Brønsted–Lowry Acids & Bases ( Proton transfer reactions)

Brønsted–Lowry Acids & Bases ( Proton transfer reactions) 






📚 Table of Contents
1. 🧪 Brønsted–Lowry Acids & Bases
Proton transfer reactions, acid/base definitions
2. 🔗 Conjugate Acids & Bases
Conjugate pairs, identifying them in reactions
3. 💧 Amphiprotic Species
Species that can both donate and accept H⁺
4. 📏 The pH Scale
What pH measures, the 0–14 scale
5. ⚖️ Ion Product of Water (Kw)
Water dissociation, Kw at 25°C
6. ⚗️ Strong vs Weak Acids & Bases
Dissociation, comparing strength
7. 🧮 Calculating pH
Step-by-step pH calculations
8. 🔁 Neutralisation Reactions
Acid + base reactions, types of salts
9. 🧂 Making Salts
Salt formation reference table
10. 📈 pH Curves
S-shaped curves, equivalence point
11. 🎯 Choosing the Right Indicator
Methyl orange vs phenolphthalein
12. ✅ Topic Summary
Key takeaways
What Are Brønsted–Lowry Acids & Bases?
🧪 Acid
A proton donor · Gives away H⁺
🧫 Base
A proton acceptor · Receives H⁺
⚗️ Example
HCl + H₂O → H₃O⁺ + Cl⁻ · HCl donates H⁺ to water
✅ Summary: Acid = H⁺ donor · Base = H⁺ acceptor · It's always about proton transfer
Brønsted–Lowry Theory — In Detail
🧪 Acid
Gives away a proton (H⁺) · Proton donor
🧫 Base
Accepts a proton (H⁺) · Uses lone pair of electrons
Acid donates H⁺ → Base accepts via lone pair

HCl(g) + NH₃(g) → NH₄Cl(s) · Gas phase reaction

Key Points

Theory applies to gas phase AND aqueous solutions
HCl(g) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq)
H₂O = base (proton acceptor) · HCl = acid (proton donor)
Ion formation = dissociation / ionisation
📝 Worked Example
Identify the correct role of each species in:
H₂PO₄⁻(aq) + H₂O(l) → HPO₄²⁻(aq) + H₃O⁺(aq)
Answer: ✅ A — H₂PO₄⁻ donates H⁺ to H₂O · H₂PO₄⁻ = acid · H₂O = base
⚠️ Examiner Tips & Tricks
H⁺ = bare proton · 1 proton, 0 electrons, 0 neutrons
H⁺ ion is also called a proton for this reason
Theory works in gas phase — not just aqueous
✅ Summary: Acid = H⁺ donor · Base = H⁺ acceptor using lone pair · Works in gas AND aqueous phase · H⁺ = bare proton · Always identify which species donates/accepts
Conjugate Acids & Bases
Every acid–base reaction produces a conjugate pair — differing by just one H⁺.
Conjugate Base
Acid loses H⁺
e.g. HCl → Cl⁻
Conjugate Acid
Base gains H⁺
e.g. NH₃ → NH₄⁺
The Rule
Conjugate pair differs by exactly ONE H⁺
HCl + H₂O ⇌ H₃O⁺ + Cl⁻
Pair 1: HCl / Cl⁻ · Pair 2: H₂O / H₃O⁺
✅ Summary: Acid → loses H⁺ → conjugate base · Base → gains H⁺ → conjugate acid · Pairs differ by exactly 1 H⁺
Conjugate Acids & Bases — In Detail
🧪 Brønsted-Lowry Acid
Donates a proton (H⁺)
Proton donor
🧫 Brønsted-Lowry Base
Accepts a proton (H⁺)
Proton acceptor
In equilibrium: products form at the same rate as reactants are used
                              CH3COOH (aq) + H2O (l)         ⇌    CH3COO- (aq)          +         H3O+ (aq)
         acid                            base                    conjugate base             conjugate acid

What is a Conjugate Pair?
Two species differing by exactly ONE H⁺
Conjugate = related by one proton
How Pairs Form
Acid loses H⁺ → conjugate base
Base gains H⁺ → conjugate acid
📝 Worked Example
CH₃CH₂CH₂COOH(aq) + H₂O(l) ⇌ CH₃CH₂CH₂COO⁻(aq) + H₃O⁺(aq)
Which species are a conjugate acid-base pair?
A — CH₃CH₂CH₂COOH and H₂O
B — H₂O and H₃O⁺ ✅
C — H₂O and CH₃CH₂CH₂COO⁻
D — CH₃CH₂CH₂COO⁻ and H₃O⁺
Answer: B — H₂O and H₃O⁺ differ by exactly one H⁺ ion
✅ Summary: Conjugate pair = two species differing by ONE H⁺ · Acid → loses H⁺ → conjugate base · Base → gains H⁺ → conjugate acid · CH₃COOH/CH₃COO⁻ and H₂O/H₃O⁺ are both conjugate pairs in the same reaction
Identifying Conjugate Pairs
How to Identify Pairs
01
Find the acid
Which species donates H⁺?
02
Find conjugate base
Remove one H⁺ from the acid.
03
Find conjugate acid
Add one H⁺ to the base.
