The pH scale is a logarithmic measure of how acidic a solution is. Strong acids ionise completely, while weak acids only ionise a little.
📏 The pH scale
pH = −log[H⁺] and pOH = −log[OH⁻]. At 25 °C the ionic product of water is Kw = [H⁺][OH⁻] = 10⁻¹⁴, so pH + pOH = 14.
[H⁺] = 10⁻³ M → pH = 3
0.01 M HCl (strong, 1 H⁺) → pH = 2
0.005 M Ba(OH)₂ → [OH⁻] = 0.01 M → pOH = 2 → pH = 12
Each pH unit is a factor of 10:
pH 3 is 100 times more acidic than pH 5
Dilute 100 times: pH 2 → pH 4
🖼️ See it
pH = −log₁₀[H⁺], so one pH unit is a tenfold change.
🧪 Strong and weak acids
The Brønsted–Lowry definition: an acid donates H⁺ and a base accepts H⁺. A weak acid HA ⇌ H⁺ + A⁻ has an acid dissociation constant Ka = [H⁺][A⁻] / [HA].
weak acid: [H⁺] ≈ √(Ka · c)
Ka = 1×10⁻⁵, c = 0.1 M
[H⁺] = √(10⁻⁶) = 10⁻³ → pH = 3
Conjugate pairs: HA / A⁻
HCl → Cl⁻, NH₄⁺ → NH₃
Strong is not the same as concentrated. Strength is how far an acid ionises, and concentration is how much is dissolved.
🧪 Worked examples: mixtures, dilution and weak acids
Excess acid after neutralisation. Work in moles first, then divide by the total volume.
50 cm³ of 0.10 M HCl + 25 cm³ of 0.10 M NaOH
n(H⁺) = 0.0050 mol n(OH⁻) = 0.0025 mol
excess H⁺ = 0.0025 mol in 75 cm³ = 0.0333 M
pH = −log 0.0333 = 1.48
Weak acid. Assume that [H⁺] = [A⁻] and that very little acid has ionised, so [HA] ≈ c.
Ka = 1.0 × 10⁻⁶, c = 0.010 M
[H⁺]² = Ka × c = 1.0 × 10⁻⁸ → [H⁺] = 1.0 × 10⁻⁴ → pH = 4.00
percentage ionised = 10⁻⁴ ÷ 10⁻² × 100 = 1%
Check the assumption. The approximation is fine when less than about 5% of the acid is ionised. For very dilute or relatively strong weak acids, solve the quadratic instead.
🎯 Exam focus: strong, weak, concentrated and dilute
Strong or weak describes the extent of ionisation. Concentrated or dilute describes the amount of solute per dm³. They are independent.
0.001 M HCl strong, dilute pH = 3
1.0 M CH₃COOH weak, concentrated pH ≈ 2.4 (Ka = 1.8 × 10⁻⁵)
Common mistakes. Using pH = 14 − pOH with [H⁺] directly, or forgetting that Ba(OH)₂ gives two OH⁻. Dividing by the volume of only one solution after mixing. Writing pH to the wrong number of decimal places: give as many decimals as there are significant figures in the concentration. Treating a weak acid as fully ionised.
Definitions. Quote them exactly: pH = −log₁₀[H⁺]; Ka = [H⁺][A⁻] ÷ [HA]; pKa = −log Ka; Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 25 °C.
✏️ Quick check · 1 of 11
Brønsted–Lowry acid
A Brønsted–Lowry acid is a:
Acid donates, base accepts.
✏️ Quick check · 2 of 11
Conjugate base
What is the conjugate base of NH₄⁺?
Remove one H⁺.
✏️ Quick check · 3 of 11
Neutral
At 25 °C a neutral solution has:
pH 7.
✍️ Type your answer · 4 of 11
A solution has [H⁺] = 10⁻³ M.
What is its pH?
Enter an exact answer, without rounding.
pH = −log[H⁺].
✍️ Type your answer · 5 of 11
A 0.005 M solution of Ba(OH)₂ gives [OH⁻] = 0.01 M.
