain to have a pH of approximately 5, but other dissolved gases can result in a much lower pH. State one environmental effect of acid rain.

Write an equation to show ammonia, NH₃, acting as a Brønsted–Lowry base and a different equation to show it acting as a Lewis base.

Determine the pH of 0.010 mol dm⁻³ 2,2-dimethylpropanoic acid solution.

K_a (2,2-dimethylpropanoic acid) = 9.333 × 10⁻⁶

Explain, using appropriate equations, how a suitably concentrated solution formed by the partial neutralization of 2,2-dimethylpropanoic acid with sodium hydroxide acts as a buffer solution.

slower rate with ethanoic acid

OR

smaller temperature rise with ethanoic acid

[H⁺] lower

OR

ethanoic acid is weak

OR

ethanoic acid is partially dissociated

Accept experimental observations such as “slower bubbling” or “feels less warm”.

[2 marks]

Any one of:

corrosion of materials/metals/carbonate materials

destruction of plant/aquatic life

«indirect» effect on human health

Accept “lowering pH of oceans/lakes/waterways”.

[1 mark]

Brønsted–Lowry base:

NH₃ + H⁺ → NH₄⁺

Lewis base:

NH₃ + BF₃ → H₃NBF₃

Accept “AlCl₃ as an example of Lewis acid”.

Accept other valid equations such as Cu²⁺ + 4NH₃ → [Cu(NH₃)₄]²⁺.

[2 marks]

[H⁺] «$=\sqrt{{\text{K}}_{\text{a}}\times \left[{\text{C}}_5{\text{H}}_{10}{\text{O}}_2\right]}=\sqrt{9.333\times {10}^{-6}\times 0.010}$» = 3.055 × 10^–4 «mol dm^–3»

«pH =» 3.51

Accept “pH = 3.52”.

Award [2] for correct final answer.

Accept other calculation methods.

[2 marks]

(CH₃)₃CCOOH(aq) + OH^–(aq) → (CH₃)₃CCOO^–(aq) + H₂O(l)

OR

(CH₃)₃CCOOH(aq) + OH^–(aq) $\rightleftharpoons$ (CH₃)₃CCOO^–(aq) + H₂O(l) AND addition of alkali causes equilibrium to move to right

(CH₃)₃CCOO^–(aq) + H⁺(aq) → (CH₃)₃CCOOH(aq)

OR

(CH₃)₃CCOO^–(aq) + H⁺(aq) $\rightleftharpoons$ (CH₃)₃CCOOH(aq) AND addition of acid causes equilibrium to move to right

Accept “HA” for the acid.

Award [1 max] for correct explanations of buffering with addition of acid AND base without equilibrium equations.

[2 marks]

State a reason why most halogenoalkanes are more reactive than alkanes.

Classify 1-bromopropane as a primary, secondary or tertiary halogenoalkane, giving a reason.

Explain the mechanism of the reaction between 1-bromopropane with aqueous sodium hydroxide using curly arrows to represent the movement of electron pairs.

State, giving your reason, whether the hydroxide ion acts as a Lewis acid, a Lewis base, or neither in the nucleophilic substitution.

Suggest two advantages of understanding organic reaction mechanisms.

polarity/polar «molecule/bond»
OR
carbon–halogen bond is weaker than C–H bond ✔

primary AND Br/bromine is attached to a carbon bonded to two hydrogens
OR
primary AND Br/bromine is attached to a carbon bonded to one C/R/alkyl «group» ✔

Accept “primary AND Br/bromine is attached to the first carbon in the chain”.

curly arrow going from lone pair/negative charge on O in HO^– to C ✔

curly arrow showing Br leaving ✔

representation of transition state showing negative charge, square brackets and partial bonds ✔

formation of organic product CH₃CH₂CH₂OH AND Br^– ✔

Do not allow curly arrow originating on H in HO^–.

Accept curly arrow either going from bond between C and Br to Br in 1-bromopropane or in the transition state.

Do not penalize if HO and Br are not at 180° to each other.

Do not award M3 if OH–C bond is represented.

«Lewis» base AND donates a pair of electrons ✔

Any two of:
choose «most» appropriate reaction «for preparing the target compound» ✔
design/discover new reactions/reagents ✔
apply this knowledge to other areas of chemistry/science ✔
«retro-»synthesis «more effective» ✔
control/predict «desired» products ✔
control rate of reaction «more effectively» ✔
satisfy intellectual curiosity ✔
predicting how changing reagents/conditions might affect reaction ✔
suggesting intermediates/transition states ✔

Accept other reasonable answers.

Formulate an equation for the reaction of one mole of phosphoric acid with one mole of sodium hydroxide.

Formulate two equations to show the amphiprotic nature of H₂PO₄⁻.