✅ Summary: Acid loses H⁺ → conjugate base · Base gains H⁺ → conjugate acid · Always identify BOTH pairs in a reaction
Amphiprotic Species
Species that can BOTH donate and accept H⁺
As an Acid
Donates H⁺ to another species
As a Base
Accepts H⁺ from another species
✅ Amphiprotic = acts as BOTH acid AND base depending on the reaction partner
Amphiprotic Species — Examples
💧 Water (H₂O)
As acid:
H₂O + NH₃ ⇌ OH⁻ + NH₄⁺
donates H⁺ to NH₃
As base:
HCl + H₂O ⇌ H₃O⁺ + Cl⁻
accepts H⁺ from HCl
🧪 Hydrogencarbonate (HCO₃⁻)
As acid:
HCO₃⁻ ⇌ CO₃²⁻ + H⁺
donates H⁺
As base:
HCO₃⁻ + H⁺ ⇌ H₂CO₃
accepts H⁺
⚗️ Hydrogensulfate (HSO₄⁻)
As acid:
HSO₄⁻ ⇌ SO₄²⁻ + H⁺
donates H⁺
As base:
HSO₄⁻ + H⁺ ⇌ H₂SO₄
accepts H⁺
✅ Summary: H₂O, HCO₃⁻ and HSO₄⁻ are all amphiprotic · Can donate OR accept H⁺ · Role depends on reaction partner · Water is the most important example
Amphiprotic Species — In Detail
🔵 Amphiprotic
Can both donate AND accept H⁺
Acts as acid OR base
🟢 Amphoteric
Can act as both acid AND base
Does NOT have to involve H⁺ transfer
💧 Water as a Brønsted-Lowry Acid
H₂O + NH₃ ⇌ OH⁻ + NH₄⁺
H₂O donates H⁺ to NH₃ → acts as acid
H₂O donates H⁺ to NH₃ · acts as Brønsted-Lowry acid
Lewis diagram: H₂O + NH₃ ⇌ OH⁻ + NH₄⁺
💧 Water as a Brønsted-Lowry Base
HCl + H₂O ⇌ H₃O⁺ + Cl⁻
H₂O accepts H⁺ from HCl → acts as base
H₂O accepts H⁺ from HCl · acts as Brønsted-Lowry base
Lewis diagram: HCl + H₂O ⇌ H₃O⁺ + Cl⁻
Amphiprotic
Donates AND accepts H⁺
All amphiprotic = also amphoteric
e.g. H₂O, HCO₃⁻, HSO₄⁻
Amphoteric
Acts as acid AND base (not necessarily via H⁺)
NOT all amphoteric = amphiprotic
e.g. Al₂O₃ — reacts with HCl AND NaOH
Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
Al₂O₃ + 2NaOH + 3H₂O → 2NaAl(OH)₄
✅ Summary: Amphiprotic = H⁺ donor AND acceptor · Amphoteric = acid AND base character · All amphiprotic substances are amphoteric · NOT all amphoteric are amphiprotic · Al₂O₃ = amphoteric but NOT amphiprotic
The pH Scale
pH < 7
Acidic · More H⁺ ions
pH = 7
Neutral · Pure water at 25°C
pH > 7
Alkaline · More OH⁻ ions
• pH = −log₁₀[H⁺]
• [H⁺] = 10⁻ᵖᴴ
✅ Summary: pH = −log[H⁺] · Each unit = ×10 change in [H⁺] · pH 3 is 10× more acidic than pH 4 · Neutral = pH 7 at 25°C
🧪 The pH Scale
What is pH?
📌 pH measures acidity
Based on H⁺ (H₃O⁺) ion concentration · Scale runs 0–14 · Below 0 possible for very strong acids
📐 Logarithmic scale (base 10)
Each pH unit = ×10 change in [H⁺] · e.g. pH 5 is 10× more acidic than pH 6
The Key Equations
pH = −log₁₀[H⁺]
Use this to find pH from [H⁺] concentration
[H⁺] = 10⁻ᵖᴴ
Use this to find [H⁺] from pH · Units: mol dm⁻³
pH & [H⁺] Table
The pH Scale
The pH scale — acidity, neutrality and alkalinity
pH of Acids, Bases & Water
 Acids (pH < 7)
More H⁺ than OH⁻ · [H⁺] > 10⁻⁷ mol dm⁻³ · Higher [H⁺] = lower pH
 Bases (pH > 7)
More OH⁻ than H⁺ · [H⁺] < 10⁻⁷ mol dm⁻³ · Higher [OH⁻] = higher pH
Water (pH = 7)
Equal H⁺ and OH⁻ · [H⁺] = [OH⁻] = 10⁻⁷ mol dm⁻³ · pH = −log(10⁻⁷) = 7
Worked Example
🧮 Worked Example
10.0 cm³ of nitric acid (pH = 1.0) is mixed with 990.0 cm³ of distilled water. What is the pH of the final solution?
