Edexcel A-Level Chemistry Paper 1, June 2018: Question 7

11 marks · Medium difficulty · Synoptic Questions

Write an equation for the reaction of borax with hydrochloric acid, draw a dot-and-cross diagram and deduce bond angles and intermolecular forces for boric acid, calculate its pH and state assumptions, and identify a conjugate acid.

Practise this question

Question

Question 7 covers boric acid, H3BO3. Part (a) asks for an equation for preparing boric acid from borax (Na2B4O7.10H2O) and hydrochloric acid. Part (b)(i) provides an overlapping circle template to complete a dot-and-cross diagram of B(OH)3 showing outer shell electrons with dots, crosses, and triangles. Part (b)(ii) is a multiple-choice question on the O-B-O and B-O-H bond angles. Part (c) is a multiple-choice question on the strongest intermolecular interactions in solid boric acid. Part (d)(i) asks to calculate the pH of 0.0500 mol dm^-3 boric acid with pKa = 9.24. Part (d)(ii) asks to state the assumptions made in the calculation. Part (e) is a multiple-choice question identifying the conjugate acid of the HBO3^2- ion.
Question text

7 Boric acid, H3BO3, is a weak acid with antiseptic properties.

(a) Boric acid can be prepared by reacting borax, Na2B4O7.10H2O, with

hydrochloric acid.

Write the equation for this reaction. State symbols are not required.

(1)

(b) The formula of boric acid can also be written as B(OH)3.

(i) Complete the dot-and-cross diagram of a molecule of boric acid.

Show the outer shell electrons only.

Use dots (•) for the hydrogen electrons, crosses (×) for the oxygen electrons

and triangles ( ) for the boron electrons.

(2)

H H

O O

B

O

H

(ii) What are the O B O and B O H bond angles in a molecule of boric acid?

(1)

O B O bond angle B O H bond angle

A 109.5° 104.5°

B 109.5° 180°

C 120° 104.5°

D 120° 180°

(c) Boric acid is a solid with melting temperature 171*P52302RA01328*°C.

What are the strongest interactions between the molecules in solid boric acid?

(1)

A covalent bonds

B hydrogen bonds

C ionic bonds

D London forces

(d) In aqueous solution, boric acid dissociates into ions in three stages.

The equation for the first dissociation is

H BO (aq) H+(aq) + H BO−(aq)

33 2 3

pKa for this dissociation is 9.24

(i) Calculate the pH of a 0.0500 mol dm−3 solution of boric acid from the pK value

a

for the first dissociation.

(3)

(ii) State any assumptions you made in your calculation in (d)(i).

(2)

(e) Boric acid can undergo further dissociation.

Which is the conjugate acid of the HBO2− ion?

(1)

A BO3−

B H BO−

C H3BO3

D H O+

(Total for Question 7 = 11 marks)

Mark scheme

Show the mark scheme Mark scheme for Question 7: 7(a) requires Na2B4O7.10H2O + 2HCl -> 4H3BO3 + 2NaCl + 5H2O (or B(OH)3). 7(b)(i) awards 1 mark for all 6 bonding pairs and 1 mark for 2 lone pairs on each O with no extra electrons on B or H. 7(b)(ii) correct answer is C (120 degrees and 104.5 degrees). 7(c) correct answer is B (hydrogen bonds). 7(d)(i) awards marks for calculating Ka = 5.7544 x 10^-10, [H+] = 5.364 x 10^-6 mol dm^-3, and pH = 5.27. 7(d)(ii) awards 2 marks for stating [H+] = [H2BO3^-] and that dissociation is negligible / [H3BO3]initial = [H3BO3]equilibrium. 7(e) correct answer is B (H2BO3^-).

Question

Acceptable Answer Additional Guidance Mark

Number

7(a) Examples of equation (1)

correct equation Na2B4O7.10H2O + 2HCl → 4H3BO3 + 2NaCl + 5H2O

or

Na2B4O7.10H2O + 2HCl → 4B(OH)3 + 2NaCl + 5H2O

Allow multiples

Allow reversible arrow provided the equation is

written in the direction shown.