Calculate the concentration of H₃PO₄ if 25.00 cm³ is completely neutralised by the addition of 28.40 cm³ of 0.5000 mol dm⁻³ NaOH.

Outline the reasons that sodium hydroxide is considered a Brønsted–Lowry and Lewis base.

H₃PO₄(aq) + NaOH (aq) → NaH₂PO₄(aq) + H₂O (l) ✔

Accept net ionic equation.

H₂PO₄⁻(aq) + H⁺(aq) → H₃PO₄(aq) ✔

H₂PO₄⁻(aq) + OH⁻(aq) → HPO₄²⁻(aq) + H₂O (l) ✔

Accept reactions of H₂PO₄⁻ with any acidic, basic or amphiprotic species, such as H₃O⁺, NH₃ or H₂O.

Accept H₂PO₄⁻ (aq) → HPO₄²⁻ (aq) + H⁺ (aq) for M2.

«$\mathrm{NaOH} \frac{28.40 {\mathrm{cm}}^3}{1000}\times 0.5000 \mathrm{mol} {\mathrm{dm}}^{-3}=0.01420 \mathrm{mol}$»

«$\frac{0.01420 \mathrm{mol}}{3}=$» 0.004733 «mol» ✔

«$\frac{0.004733 \mathrm{mol}}{{\displaystyle \frac{25.00 {\mathrm{cm}}^3}{1000}}}=$» 0.1893 «mol dm⁻³» ✔

Award [2] for correct final answer.

Brønsted–Lowry base:
proton acceptor

AND

Lewis Base:
e^– pair donor/nucleophile ✔

Distinguish between a sigma and pi bond.

Identify the hybridization of carbon in ethane, ethene and ethyne.

State, giving a reason, if but-1-ene exhibits cis-trans isomerism.

State the type of reaction which occurs between but-1-ene and hydrogen iodide at room temperature.

Explain the mechanism of the reaction between but-1-ene with hydrogen iodide, using curly arrows to represent the movement of electron pairs.

State, giving a reason, if the product of this reaction exhibits stereoisomerism.

Deduce the rate expression for this reaction.

Deduce the units of the rate constant.

Determine the initial rate of reaction in experiment 4.

Deduce, with a reason, the mechanism of the reaction between 2-chloropentane and sodium hydroxide.

Discuss the reason benzene is more reactive with an electrophile than a nucleophile.

Sigma (σ) bond:

overlap «of atomic orbitals» along the axial / intermolecular axis / electron density is between nuclei
OR
head-on/end-to-end overlap «of atomic orbitals» ✔

Pi (π) bond:

overlap «of p-orbitals» above and below the internuclear axis/electron density above and below internuclear axis
OR
sideways overlap «of p-orbitals» ✔

Accept a suitable diagram.

All 3 required for mark.

no AND 2 groups on a carbon «in the double bond» are the same/hydrogen «atoms»

OR

no AND molecule produced by rearranging atoms bonded on a carbon «in the double bond» is the same as the original ✔

«electrophilic» addition ✔

Do not allow nucleophilic addition.

curly arrow going from C=C to H of HI AND curly arrow showing I leaving ✔

representation of carbocation ✔

curly arrow going from lone pair/negative charge on I^– to C⁺ ✔

2-iodobutane formed ✔

Penalize incorrect bond, e.g. –CH–H₃C or –CH₃C once only.

yes AND has a carbon attached to four different groups
OR
yes AND it contains a chiral carbon ✔

Accept yes AND mirror image of molecule different to original/non-superimposable on original.

«rate =» k[NaOH][C₅H₁₁Cl] ✔

mol^–1dm³s^–1 ✔

ALTERNATIVE 1:

«k = » 1.25 «mol^–1dm³s^–1» ✔

«rate = 1.25 mol^–1dm³s^–1 × 0.60 mol dm^–3 × 0.25 mol dm^–3»

1.9 x 10^–1 «mol dm^–3s^–1» ✔

ALTERNATIVE 2:

«[NaOH] exp. 4 is 3 × exp. 1»

«[C₅H₁₁Cl] exp. 4 is 2.5 × exp. 1»

«exp. 4 will be » 7.5× faster ✔

1.9 x 10^–1 «mol dm^–3s^–1» ✔

Award [2] for correct final answer.

S_N2 AND rate depends on both OH^– and 2-chloropentane ✔

Accept E2 AND rate depends on both OH^– and 2-chloropentane.

delocalized electrons/pi bonds «around the ring»
OR
molecule has a region of high electron density/negative charge ✔

electrophiles are attracted/positively charged AND nucleophiles repelled/negatively charged ✔

Do not accept just “nucleophiles less attracted” for M2.

Accept “benzene AND nucleophiles are both electron rich” for “repels nucleophiles”.