Options: A. 1 · B. 2 · C. 3 · D. 10
✅ Answer: C — pH 3
Total volume = 1000.0 cm³
[H⁺] reduced by factor of 100 (= 10⁻²)
pH increases by 2 units
Final pH = 1 + 2 = pH 3
How to Measure pH
🔬 pH Meter (Most Accurate)
Connect pH electrode to meter · Read pH value directly · Most reliable method
🌈 Universal Indicator Paper (Less Accurate)
Dip paper into solution · Paper changes colour · Compare to colour chart · Gives approximate pH
Digital pH meter — measures pH via a pH electrode
Universal indicator paper — colour compared to chart · pH 1–2 = strong acid · pH 3–4 = weak acid
Examiner Tips & Tricks
⚠️ Examiner Tips & Tricks
Use the antilog (10ˣ) button on your calculator to find [H⁺] from pH
On most calculators: 10ˣ button · Others: LOG⁻¹, ALOG, or INV + LOG
pH values are usually given to 2 decimal places
pH scale: 0–14 (values below 0 exist for very concentrated acids)
Lower pH = more acidic · Higher pH = more alkaline
Summary
✅ Key Takeaways — The pH Scale
pH = −log₁₀[H⁺] · [H⁺] = 10⁻ᵖᴴ
Logarithmic scale — each unit = ×10 change in [H⁺]
pH < 7 = acidic · pH = 7 = neutral · pH > 7 = alkaline
Water at 298K: [H⁺] = [OH⁻] = 10⁻⁷ mol dm⁻³ → pH = 7
Most accurate measurement: pH meter · Less accurate: universal indicator
pH values given to 2 decimal places
The Ion Product of Water (Kw)
The Equilibrium
H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)
Kw = [H⁺][OH⁻]
At 25°C: Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶
Neutral: [H⁺] = [OH⁻] = 1×10⁻⁷ → pH = 7
🌡️ Temperature
Kw increases with temp · Dissociation is endothermic
💧 Neutral Water
[H⁺] = [OH⁻] · At 25°C: pH = 7
⚠️ Exam Trap
At higher temps, neutral pH < 7 · Water is still neutral!
✅ Summary: Kw = [H⁺][OH⁻] = 1×10⁻¹⁴ at 25°C · Use Kw to find [OH⁻] from [H⁺] · Higher temp → higher Kw → lower neutral pH
💧 Ion Product of Water (Kw)
⚖️ Why Does Water Dissociate?
⚖️ Water Dissociation
Water is NOT just a liquid — it partially ionises
A tiny fraction of molecules split into ions
H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)
Equilibrium lies FAR to the LEFT — very few ions form
🔍 Why It Matters
Even pure water contains H⁺ and OH⁻ ions
This is why water can conduct a tiny amount of electricity
The concentration of these ions determines if a solution is acidic, neutral or alkaline
📐 Where Does Kw Come From?
01
Write the Equilibrium Expression
For any equilibrium: Kc = [products] / [reactants]
So: Kc = [H⁺][OH⁻] / [H₂O]
This is the standard equilibrium constant expression
02
Why Remove [H₂O]?
[H₂O] is HUGE compared to [H⁺] and [OH⁻]
It barely changes — treated as a constant
So we absorb it into a new constant: Kw = Kc × [H₂O]
03
The Result — Kw
Kw = [H⁺][OH⁻]
This is called the ion product of water
It has a fixed value at a given temperature
04
The Value at 298K
Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶
[H⁺] × [OH⁻] ALWAYS = 10⁻¹⁴ at 298K
If one goes up, the other MUST go down
Kw = [H⁺][OH⁻]
The ion product of water — always true at any temperature
Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K
[H⁺] × [OH⁻] is always this value at 298K (25°C)
✅ Remember: If [H⁺] = 10⁻³, then [OH⁻] = 10⁻¹⁴ ÷ 10⁻³ = 10⁻¹¹ · They always multiply to give 10⁻¹⁴
[H⁺] & [OH⁻] — Always Multiply to 10⁻¹⁴
Notice how [H⁺] × [OH⁻] always = 1.00 × 10⁻¹⁴. When [H⁺] is large → acidic. When [OH⁻] is large → alkaline. When equal → neutral.
✅ Key pattern:
[H⁺] × [OH⁻] = 1.00 × 10⁻¹⁴ ALWAYS at 298K
Neutral: [H⁺] = [OH⁻] = 1 × 10⁻⁷ mol dm⁻³
Acidic: [H⁺] > 10⁻⁷ · Alkaline: [H⁺] < 10⁻⁷
🧮 Worked Example — Finding pH of a Base
🧮 Question: What is the pH of KOH(aq) at concentration 1.0 × 10⁻³ mol dm⁻³? (Kw = 1.0 × 10⁻¹⁴ at 298K)
Options: A. 3 · B. 4 · C. 10 · D. 11
✅ Answer: D — pH 11
Step 1 — Identify what you know:
KOH is a strong base → fully dissociates
KOH → K⁺ + OH⁻ (1:1 ratio)
So [OH⁻] = [KOH] = 1.0 × 10⁻³ mol dm⁻³
We know Kw = 1.0 × 10⁻¹⁴
Step 2 — Rearrange Kw to find [H⁺]:
Start with: Kw = [H⁺][OH⁻]
Rearrange: [H⁺] = Kw ÷ [OH⁻]
Why? Because we know Kw and [OH⁻], but need [H⁺]
Step 3 — Substitute and calculate:
[H⁺] = (1.0 × 10⁻¹⁴) ÷ (1.0 × 10⁻³)
[H⁺] = 1.0 × 10⁻¹¹ mol dm⁻³
(Subtract the powers: −14 − (−3) = −11)
Step 4 — Calculate pH:
pH = −log₁₀[H⁺]
pH = −log₁₀(1.0 × 10⁻¹¹)
pH = −(−11) = 11
High pH confirms it's alkaline ✓
[H⁺] = Kw ÷ [OH⁻]
Rearrangement of Kw = [H⁺][OH⁻]
pH = −log₁₀(1.0 × 10⁻¹¹) = 11
Use the log button on your calculator
🌡️ How Temperature Affects Kw
01
The Reaction is ENDOTHERMIC
2H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)
Energy is ABSORBED to break bonds in water
Heat is a "reactant" in this process
02
Le Chatelier's Principle
Adding heat = adding a reactant
Equilibrium shifts to the RIGHT to oppose the change
More H₃O⁺ and OH⁻ produced → Kw INCREASES
03
Effect on pH
More H⁺ ions in solution
pH = −log₁₀[H⁺] → as [H⁺] increases, pH DECREASES
Pure water pH drops below 7 at higher temperatures
↑ Temperature → ↑ Kw → ↑ [H⁺] → ↓ pH
Each arrow follows logically — learn the chain!