Ignore state symbols, even if incorrect

Question

Acceptable Answer Additional Guidance Mark

Number

7(b)(i) Example of diagram (2)

all 6 bonding pairs correct (1)

2 lone pairs on each O and no

additional electrons on boron or

hydrogen (1)

Non-bonding electrons on O can be shown as pairs, all 4

together or as 3 and 1

Electrons in overlap regions can be on the lines or the gaps

between the lines

Allow (1) for electrons in correct places but incorrect

symbols for electrons

Ignore inner shell electrons shown on B and/or O

Note

If any double bonds are shown the answer scores (0)

Question Answer

Mark

Number

7(b)(ii) The only correct answer is C (1)

A is incorrect because 109.5o is incorrect

B is incorrect because 109.5o and 180o are incorrect

D is incorrect because 180o is incorrect

Mark

Number

7(c) The only correct answer is B (1)

A is incorrect because covalent bonds are within molecules not between molecules

C is incorrect because there are no ionic bonds

D is incorrect because London forces are not the strongest force

Question

Acceptable Answer Additional Guidance Mark

Number

7(d)(i) Example of calculation (3)

calculation of K (1) K = 10−pKa = 10−9.24 = 5.7544 x 10−10 (mol dm−3)

a a

calculation of [H+] (1) [H+] = √K [H BO ] = √5.7544 x 10−10 x 0.05

a 3 3

= 5.364 x 10−6 (mol dm−3)

TE on Ka

calculation of pH (1) pH = −log [H+] = −log 5.364 x 10−6

10 10

= 5.2705 / 5.271 / 5.27 / 5.3

TE on [H+] provided pH is >2 and <7

Accept alternative methods, for example

[H+] = √K [H BO ]

a 3 3

pH = ½pKa – ½log[H3BO3] (1)

= ½9.24 – ½log0.05 (1)

= 5.2705 / 5.271 / 5.27 / 5.3 (1)

Alternative method:

K = 10−pKa = 10−9.24 = 5.7544 x 10−10 (mol dm−3) (1)

a

[H+]2 = K ([H BO ] − [H+])

a 3 3

= 5.7544 x 10−10 x (0.05 − [H+])

[H+] = 5.135 x 10−6 (1)

pH = 5.29 (1)

Ignore SF except 1SF

Correct answer without working scores 3 marks

Question

Acceptable Answer Additional Guidance Mark

Number

7(d)(ii) An answer that makes reference to the following Allow [A−] for [H BO −] / [HA] for [H BO ] (2)

23 3 3

points: Allow any of the expressions described in words

Allow approximately equal to for = (in symbols

or words)

Ignore reference to standard conditions

Do not award two marks from the same marking

point

[H+] = [H BO −]

or Allow the effect of the third ionisation is

no H+ from the (ionisation of) water / ionisation negligible

of water is negligible

or

H+ is only from the acid

or

no H+ from ionisation of H2BO3− (1)

ionisation / dissociation of the acid is negligible / Ignore partial dissociation / not completely

very small / insignificant dissociated

or

[H3BO3]initial = [H3BO3]equilibrium Do not award H3BO3 / [HA]is completely

or dissociated

[H BO ] = 0.05 (mol dm−3)

33 equilibrium

or

[H+]/[H BO ] << [H BO ]

23 3 3

−

or

[H3BO3] / acid concentration remains constant

or

[H BO ] = [H BO ] − [H+] used in

33 equilibrium 3 3 initial

calculation in (i) (1)

Mark

Number

7(e) The only correct answer is B (1)

A is not correct because it is the conjugate base not acid

C is not correct because it is not the conjugate acid

D is not correct because it is not the conjugate acid

(Total for Question 7 = 11 marks)

How to answer it

Boric Acid & Weak Acid Equilibria

📋 WHAT THIS QUESTION TESTS

This question covers core Inorganic and Physical Chemistry concepts: balancing stoichiometric equations with hydrated salts, constructing specific-symbol Lewis dot-and-cross diagrams (Group 3 electron deficiency), deducing molecular geometries via VSEPR, identifying intermolecular forces, calculating weak acid pH from pKa, justifying equilibrium assumptions, and identifying Brønsted–Lowry conjugate acid–base pairs.