What is its pH?
Enter an exact answer, without rounding.
Convert via pOH.
✍️ Type your answer · 6 of 11
A solution of pH 3 is compared with one of pH 5.
How many times greater is [H⁺] in the pH 3 solution?
Enter an exact answer, without rounding.
Two pH units apart.
✍️ Type your answer · 7 of 11
A weak acid has Ka = 1×10⁻⁵ and a concentration of 0.1 M. Use [H⁺] = √(Ka·c).
What is its pH?
Enter an exact answer, without rounding.
Take the square root, then −log.
✍️ Type your answer · 8 of 11
A solution of 0.020 M Ba(OH)₂ is fully ionised.
What is its pH, to 2 decimal places?
Round your answer as the question asks.
Each unit of Ba(OH)₂ gives two OH⁻.
✍️ Type your answer · 9 of 11
50 cm³ of 0.10 M HCl is mixed with 25 cm³ of 0.10 M NaOH.
What is the pH of the mixture, to 2 decimal places?
Round your answer as the question asks.
Work in moles before dividing by the total volume.
✍️ Type your answer · 10 of 11
A weak acid has Ka = 1.0 × 10⁻⁶ and a concentration of 0.010 M.
What is the pH of the solution?
Enter an exact answer, without rounding.
Use [H⁺] = √(Ka c).
✏️ Quick check · 11 of 11
Strong or concentrated
Which statement about 0.001 M HCl is correct?
Separate strength from concentration.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
A solution with pH 9 is:
Question 2 of 5
A strong acid:
Question 3 of 5
Diluting an acid 10 times changes its pH by:
Question 4 of 5
The larger the value of Ka, the:
Question 5 of 5
In the reaction HCl + H₂O → H₃O⁺ + Cl⁻, water acts as a:
Lesson 2 · Equilibrium
🧫 Titration & Buffers
A titration finds an unknown concentration by reacting it exactly with a standard solution, and a buffer resists changes in pH when acid or base is added.
⚖️ Titration calculations
At the equivalence point the moles of acid and base have reacted exactly in the ratio of the equation. The indicator changes colour at the end point, which is chosen to be close to it.
n = c · V (V in litres)
20.0 mL of 0.100 M HCl neutralises 25.0 mL of NaOH
n(HCl) = 0.100 · 0.0200 = 0.00200 mol
HCl : NaOH = 1 : 1
c(NaOH) = 0.00200 / 0.0250 = 0.08 M
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
10 mL of 0.1 M H₂SO₄ = 0.001 mol → 0.002 mol NaOH
V = 0.002 / 0.1 = 0.02 L = 20 mL
Equivalence pH: strong acid + strong base is 7; weak acid + strong base is above 7; strong acid + weak base is below 7.
🖼️ See it
25 mL of 0.1 M HCl. The pH jumps steeply near equivalence, which is why an indicator changes colour so sharply.
🛡️ Buffers
A buffer is a weak acid with its conjugate base (or the reverse). Added acid is used up by the base, and added base by the acid, so the pH barely changes. The Henderson–Hasselbalch equation gives its pH.
Choosing an indicator. The indicator’s range must lie on the steep part of the curve.
strong acid + strong base steep section pH 3–11 methyl orange or phenolphthalein
weak acid + strong base equivalence above 7 phenolphthalein (8.3–10)
strong acid + weak base equivalence below 7 methyl orange (3.1–4.4)
🎯 Exam focus: titration technique and buffer explanations
To explain how a buffer works, use both directions:
adding H⁺: A⁻ + H⁺ → HA the conjugate base removes the added acid
adding OH⁻: HA + OH⁻ → A⁻ + H₂O the weak acid removes the added base
so [A⁻] ÷ [HA] hardly changes, and neither does the pH
Common mistakes. Averaging all titres rather than only concordant ones (within 0.10 cm³). Mixing up the volume in cm³ and dm³. Saying that a buffer “stops” the pH from changing; it resists change, and fails when one component is used up. Using phenolphthalein for a strong acid and weak base titration. Forgetting that pH = pKa at the half-equivalence point.