As temperature increases, Kw increases so pH decreases
💡 But wait — is hot water acidic?
NO! Hot water is still NEUTRAL
Neutral means [H⁺] = [OH⁻] — this is ALWAYS true for pure water
Both [H⁺] AND [OH⁻] increase equally with temperature
So the solution stays neutral — just with a lower pH number
"pH 7 = neutral" is ONLY true at 298K
✅ Summary — Temperature Effect:
Kw increases as temperature increases
At 298K: Kw = 10⁻¹⁴ → neutral pH = 7
At higher temps: Kw > 10⁻¹⁴ → neutral pH < 7
Pure water is ALWAYS neutral regardless of temperature
Neutral ≠ pH 7 (except at 298K)
⚠️ Examiner Tips & Tricks
⚠️ Examiner Tips & Tricks
Hot water pH < 7 but it is still NEUTRAL — [H⁺] = [OH⁻]
"pH 7 = neutral" ONLY at 298K (25°C)
Always use Kw = 1.00 × 10⁻¹⁴ unless the question gives a different value
To find [H⁺] from [OH⁻]: divide Kw by [OH⁻]
To find [OH⁻] from [H⁺]: divide Kw by [H⁺]
Subtract powers of 10 when dividing: 10⁻¹⁴ ÷ 10⁻³ = 10⁻¹¹
✅ Summary
✅ Key Takeaways — Ion Product of Water
Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 298K
Kw is temperature dependent — increases with temperature
Neutral: [H⁺] = [OH⁻] (not necessarily pH 7)
At 298K: neutral pH = 7 · Above 298K: neutral pH < 7
To find [H⁺]: [H⁺] = Kw ÷ [OH⁻]
To find [OH⁻]: [OH⁻] = Kw ÷ [H⁺]
Higher temperature → higher Kw → lower pH of pure water
Strong vs Weak Acids
Strength = degree of dissociation in water — NOT concentration
Strong Acid
Fully dissociates
[H⁺] = [acid]
e.g. HCl, H₂SO₄, HNO₃
Weak Acid
Partially dissociates
Equilibrium exists
e.g. CH₃COOH, HF
✅ Summary: Strong = complete dissociation · Weak = partial · Strength ≠ concentration · Ka measures acid strength
⚗️ Strong & Weak Acids and Bases
💡 What does "strong" mean?
NOTHING to do with concentration
Strong = how completely it splits apart in water
Strong acid → splits 100%
Weak acid → only splits a little bit
⚠️ Don't confuse these!
Strong/Weak = degree of splitting (dissociation)
Concentrated/Dilute = how much acid is dissolved
A weak acid CAN be concentrated!
🔴 Strong Acids — Split Completely
✅ Fully Dissociate
Every molecule splits into H⁺ and A⁻
No molecules left over
Arrow goes ONE WAY only (→)
📊 High [H⁺]
Lots of H⁺ ions in solution
Very low pH
Highly acidic
🧪 Examples
HCl — hydrochloric acid
HNO₃ — nitric acid
H₂SO₄ — sulfuric acid (1st ionisation)
HCl → H⁺ + Cl⁻
One-way arrow (→) = complete dissociation · ALL HCl splits up
Strong acid — complete dissociation in aqueous solution
✅ Strong Acid Summary:
Fully dissociates → one-way reaction (→)
[H⁺] = [acid] (they're equal!)