Part (a) • 1 Mark

Preparation of Boric Acid from Borax

Writing a balanced stoichiometric equation

✅ Correct Answer

Na₂B₄O₇·10H₂O + 2HCl → 4H₃BO₃ + 2NaCl + 5H₂O

Alternative: Writing boric acid as 4B(OH)₃ is fully accepted.

🧠 Exam Technique

  • State symbols were explicitly not required, so skip them to save time and avoid introducing errors.
  • Balance boron first: 4 boron atoms in borax means 4H₃BO₃ must form.
  • Balance sodium next: 2 sodium atoms in borax require 2NaCl , fixing the acid at 2HCl .
  • Finally count O and H to determine water: 17 O on the left minus 12 O in 4H₃BO₃ leaves 5 O, giving 5H₂O .
Mark scheme guidance: 1 mark for the correctly balanced equation. Reversible arrow accepted. Allow integer multiples.
Part (b)(i) • 2 Marks

Dot-and-Cross Diagram of B(OH)₃

Representing covalent bonding and lone pairs with specified symbols

✅ Diagram Layout & Requirements

  • Symbol Key: Hydrogen = dot (•), Oxygen = cross (×), Boron = triangle (Δ).
  • B–O Bonds (3 bonds): In each B–O overlap, place 1 triangle (Δ) from B and 1 cross (×) from O.
  • O–H Bonds (3 bonds): In each O–H overlap, place 1 cross (×) from O and 1 dot (•) from H.
  • Oxygen Lone Pairs: 2 lone pairs (4 crosses, ×× ××) in the non-bonding area of each of the 3 oxygen atoms.
  • Boron & Hydrogen: No extra electrons outside the overlap regions.

💡 Key Knowledge: Incomplete Octet

Boron is in Group 3, having only 3 outer-shell electrons. When forming B(OH)₃, it shares all 3 electrons to form 3 single covalent bonds, leaving it with 6 electrons in its valence shell. It does not obey the standard octet rule here.

❌ Common Errors

  • Forgetting O lone pairs: Oxygen must have an octet (2 bonding pairs + 2 lone pairs = 8 electrons).
  • Wrong electron symbols: Using generic dots/crosses instead of the assigned triangles (Δ) for boron loses a mark.
  • Drawing double bonds: The mark scheme explicitly states that showing any double bonds scores 0/2!
Mark allocation: [1] All 6 bonding pairs correct with assigned symbols. [1] 2 lone pairs on each O atom and no excess electrons on B or H.
Part (b)(ii) • 1 Mark

Bond Angles in B(OH)₃

Deducing molecular geometry from electron pair repulsion (VSEPR)

✅ Correct Answer: C

O–B–O angle: 120°

B–O–H angle: 104.5°

💡 VSEPR Analysis

  • Around Central Boron: 3 bond pairs, 0 lone pairs. Regions of electron density repel equally into a trigonal planar shape → ideal angle = 120°.
  • Around Oxygen: 2 bond pairs, 2 lone pairs (total 4 pairs, tetrahedral arrangement). Lone pair–lone pair repulsion is greater than bond pair repulsion, reducing the angle from 109.5° by ~2.5° per lone pair → bent / non-linear shape → 104.5° (identical to H₂O).

❌ Why Other Options Are Incorrect

  • A (109.5°, 104.5°): Incorrectly assumes boron has a tetrahedral arrangement (it has no lone pair).
  • B & D (180° for B–O–H): Mistakenly assumes oxygen is linear. Two lone pairs force a bent geometry.
Part (c) • 1 Mark

Intermolecular Forces in Solid Boric Acid

Distinguishing intermolecular forces from intramolecular bonds

✅ Correct Answer: B (Hydrogen bonds)

Boric acid contains polar O–H bonds where hydrogen is bonded to a highly electronegative oxygen atom. The lone pairs on oxygen act as hydrogen bond acceptors, forming an extensive hydrogen-bonded network in the solid state (hence the relatively high melting temperature of 171 °C).