Technique marks. Rinse the burette with the solution it will contain, remove the funnel before reading, swirl the flask, add the last drops one at a time, and read the bottom of the meniscus at eye level.
✏️ Quick check · 1 of 11
Equivalence
At the equivalence point of a titration:
Stoichiometric amounts.
✏️ Quick check · 2 of 11
Buffer
A buffer solution contains:
A conjugate pair.
✏️ Quick check · 3 of 11
Weak acid titration
The pH at the equivalence point when a weak acid is titrated with NaOH is:
The salt hydrolyses.
✍️ Type your answer · 4 of 11
20.0 mL of 0.100 M HCl neutralises 25.0 mL of NaOH solution (1 : 1).
What is the NaOH concentration, in M?
Enter an exact answer, without rounding.
M
Find moles of HCl first.
✍️ Type your answer · 5 of 11
10 mL of 0.1 M H₂SO₄ is titrated with 0.1 M NaOH (H₂SO₄ + 2 NaOH → …).
What volume of NaOH is needed, in mL?
Enter an exact answer, without rounding.
mL
Use the 1 : 2 ratio.
✍️ Type your answer · 6 of 11
A buffer has pKa = 4.75 and equal concentrations of acid and conjugate base.
What is its pH?
Enter an exact answer, without rounding.
Henderson–Hasselbalch.
✍️ Type your answer · 7 of 11
A buffer has pKa = 5 and [A⁻]/[HA] = 100.
What is its pH?
Enter an exact answer, without rounding.
log 100 = 2.
✍️ Type your answer · 8 of 11
25.0 cm³ of 0.100 M Na₂CO₃ reacts with HCl according to Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂.
What volume of 0.200 M HCl is needed, in cm³?
Enter an exact answer, without rounding.
cm³
Use the 1 : 2 ratio.
✍️ Type your answer · 9 of 11
0.50 g of impure CaCO₃ (M = 100) reacts exactly with 40.0 cm³ of 0.20 M HCl.
What is the percentage purity of the sample?
Enter an exact answer, without rounding.
%
Find the mass of pure CaCO₃ first.
✍️ Type your answer · 10 of 11
A buffer contains 0.20 M ethanoate and 0.10 M ethanoic acid (pKa = 4.75). Use log 2 = 0.30.
What is the pH of the buffer?
Enter an exact answer, without rounding.
Use the Henderson–Hasselbalch equation.
✏️ Quick check · 11 of 11
Indicator choice
Which indicator is most suitable for the titration of a weak acid with a strong base?
Where is the equivalence point?
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The indicator for a titration is chosen so that:
Question 2 of 5
Adding a small amount of strong acid to a buffer:
Question 3 of 5
A burette is used to:
Question 4 of 5
At the half-equivalence point of a weak acid titration:
Question 5 of 5
Blood is kept at about pH 7.4 mainly by the:
Lesson 3 · Equilibrium
⚖️ Equilibrium & Solubility
Reversible reactions settle at a balance point. The equilibrium constant tells us where, and Le Châtelier’s principle tells us how it moves.
🧮 The equilibrium constant
For aA + bB ⇌ cC + dD, Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ) using equilibrium concentrations. Large K favours the products and small K favours the reactants. Comparing the reaction quotient Q to K shows which way the system moves: Q < K shifts right, Q > K shifts left.
A + B ⇌ 2C
[A] = 0.2, [B] = 0.5, [C] = 0.4 mol/L
Kc = 0.4² / (0.2 · 0.5) = 0.16 / 0.1 = 1.6
A + B ⇌ C + D, Kc = 9, start 1 M of A and B
x² / (1 − x)² = 9 → x / (1 − x) = 3
x = 0.75 M of C and D at equilibrium
🖼️ See it
The concentrations stop changing once the forward and backward rates are equal. The reaction has not stopped: both reactions continue at the same speed.
🔄 Le Châtelier and solubility
Le Châtelier: if you disturb a system at equilibrium it shifts to partly undo the change. Adding reactant shifts right, raising the temperature favours the endothermic direction, and raising the pressure favours the side with fewer gas moles. A catalyst does not change K.