Very low pH · Very acidic
Examples: HCl, HNO₃, H₂SO₄
🟡 Weak Acids — Split Partially
⚖️ Partially Dissociate
Only SOME molecules split
Most stay as HA
Equilibrium arrow (⇌) — goes both ways
📊 Lower [H⁺]
Fewer H⁺ ions in solution
Higher pH than strong acid
Less acidic
🧪 Examples
CH₃COOH — ethanoic acid
HCN — hydrocyanic acid
H₂CO₃ — carbonic acid
H₂S — hydrogen sulfide
CH₃COOH ⇌ H⁺ + CH₃COO⁻
Double arrow (⇌) = partial dissociation · Most CH₃COOH stays as molecules
Weak acid — partial dissociation, equilibrium established
✅ Weak Acid Summary:
Partially dissociates → equilibrium (⇌)
[H⁺] < [acid] (less H⁺ than acid molecules)
Higher pH than strong acid at same concentration
Use Ka to calculate [H⁺] for weak acids
⚡ Strong vs Weak — Side by Side
📉 Acid Strength Down Group 17
01
Bond length increases
As you go down Group 17 (F→Cl→Br→I)
The halogen atom gets BIGGER
The H–X bond gets LONGER
02
Longer bond = weaker bond
A longer bond is easier to break
Easier to break = easier to release H⁺
More H⁺ released = stronger acid
03
Acid strength order
HF < HCl < HBr < HI
HF is weakest (short, strong bond)
HI is strongest (long, weak bond)
HF < HCl < HBr < HI
Acid strength INCREASES down Group 17 — bond gets longer and weaker
🔵 Strong Bases — Split Completely
✅ Fully Dissociate
Every molecule splits into metal ion + OH⁻
One-way reaction (→)
High [OH⁻] concentration · Very alkaline
🧪 Examples
NaOH — sodium hydroxide
KOH — potassium hydroxide
Group 1 metal hydroxides
NaOH → Na⁺ + OH⁻
Complete dissociation — all NaOH splits up
Strong base — complete dissociation in aqueous solution
🟢 Weak Bases — Split Partially
⚖️ Partially Dissociate
Only some molecules accept H⁺
Equilibrium established (⇌)
Lower [OH⁻] · Less alkaline
🧪 Examples
NH₃ — ammonia
Amines (organic compounds)
Some transition metal hydroxides
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
Partial reaction — most NH₃ stays as molecules
Weak base — partial dissociation, equilibrium established
✅ Bases Summary:
Strong base: fully dissociates → high [OH⁻] · e.g. NaOH, KOH
Weak base: partially dissociates → low [OH⁻] · e.g. NH₃, amines
Strong base → weak conjugate acid
Weak base → strong conjugate acid
🔄 Conjugate Pairs & Strength
🔴 Strong acid → Weak conjugate base
HCl fully dissociates → Cl⁻ is very weak base
Cl⁻ has almost NO tendency to accept H⁺ back
Strong acid = reverse reaction barely happens
🔵 Strong base → Weak conjugate acid
OH⁻ readily accepts H⁺
OH⁻ + H⁺ ⇌ H₂O
H₂O is a very weak conjugate acid
Strong acid ↔ Weak conjugate base
Weak acid ↔ Strong conjugate base
The stronger the acid, the weaker its conjugate base — and vice versa!
🔬 Distinguishing Strong vs Weak Acids
Method 1: pH
Same concentration, different pH
Strong acid: pH 1
Weak acid: pH ~2.9
Lower pH = stronger acid
Use pH meter or universal indicator
Method 2: Electrical Conductivity
More H⁺ ions = conducts electricity better
Strong acid: HIGH conductivity
Weak acid: LOWER conductivity
Measure with a conductivity meter
Method 3: Reactivity with Metals
Strong acid + Mg → LOTS of bubbles (H₂ gas)
Weak acid + Mg → fewer bubbles
More H⁺ = faster reaction = more gas
Digital conductivity meter — stronger acid = higher reading
Strong acid (HCl) + Mg → lots of H₂ bubbles — high [H⁺]
Weak acid (CH₃COOH) + Mg → fewer bubbles — lower [H⁺]
✅ How to distinguish strong vs weak (same concentration):
Strong acid: lower pH · higher conductivity · more vigorous reaction
Weak acid: higher pH · lower conductivity · slower reaction
All due to MORE H⁺ ions in strong acids
💡 Special Cases to Know
💡 Dibasic Acids (H₂SO₄)
H₂SO₄ has TWO ionisations:
1st: H₂SO₄ → H⁺ + HSO₄⁻ (STRONG — complete)
2nd: HSO₄⁻ ⇌ H⁺ + SO₄²⁻ (WEAK — partial)
So 1 mol dm⁻³ H₂SO₄ does NOT give 2 mol dm⁻³ H⁺
💡 Writing H⁺ vs H₃O⁺
Both are correct! H⁺ is shorthand for H₃O⁺
If you write H₃O⁺, include H₂O in the equation:
HCl → H⁺ + Cl⁻ OR HCl + H₂O → H₃O⁺ + Cl⁻
⚠️ Examiner Tips & Tricks
⚠️ Examiner Tips & Tricks
Strong/Weak = degree of dissociation — NOT concentration
Concentrated/Dilute = amount dissolved — NOT strength
A weak acid CAN be concentrated · A strong acid CAN be dilute
Strong acid → [H⁺] = [acid] (use directly for pH)
Weak acid → [H⁺] ≠ [acid] (must use Ka)
HF is a WEAK acid despite being a hydrogen halide (very short, strong bond)
📋 Full Summary
✅ Key Takeaways — Strong & Weak Acids/Bases
Strong acid: fully dissociates (→) · high [H⁺] · low pH · e.g. HCl, HNO₃, H₂SO₄
Weak acid: partially dissociates (⇌) · low [H⁺] · higher pH · e.g. CH₃COOH
Strong base: fully dissociates (→) · high [OH⁻] · e.g. NaOH, KOH
Weak base: partially dissociates (⇌) · low [OH⁻] · e.g. NH₃
Acid strength increases down Group 17: HF < HCl < HBr < HI
Strong acid → weak conjugate base · Weak acid → strong conjugate base
Distinguish by: pH · conductivity · reactivity with metals
Strong & Weak Acids Compared
Strong Acids
Fully dissociate in water
[H⁺] = [acid]
e.g. HCl, H₂SO₄, HNO₃
Low pH
High conductivity
Weak Acids
Partially dissociate
Equilibrium exists
Use Ka to find [H⁺]
e.g. CH₃COOH, HF, H₂CO₃
Higher pH than strong at same concentration
Ka = [H⁺][A⁻] / [HA] — larger Ka = stronger weak acid
✅ Summary: Strong = full dissociation · Weak = partial · Ka = acid dissociation constant · Larger Ka → stronger weak acid
🧮 Calculating pH — Step by Step
The Two Formulas
Formula 1: pH = −log₁₀[H⁺]
Use when you KNOW [H⁺] and want pH
−log = take the power of 10 and flip the sign
e.g. [H⁺] = 0.01 → pH = −log(0.01) = 2
Formula 2: [H⁺] = 10⁻ᵖᴴ
Use when you KNOW pH and want [H⁺]
Put pH as a negative power of 10
e.g. pH = 3 → [H⁺] = 10⁻³ mol dm⁻³
pH = −log₁₀[H⁺]
Use when you KNOW [H⁺] and want pH
[H⁺] = 10⁻ᵖᴴ
Use when you KNOW pH and want [H⁺]
🔴 Strong Acid — Finding pH
Why easy? Strong acid = fully dissociates · Every molecule breaks apart → one H⁺ · So [H⁺] = [acid] · No Ka needed — no extra steps!