❌ Distractor Breakdown

  • A (covalent bonds): Covalent bonds are intramolecular (within molecules), not interactions between molecules.
  • C (ionic bonds): Boric acid is a simple covalent molecular compound; there are no ions in pure solid boric acid.
  • D (London forces): London dispersion forces are present, but they are not the strongest intermolecular interactions.
Part (d)(i) • 3 Marks

Weak Acid pH Calculation

Step-by-step determination of pH from pKa

📐 Step-by-Step Calculation

Given: pKa = 9.24, [H₃BO₃] = 0.0500 mol dm⁻³

Step 1: Calculate Ka
Ka = 10−pKa = 10−9.24 = 5.7544 × 10⁻¹⁰ mol dm⁻³ [1 Mark]

Step 2: Calculate [H⁺]
Using the weak acid expression: Ka = [H⁺]² / [H₃BO₃]
[H⁺]² = Ka × [H₃BO₃] = (5.7544 × 10⁻¹⁰) × 0.0500 = 2.8772 × 10⁻¹¹
[H⁺] = √(2.8772 × 10⁻¹¹) = 5.364 × 10⁻⁶ mol dm⁻³ [1 Mark]

Step 3: Calculate pH
pH = −log₁₀[H⁺] = −log₁₀(5.364 × 10⁻⁶) = 5.27 (or 5.271) [1 Mark]

🧠 Shortcut Formula: Half-pKa Method

For a monoprotic weak acid:
pH = ½pKa − ½log₁₀[HA]

pH = ½(9.24) − ½log₁₀(0.0500) = 4.62 − ½(−1.301) = 4.62 + 0.6505 = 5.27

❌ Calculation Traps

  • Forgetting the square root: Finding [H⁺]² and calculating −log₁₀([H⁺]²) directly gives pH ~ 10.54, which makes no sense for an acid!
  • Significant figures: The mark scheme accepts 2 to 4 sig figs (5.27 to 5.2705), but never round to 1 SF (e.g. 5) as 1 SF is specifically penalised.
Transfer of Error (TE): TE applies from Ka to [H⁺], and from [H⁺] to pH (provided 2 < pH < 7). Correct final answer without working scores all 3 marks.
Part (d)(ii) • 2 Marks

Assumptions Made in the Weak Acid Calculation

Explaining approximations in aqueous acid equilibria

✅ Acceptable Points (Any TWO needed for 2 marks)

  • Assumption 1: [H⁺] = [H₂BO₃⁻]
    Meaning: All H⁺ ions originate solely from the dissociation of the acid (the ionisation of water is negligible; negligible H⁺ from subsequent dissociations).
  • Assumption 2: [H₃BO₃]equilibrium ≈ [H₃BO₃]initial (or [H⁺] << [H₃BO₃] )
    Meaning: The dissociation of the acid is so small/negligible that the concentration of undissociated acid remains constant at 0.0500 mol dm⁻³.

❌ Common Misconceptions

  • Standard conditions: Stating "reaction takes place at 298 K / standard conditions" scores 0 marks. That is an experimental condition, not an equilibrium calculation assumption.
  • Saying "acid is completely dissociated": Completely wrong for a weak acid—it barely dissociates at all!
  • Vague phrasing: Simply writing "it's a weak acid" without referencing concentrations or species will not earn credit.
Mark guidance: 1 mark per distinct assumption. Do not award two marks from the same marking point.
Part (e) • 1 Mark

Conjugate Acid-Base Pairs

Identifying conjugate species in stepwise dissociation

✅ Correct Answer: B (H₂BO₃⁻)

By Brønsted–Lowry definition, a conjugate acid is formed when a base gains a proton (H⁺):

HBO₃²⁻ + H⁺ ⇌ H₂BO₃⁻

Therefore, the conjugate acid of HBO₃²⁻ is H₂BO₃⁻.

🧠 Golden Rule for Conjugate Pairs

A conjugate acid has one MORE H⁺ (increase H count by 1, increase charge by +1).

A conjugate base has one LESS H⁺ (decrease H count by 1, decrease charge by −1).

  • Conjugate acid of HBO₃²⁻ → add H⁺ → H₂BO₃⁻ (Option B)
  • Conjugate base of HBO₃²⁻ → remove H⁺ → BO₃³⁻ (Option A)

Topics

Physical Chemistry · Topic 2: Bonding and Structure · Topic 5: Formulae, Equations and Amounts of Substance · Topic 12: Acid-base Equilibria

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.