For a sparingly soluble salt, the solubility product is Ksp. For AgCl ⇌ Ag⁺ + Cl⁻, Ksp = [Ag⁺][Cl⁻].
Ksp(AgCl) = 1×10⁻¹⁰
[Ag⁺] = [Cl⁻] = s → s² = 10⁻¹⁰ → s = 10⁻⁵ M
Common-ion effect: adding NaCl raises [Cl⁻],
so less AgCl dissolves.
🧪 Worked examples: Kc, Kp and solubility products
Kc from amounts. Convert to concentrations; if the number of moles is the same on both sides, the volume cancels.
H₂ + I₂ ⇌ 2HI in a 2.0 dm³ flask
at equilibrium: 0.20 mol H₂, 0.20 mol I₂, 1.60 mol HI
Kc = [HI]² ÷ ([H₂][I₂]) = 1.60² ÷ (0.20 × 0.20) = 2.56 ÷ 0.04 = 64 (no units)
Kp from partial pressures. Only gases appear, and the partial pressure is mole fraction × total pressure.
Common-ion effect. Adding a soluble salt that shares an ion with a sparingly soluble salt shifts the equilibrium back towards the solid, so the solubility falls.
🎯 Exam focus: effects of conditions on K and the position
Only temperature changes the value of K. Concentration, pressure and catalysts change how fast equilibrium is reached or where the position lies, but not K.
N₂ + 3H₂ ⇌ 2NH₃ ΔH = −92 kJ mol⁻¹
↑ temperature shifts LEFT (endothermic direction) K decreases yield falls
↑ pressure shifts RIGHT (fewer gas moles) K unchanged yield rises
catalyst no shift, K unchanged, equilibrium reached faster
Common mistakes. Including solids or pure liquids in Kc or Kp. Writing the powers from the wrong side of the equation. Giving units for K without working them out from the expression. Forgetting that Ksp changes with temperature but not with concentration. Saying that a catalyst increases the yield.
Q versus K. If Q < K, the reaction proceeds to the right until Q = K. If Q > K, it proceeds to the left. For a precipitation test, a precipitate forms when the ionic product exceeds Ksp.
✏️ Quick check · 1 of 11
Catalyst
A catalyst added to a system at equilibrium will:
It only affects the rate.
✏️ Quick check · 2 of 11
Q greater than K
If Q is larger than K, the reaction will:
Compare Q to K.
✏️ Quick check · 3 of 11
Pressure
For N₂ + 3H₂ ⇌ 2NH₃, increasing the pressure shifts the equilibrium:
Count gas moles.
✍️ Type your answer · 4 of 11
For A + B ⇌ 2C the equilibrium concentrations are [A] = 0.2, [B] = 0.5 and [C] = 0.4 mol/L.
What is Kc?
Enter an exact answer, without rounding.
The coefficient 2 becomes a power.
✍️ Type your answer · 5 of 11
For A + B ⇌ C + D, Kc = 9 and you start with 1 M each of A and B and nothing else.
What is [C] at equilibrium, in M?
Enter an exact answer, without rounding.
M
Take the square root of both sides.
✍️ Type your answer · 6 of 11
Kc for 2SO₂ + O₂ ⇌ 2SO₃ is 4 when [SO₂] = 0.5 M and [O₂] = 1.0 M. Kc = [SO₃]²/([SO₂]²[O₂]).
What is [SO₃], in M?
Enter an exact answer, without rounding.
M
Rearrange for [SO₃]².
✍️ Type your answer · 7 of 11
For AgCl, Ksp = 1×10⁻¹⁰ and [Ag⁺] = [Cl⁻] = s, so s = 10⁻ˣ.
What is x?
Enter an exact answer, without rounding.
Square root of 10⁻¹⁰.
✍️ Type your answer · 8 of 11
In a 2.0 dm³ flask at equilibrium there are 0.20 mol H₂, 0.20 mol I₂ and 1.60 mol HI for H₂ + I₂ ⇌ 2HI.