[H⁺] = [acid concentration]
For strong acids ONLY — they fully dissociate in water
✏️ Worked Example 1 — Find pH of 0.1 mol dm⁻³ HCl
01
Step 1: Identify the acid
HCl is a STRONG acid
Fully dissociates: HCl → H⁺ + Cl⁻
So [H⁺] = 0.1 mol dm⁻³
02
Step 2: Write as a power of 10
0.1 = 10⁻¹
So [H⁺] = 10⁻¹ mol dm⁻³
This makes the log step easier to see!
03
Step 3: Apply the formula
pH = −log₁₀(10⁻¹)
pH = −(−1)
pH = 1 ✓
pH = −log₁₀(0.1) = −(−1) = 1
Negative of a negative = positive! That's why pH is always a positive number here.
✅ Strong Acid Pattern:
· 0.1 mol dm⁻³ → pH = 1
· 0.01 mol dm⁻³ → pH = 2
· 0.001 mol dm⁻³ → pH = 3
· Divide concentration by 10 → pH goes UP by 1
🟡 Weak Acid — Finding pH
Why harder? Weak acid = only PARTIALLY dissociates · [H⁺] ≠ [acid] · You need Ka to find [H⁺] · More steps, but still very doable!
Ka = [H⁺][A⁻] / [HA]
Ka = acid dissociation constant · Bigger Ka = stronger weak acid · Smaller Ka = weaker acid
KEY ASSUMPTION (write this in every exam!)
HA ⇌ H⁺ + A⁻
· Assumption 1: [H⁺] = [A⁻] always (1:1 ratio from equation) → so Ka = [H⁺]² / [HA]
· Assumption 2: [HA]equilibrium ≈ [HA]initial (so little dissociates that concentration barely changes)
[H⁺] = √(Ka × [HA])
Square root of (Ka × concentration) · Rearranged from Ka = [H⁺]² / [HA]
✏️ Worked Example 2 — Find pH of 0.1 mol dm⁻³ CH₃COOH, Ka = 1.8 × 10⁻⁵
01
Step 1: Write what you know
[HA] = 0.1 mol dm⁻³
Ka = 1.8 × 10⁻⁵ mol dm⁻³
CH₃COOH ⇌ H⁺ + CH₃COO⁻
02
Step 2: Substitute into formula
[H⁺] = √(Ka × [HA])
[H⁺] = √(1.8 × 10⁻⁵ × 0.1)
[H⁺] = √(1.8 × 10⁻⁶)
03
Step 3: Calculate the square root
[H⁺] = √(1.8 × 10⁻⁶)
[H⁺] = 1.342 × 10⁻³ mol dm⁻³
Calculator: press √ then type the number
04
Step 4: Calculate pH
pH = −log₁₀(1.342 × 10⁻³)
pH = 2.87 ✓
Always give pH to 2 decimal places!
[H⁺] = √(1.8 × 10⁻⁵ × 0.1) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³
Multiply Ka × [HA] first, THEN press √ on your calculator
pH = −log₁₀(1.34 × 10⁻³) = 2.87
Type into calculator: log(answer) then press the minus key in front
✅ Weak Acid Steps (in order):
1. Write Ka = [H⁺]² / [HA]
2. Rearrange: [H⁺] = √(Ka × [HA])
3. Calculate [H⁺]
4. pH = −log₁₀[H⁺]
5. State your assumptions in the exam!
🔁 pH → [H⁺] — Working Backwards
✏️ Worked Example 3 — Find [H⁺] when pH = 3.5
01
Step 1: Write the formula
[H⁺] = 10⁻ᵖᴴ
pH = 3.5
02
Step 2: Substitute pH in
[H⁺] = 10⁻³·⁵
Put the pH value as the negative power of 10
03
Step 3: Use your calculator
Press the 10ˣ button
Type −3.5
[H⁺] = 3.16 × 10⁻⁴ mol dm⁻³ ✓
[H⁺] = 10⁻³·⁵ = 3.16 × 10⁻⁴ mol dm⁻³
Calculator: press 10ˣ then type −3.5 (or press INV → LOG → type −3.5)
📋 Quick Reference Table
⚠️ Examiner Tips
⚠️ Don't lose marks — remember these:
· pH always to 2 decimal places
· State assumptions for weak acids in the exam
· Assumption 1: [H⁺] = [A⁻] (from 1:1 dissociation ratio)
· Assumption 2: [HA]equilibrium ≈ [HA]initial (negligible dissociation)
· H₂SO₄ is dibasic — be careful with concentration (multiply by 2!)
