Edexcel A-Level Chemistry Paper 1, June 2019: Question 6

Properties of Water and Ionic Product Calculations

📋 What this question tests

This question assesses core physical chemistry concepts including intermolecular forces (London forces vs hydrogen bonding), molecular shapes and polarity using dipole moment vectors, dot-and-cross bonding diagrams for complex ions, calculations involving the ionic product of water ( K_w ), pH determination to specified decimal places, and the application of Le Chatelier's principle to temperature-dependent equilibria.

Part (a): Boiling Temperatures of H₂O and H₂S

Comment on boiling temperature differences using intermolecular forces (4 marks)

✅ Correct Answer Breakdown

(M1): Water is expected to have a lower boiling temperature because it has fewer electrons than hydrogen sulfide (10 vs 18 electrons per molecule).
(M2): Therefore, water has weaker/less extensive London forces.
(M3): However, water actually has a higher boiling temperature because it exhibits hydrogen bonding.
(M4): Hydrogen bonding is significantly stronger than London forces and requires more energy to overcome/break.

❌ Common Errors & Examiner Pitfalls

Vague terminology: Referring to "molecular forces" or "intermolecular bonds" without specifying London forces or hydrogen bonds loses marks.
Incorrect comparisons: Mentioning "smaller relative molecular mass (Mr)" instead of counting electrons or electron shells directly can be penalised if imprecise.

Mark allocation: 1 mark for each valid point up to a maximum of 4 marks.

Part (b): Bond Polarity vs. Molecular Polarity

Explain why H₂O and CO₂ have polar bonds, but only water is a polar molecule (4 marks)

💡 Key Knowledge

(M1): Oxygen is significantly more electronegative than both hydrogen and carbon, pulling shared electron density towards itself.
(M2): This creates individual polar bonds with oxygen carrying a partial negative charge ( δ⁻ ) and carbon/hydrogen carrying a partial positive charge ( δ⁺ ).
(M3): Carbon dioxide ( CO₂ ) is a symmetrical, linear molecule, meaning individual dipole moments/vectors act in opposite directions and cancel out completely.
(M4): Water ( H₂O ) is a non-linear (V-shaped/bent) molecule due to its two lone pairs on oxygen. Its bond dipoles do not cancel, resulting in a net dipole moment.

🧠 Exam Technique

Always explicitly mention symmetry and whether dipole moments cancel or do not cancel when explaining molecular polarity. Simply stating "water is bent" is insufficient without linking it to vector addition of dipoles.

Mark allocation: 4 marks total (1 mark per valid point addressing electronegativity, bond dipoles, CO₂ symmetry/cancellation, and H₂O V-shape/non-cancellation).

Part (c): Dot-and-Cross Diagrams for Ions

Draw dot-and-cross diagrams for the oxonium ion (H₃O⁺) and hydroxide ion (OH⁻) (2 marks)

✅ Correct Answer Description

Oxonium ion ( H₃O⁺ ): Central oxygen surrounded by 3 single covalent bonds to hydrogen atoms. Oxygen must show 1 lone pair (2 electrons) remaining. Enclose the entire structure in square brackets with a ⁺ charge outside.
Hydroxide ion ( OH⁻ ): Oxygen single-bonded to one hydrogen atom. Oxygen must display 3 lone pairs (6 electrons). Enclose in square brackets with a ⁻ charge outside.

❌ Common Errors

Omitting square brackets or overall ionic charges.
Forgetting to include the lone pair on the oxygen in the oxonium ion, or drawing too many/few lone pairs on the hydroxide ion.
Mixing up dot (•) and cross (×) designations if explicitly requested in the prompt.

Mark allocation: 1 mark for oxonium ion diagram; 1 mark for hydroxide ion diagram.

Part (d)(i): Calculating pH of Water at Elevated Temperature

Calculate the pH of water at 310 K given K_w = 2.40 × 10⁻¹⁴ mol² dm⁻⁶ (2 marks)

📐 Step-by-Step Calculation

Recognise the relationship: In pure water, [H⁺(aq)] = [OH⁻(aq)] , therefore K_w = [H⁺]² .
Rearrange to find hydrogen ion concentration:
[H⁺] = √(K_w) = √(2.40 × 10⁻¹⁴) = 1.549 × 10⁻⁷ mol dm⁻³
Calculate pH:
pH = -log₁₀[H⁺] = -log₁₀(1.549 × 10⁻⁷) = 6.80989...
Apply significant figures / decimal places:
Round to two decimal places as requested: pH = 6.81

❌ Calculation Traps

Failing to take the square root of K_w before applying the negative logarithm.
Rounding to an incorrect number of decimal places (remember: decimal places in a pH value correspond to significant figures in the concentration).

Mark allocation: 1 mark for taking the square root to find [H⁺] ; 1 mark for the correct pH value to 2 decimal places.

Part (d)(ii): Acid-Base Nature of Pure Water at 310 K

Predict, with a reason, whether water is acidic, alkaline or neutral at 310 K (2 marks)

✅ Correct Answer

(M1): Neutral.
(M2): Because [H⁺(aq)] = [OH⁻(aq)] (equal amounts of hydrogen and hydroxide ions are present).

❌ Common Misconceptions

Many students automatically assume a pH below 7 (such as 6.81) means a solution is acidic. However, water remains neutral because the concentrations of H⁺ and OH⁻ remain perfectly balanced, even though K_w changes with temperature.

Mark allocation: 1 mark for stating neutral; 1 mark for the correct justification referencing equal concentrations of H⁺ and OH⁻ .

Part (d)(iii): Enthalpy Change of Ionisation

Predict, with a reason, the sign of the enthalpy change for the ionisation of water (1 mark)

💡 Key Knowledge & Answer

Sign: Positive (or endothermic).
Reason: As temperature increases, the value of K_w increases, indicating that the forward ionisation reaction is endothermic and absorbs heat to shift the equilibrium to the right in accordance with Le Chatelier's principle.

Mark allocation: 1 mark for predicting a positive sign with a valid supporting reason (e.g. K_w increases with temperature, or bond breaking requires energy).