What is the value of Kc?
Enter an exact answer, without rounding.
Both sides have the same number of moles.
✍️ Type your answer · 9 of 11
N₂O₄ ⇌ 2NO₂ has partial pressures of 20 kPa for NO₂ and 10 kPa for N₂O₄.
What is the value of Kp, in kPa?
Enter an exact answer, without rounding.
kPa
Square the pressure of NO₂.
✍️ Type your answer · 10 of 11
PbI₂ has Ksp = 4s³ = 3.2 × 10⁻⁸ mol³ dm⁻⁹.
What is the solubility s, in mol dm⁻³?
Enter an exact answer, without rounding.
Divide by 4, then take the cube root.
✏️ Quick check · 11 of 11
Effect of temperature
The forward reaction in N₂ + 3H₂ ⇌ 2NH₃ is exothermic. What happens to Kc when the temperature is raised?
Only temperature changes K.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The equilibrium constant is changed only by a change in:
Question 2 of 5
At equilibrium:
Question 3 of 5
Heating an exothermic reaction at equilibrium shifts it:
Question 4 of 5
The common-ion effect makes a sparingly soluble salt:
Question 5 of 5
A large Kc (for example 10⁶) means that at equilibrium:
Lesson 4 · Redox
🔋 Electrochemistry
Redox reactions move electrons. A galvanic cell uses a spontaneous reaction to produce electricity, and electrolysis uses electricity to force a reaction.
⚡ Galvanic cells
In a galvanic cell oxidation happens at the anode (−) and reduction at the cathode (+). Electrons flow through the wire from anode to cathode, and a salt bridge completes the circuit. The cell voltage from standard electrode potentials is E°cell = E°cathode − E°anode.
Zn²⁺/Zn E° = −0.76 V Cu²⁺/Cu E° = +0.34 V
Ag⁺/Ag E° = +0.80 V
Zn–Cu cell: 0.34 − (−0.76) = 1.10 V
Cu–Ag cell: 0.80 − 0.34 = 0.46 V
Oxidation numbers: Mn in MnO₄⁻ = +7
Cr in Cr₂O₇²⁻ = +6
Mnemonic: OIL RIG, Oxidation Is Loss of electrons and Reduction Is Gain. AN OX and RED CAT: anode oxidation and reduction at the cathode.
🖼️ See it
Electrons flow through the wire from anode to cathode; the salt bridge balances charge.
🧲 Electrolysis and Faraday
Electrolysis uses a current to drive a non-spontaneous reaction, so now the cathode is negative. The charge passed is Q = It, and moles of electrons = Q / F with F = 96500 C/mol.
2 A for 965 s → Q = 1930 C
mol e⁻ = 1930 / 96500 = 0.02 mol
Cu²⁺ + 2e⁻ → Cu
mol Cu = 0.02 / 2 = 0.01 mol
mass = 0.01 · 63.5 = 0.635 g
🧪 Worked examples: cell voltages and electrolysis calculations
E°cell and feasibility. The more positive electrode potential is the cathode (reduction). A positive E°cell means the reaction is feasible.
Zn²⁺ + 2e⁻ ⇌ Zn E° = −0.76 V
Ag⁺ + e⁻ ⇌ Ag E° = +0.80 V
E°cell = 0.80 − (−0.76) = +1.56 V
Zn + 2Ag⁺ → Zn²⁺ + 2Ag
Mass deposited. Q = I t, moles of electrons = Q ÷ 96 500, then use the half-equation.
0.50 A for 1930 s → Q = 965 C → n(e⁻) = 0.0100 mol
Ag⁺ + e⁻ → Ag so n(Ag) = 0.0100 mol
mass = 0.0100 × 108 = 1.08 g
Time needed, and gas volume.