· Calculator: use 10ˣ or INV + LOG for antilog
✅ Full Summary
✅ Key Takeaways:
· Strong acid: [H⁺] = [acid] → pH = −log₁₀[H⁺]
· Weak acid: [H⁺] = √(Ka × [HA]) → pH = −log₁₀[H⁺]
· Reverse: [H⁺] = 10⁻ᵖᴴ
· pH to 2 decimal places always
· Weak acid pH > strong acid pH at the same concentration
· Always state assumptions for weak acid calculations in the exam
Neutralisation Reactions
Acid + Base → Salt + Water
The Reaction
Acid + Base → Salt + Water
Ionic Equation
H⁺(aq) + OH⁻(aq) → H₂O(l)
Salt Formed
Depends on acid and base used
✅ Summary: Neutralisation = acid + base → salt + water · Net ionic equation: H⁺ + OH⁻ → H₂O · Salt pH depends on acid/base strengths
Types of Neutralisation
1
Strong Acid + Strong Base
e.g. HCl + NaOH → NaCl + H₂O
Salt pH = 7 (neutral)
2
Strong Acid + Weak Base
e.g. HCl + NH₃ → NH₄Cl
Salt pH < 7 (acidic)
3
Weak Acid + Strong Base
e.g. CH₃COOH + NaOH → CH₃COONa + H₂O
Salt pH > 7 (alkaline)
4
Weak Acid + Weak Base
e.g. CH₃COOH + NH₃
pH depends on relative strengths
✅ Summary: Strong+Strong → pH 7 · Strong acid+Weak base → pH < 7 · Weak acid+Strong base → pH > 7 · Weak+Weak → unpredictable
⚗️ Neutralisation Reactions
🧪 What is it?
Acid (pH <7) + Base (pH >7) → Salt + Water (pH = 7)
Spectator ions form the salt
⚡ Ionic Equation
H⁺(aq) + OH⁻(aq) → H₂O(l)
Only H⁺ and OH⁻ react — spectator ions don't
acid + base (alkali) → salt + water
Step 1
HCl + NaOH →
Step 2
H₂O + NaCl
Ionic Steps
H⁺ + OH⁻ → H₂O
Na⁺ + Cl⁻ → NaCl
✅ Key Summary
Neutralisation = acid + base → salt + water
H⁺ reacts with OH⁻ to form water
Spectator ions form the salt
Salt name comes from the acid used: HCl → chloride, HNO₃ → nitrate, H₂SO₄ → sulfate
Enthalpy of neutralisation ≈ −57 kJ mol⁻¹ for strong acid + strong base
💡 Examiner Tip
The enthalpy of neutralisation is always ~−57 kJ mol⁻¹ for strong acid + strong base reactions — it's the same because the net ionic equation is always H⁺ + OH⁻ → H₂O
🔩 Acid + Metal + Metal Oxide
Section 1 — Metals + Acids
⚗️ General Rule
acid + metal → salt + hydrogen
⚡ Rate depends on:
Reactivity of the metal
Strength of the acid
Highly reactive metals = dangerous!
Unreactive metals (e.g. Cu) = no reaction
2HCl (aq) + Zn (s) → ZnCl₂ (aq) + H₂ (g)
hydrochloric acid + zinc → zinc chloride + hydrogen
💨 More Effervescence
⚡ Faster Dissolution
🔥 Greater Exothermic Effect
Section 2 — Metals + Oxides
⚗️ General Rule
acid + metal oxide → salt + water
📝 Example
2HCl (aq) + CaO (s) → CaCl₂ (aq) + H₂O (l)
hydrochloric acid + calcium oxide → calcium chloride + water
✅ Key Summary
Metals + acids → salt + hydrogen gas (H₂)
Metal oxides + acids → salt + water only
Stronger acid = more vigorous reaction (more effervescence, faster, more exothermic)
Unreactive metals (Cu, Au) do NOT react with dilute acids
Highly reactive metals (Na, K) are too dangerous to use with acids
🧱 Acid + Metal Hydroxide, Carbonate & Hydrogencarbonate
Section 1 — Metals + Hydroxides
⚗️ General Rule
acid + metal hydroxide → salt + water
📝 Example
H₂SO₄ (aq) + Mg(OH)₂ (s) → MgSO₄ (aq) + 2H₂O (l)
sulfuric acid + magnesium hydroxide → magnesium sulfate + water
Section 2 — Metals + Carbonates
⚗️ General Rule
acid + metal carbonate → salt + water + carbon dioxide
📝 Example
2HNO₃ (aq) + CuCO₃ (s) → Cu(NO₃)₂ (aq) + H₂O (l) + CO₂ (g)
nitric acid + copper carbonate → copper nitrate + water + carbon dioxide
Section 3 — Metals + Hydrogencarbonates
⚗️ General Rule
acid + metal hydrogencarbonate → salt + water + carbon dioxide
📝 Example
HCl (aq) + NaHCO₃ (s) → NaCl (aq) + H₂O (l) + CO₂ (g)
hydrochloric acid + sodium hydrogencarbonate → sodium chloride + water + carbon dioxide
✅ Key Summary
Hydroxides + acid → salt + water (no gas)
Carbonates + acid → salt + water + CO₂ gas
Hydrogencarbonates + acid → salt + water + CO₂ gas (same as carbonates, slightly different stoichiometry)
CO₂ gas produced = fizzing/effervescence observed
Note: metals used to make salts are NOT classified as bases (water is not a product)
🧂 Making Salts — Reference Table & Worked Example
The salt produced depends on the acid used and the base reacted with.