3.175 g Cu = 0.0500 mol → n(e⁻) = 0.100 mol
Q = 0.100 × 96 500 = 9650 C t = 9650 ÷ 5.0 A = 1930 s
O₂ at the anode: 4OH⁻ → O₂ + 2H₂O + 4e⁻
0.040 mol e⁻ → 0.010 mol O₂ → 0.010 × 24 000 = 240 cm³
Products in aqueous electrolysis. At the cathode, a metal below hydrogen in the series is deposited, but H₂ forms for reactive metals such as sodium. At the anode, halide ions (Cl⁻, Br⁻, I⁻) are discharged in concentrated solution, otherwise oxygen forms from water or OH⁻.
🎯 Exam focus: the standard hydrogen electrode and conventions
Standard conditions: 298 K, 1.0 mol dm⁻³ for ions and 100 kPa for gases, with a platinum electrode in the standard hydrogen electrode (E° = 0.00 V).
cell notation Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s)
left = anode (oxidation, negative) right = cathode (reduction, positive)
electrons flow anode → cathode through the external circuit
Common mistakes. Doubling the E° value when you double the half-equation (it does not change). Swapping the polarity in electrolysis, where the cathode is negative. Forgetting the number of electrons in the half-equation when finding moles. Quoting “feasible” as “fast”: E° says nothing about the rate. Using Q = I t with the current in mA or the time in minutes.
Fuel cells. In a hydrogen–oxygen fuel cell, H₂ is oxidised at the anode and O₂ is reduced at the cathode. The only product is water, so it is clean at the point of use.
✏️ Quick check · 1 of 11
Anode
In any electrochemical cell, oxidation occurs at the:
AN OX.
✏️ Quick check · 2 of 11
Salt bridge
The purpose of a salt bridge is to:
It carries ions.
✏️ Quick check · 3 of 11
Galvanic or electrolytic
In an electrolytic cell:
Think of electroplating.
✍️ Type your answer · 4 of 11
E°(Cu²⁺/Cu) = +0.34 V (cathode) and E°(Zn²⁺/Zn) = −0.76 V (anode).
What is the cell voltage, in V?
Enter an exact answer, without rounding.
V
cathode − anode.
✍️ Type your answer · 5 of 11
E°(Ag⁺/Ag) = +0.80 V (cathode) and E°(Cu²⁺/Cu) = +0.34 V (anode).
What is the cell voltage, in V?
Enter an exact answer, without rounding.
V
cathode − anode.
✍️ Type your answer · 6 of 11
What is the oxidation number of manganese in MnO₄⁻?
Give the number.
Enter an exact answer, without rounding.
The charges must add up to −1.
✍️ Type your answer · 7 of 11
A current of 2 A flows for 965 s (F = 96500 C/mol).
How many moles of electrons are transferred, as a decimal?
Enter an exact answer, without rounding.
mol
n = It / F.
✍️ Type your answer · 8 of 11
Zn²⁺/Zn has E° = −0.76 V and Ag⁺/Ag has E° = +0.80 V.
What is E°cell for a zinc–silver cell, in volts?
Enter an exact answer, without rounding.
V
E°cathode − E°anode.
✍️ Type your answer · 9 of 11
A current of 0.50 A passes for 1930 s through silver nitrate solution (Ag⁺ + e⁻ → Ag).
What mass of silver is deposited, in g? (Ag = 108)
Enter an exact answer, without rounding.
g
One mole of electrons per mole of silver.
✍️ Type your answer · 10 of 11
A current of 5.0 A is used to deposit 3.175 g of copper from Cu²⁺ (Cu = 63.5).
How long does it take, in seconds?
Enter an exact answer, without rounding.
s
Two electrons per copper atom.
✏️ Quick check · 11 of 11
Electrolysis of dilute acid
What is formed at the cathode and the anode during the electrolysis of dilute sulfuric acid?
Which ion is attracted to the negative electrode?
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The standard hydrogen electrode has E° equal to:
Question 2 of 5
A positive E°cell means the reaction is:
Question 3 of 5
Electroplating a spoon with silver, the spoon is the:
Question 4 of 5
In a galvanic cell electrons flow:
Question 5 of 5
The oxidising agent in a redox reaction is the species that:
Lesson 5 · Organic
🧬 Organic Chemistry: Structures
Carbon forms four bonds and chains of every length, so organic chemistry is organised by functional groups, the parts of a molecule that decide how it reacts.