🔬 Worked Example
❓ Question
Products of zinc oxide + hydrochloric acid?
🔍 Reasoning
Metal oxide + acid → salt + water ONLY (no H₂, no CO₂)
✅ Answer: D
ZnO + 2HCl → ZnCl₂ + H₂O
zinc chloride + water
✅ Key Summary
HCl → chloride salts
HNO₃ → nitrate salts
H₂SO₄ → sulfate salts
CH₃COOH → ethanoate salts
Any acid + ammonia → ammonium salts
Always use: metal oxide, hydroxide, carbonate or hydrogencarbonate as the base
Metal oxides + acid → salt + water ONLY (no gas produced)
💡 Examiner Tip
Learn the formulae of common acids and bases. You must be able to write balanced equations for all characteristic reactions — this is a common exam question!
pH Curves
Key Features to Know
Equivalence Point
Moles acid = moles base
Vertical Section
Only with strong acid OR strong base
Half-Equivalence Point
For weak acids: pH = pKa
Buffer Region
Flat section · Resists pH change
✅ Summary: Know all 4 curve shapes · Strong+Strong → vertical at pH 7 · Weak acid+Strong base → vertical above pH 7 · pH = pKa at half-equivalence
Choosing the Right Indicator
Indicator must change colour within the vertical section of the pH curve.
Methyl Orange
Range: pH 3.1–4.4
Strong acid + strong base
Strong acid + weak base
Phenolphthalein
Range: pH 8.2–10.0
Strong acid + strong base
Weak acid + strong base
Weak + Weak
❌ No suitable indicator
No sharp vertical section exists
✅ Summary: Match indicator range to vertical section · Methyl orange → acidic equivalence · Phenolphthalein → alkaline equivalence · Weak+Weak = no indicator works
📈 pH Curves
How pH changes during a titration
🧪 What is a pH Curve?
A graph showing how pH changes as acid or base is gradually added during a titration. Recorded using a pH meter.
📐 Shape
All pH curves show a characteristic S-shaped curve with a steep vertical section at the equivalence point.
HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l)
Example: Strong Acid + Strong Base
✅ Key Facts
S-shaped curve
Midpoint of vertical section = equivalence point
Start of curve = initial pH of acid
End of curve = pH of excess base
📊 pH Curve — Key Characteristics
Diagram: Characteristics of a strong acid–strong base pH curve
📍 Start of Curve
Initial pH of the acid (low for strong acid)
⬆️ Vertical Section
Sharp pH change — this is where neutralisation occurs
⚖️ Equivalence Point
Midpoint of vertical section — acid fully neutralised
📍 End of Curve
pH of excess base (high for strong base)
✅ From a pH Curve You Can Read:
Initial pH of the acid
pH at the equivalence point
Volume of base at equivalence point
pH range of the sharp vertical section
🧮 Calculating pH — Step by Step
Example: 50 cm³ of 0.10 mol dm⁻³ HCl + 0.10 mol dm⁻³ NaOH
n = c (mol dm⁻³) × v (dm³)
Moles Formula — use this at every step
01
🟢 Start: 0 cm³ NaOH added
Only HCl present
 = 0.10 mol dm⁻³

✅ pH = 1.0
02
🟡 25 cm³ NaOH added (HCl in excess)
 mol
 mol
 mol
Total volume =  dm³
 mol dm⁻³

✅ pH = 1.5
03
🟠 49 cm³ NaOH added (HCl just in excess)
 mol
 mol
 mol
Total volume =  dm³
 mol dm⁻³

✅ pH = 3.0
04
⚖️ 50 cm³ NaOH added (Equivalence Point)
Acid completely neutralised
Solution =  only
✅ pH = 7.0
05
🔵 51 cm³ NaOH added (NaOH in excess)
 mol
 mol
Total volume =  dm³
 mol dm⁻³


✅ pH = 11.0
✅ Summary — pH Calculation Rules
Before equivalence: use  from excess acid → 
At equivalence:  (strong acid + strong base)
After equivalence: use  from excess base →  → 
Topic Summary — Key Takeaways
01
Brønsted–Lowry
Acid = H⁺ donor · Base = H⁺ acceptor
02
Conjugate Pairs
Differ by one H⁺ · Identify both pairs in any equation
03
Amphiprotic
Can donate AND accept H⁺ · e.g. H₂O, HCO₃⁻, HSO₄⁻
04
pH & Kw
pH = −log[H⁺] · Kw = 1×10⁻¹⁴ at 25°C
05
Strong vs Weak
Strong = fully dissociates · Weak uses Ka · [H⁺] = √(Ka × [HA])
06
pH Curves & Indicators
Know all 4 shapes · Match indicator to vertical section
✅ You've covered the full Brønsted–Lowry topic · Revise conjugate pairs, Ka calculations, pH curves and indicator choice for the exam