⛓️ Hydrocarbons
Alkanes (single bonds): CₙH₂ₙ₊₂. Alkenes (one C=C): CₙH₂ₙ. Alkynes (one C≡C): CₙH₂ₙ₋₂. Names come from the carbon count: meth, eth, prop, but, pent, hex.
C₅H₁₂ is pentane C₄H₈ is butene C₆H₁₄ is hexane
Isomers: same formula, different structure
C₄H₁₀: butane, 2-methylpropane (2 isomers)
C₅H₁₂: pentane, 2-methylbutane, 2,2-dimethylpropane (3 isomers)
🖼️ See it
The functional group decides how a molecule reacts.
🧩 Functional groups
A functional group gives a family its characteristic reactions.
Count the carbon chain first (longest chain), then number from the end that gives the functional group the lowest number.
🧪 Worked examples: isomers and stereochemistry
Structural isomers. Same molecular formula, different arrangement of atoms. They may differ in the chain, the position of a group or the functional group.
C₆H₁₄ has 5 chain isomers:
hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, 2,3-dimethylbutane
C₂H₆O: ethanol CH₃CH₂OH and methoxymethane CH₃OCH₃ (functional group isomers)
E/Z isomerism. It needs a C=C double bond (which cannot rotate) and two different groups on each carbon.
but-2-ene CH₃CH=CHCH₃ shows E/Z isomerism
but-1-ene CH₂=CHCH₂CH₃ does not (two H on one carbon)
Optical isomerism. A carbon with four different groups is a chiral centre and gives two non-superimposable mirror images (enantiomers). A molecule with n chiral centres can have up to 2ⁿ stereoisomers.
lactic acid CH₃CH(OH)COOH the central carbon has H, OH, CH₃ and COOH
one chiral centre → 2 enantiomers
with 2 chiral centres → up to 4 stereoisomers
Formula pattern. An alkane CₙH₂ₙ₊₂ with n = 8 has 18 hydrogens. A ring or double bond removes 2 hydrogens.
🎯 Exam focus: naming and drawing
Name in order: longest chain containing the functional group → number to give it the lowest locant → prefixes in alphabetical order.
Common mistakes. Choosing a chain that is not the longest one. Numbering from the wrong end. Drawing a displayed formula with the wrong number of bonds to carbon (always four). Showing the 3D shape of a chiral centre without the wedge and dash bonds. Calling two identical molecules drawn at different angles “isomers”.
Skeletal formula. Each corner or line end is a carbon; hydrogens on carbon are not drawn, but those on O or N are shown.
✏️ Quick check · 1 of 11
Alkene
Which formula is an alkene?
CₙH₂ₙ.
✏️ Quick check · 2 of 11
Alcohol group
The functional group in an alcohol is:
The name ends in -ol.
✏️ Quick check · 3 of 11
Isomers
Isomers have the same:
Same atoms, different arrangement.
✍️ Type your answer · 4 of 11
An alkane has 5 carbon atoms (CₙH₂ₙ₊₂).
How many hydrogen atoms does it have?
Enter an exact answer, without rounding.
Use CₙH₂ₙ₊₂.
✍️ Type your answer · 5 of 11
An alkene has 4 carbon atoms (CₙH₂ₙ).
How many hydrogen atoms does it have?
Enter an exact answer, without rounding.
Use CₙH₂ₙ.
✍️ Type your answer · 6 of 11
How many structural isomers does C₄H₁₀ have?
Give the number.
Enter an exact answer, without rounding.
A straight chain and a branched chain.
✍️ Type your answer · 7 of 11
Ethanol is C₂H₅OH (C = 12, H = 1, O = 16).
What is its molar mass, in g/mol?
Enter an exact answer, without rounding.
g/mol
Add up the atomic masses.
✍️ Type your answer · 8 of 11
Alkanes have the formula CₙH₂ₙ₊₂.
How many hydrogen atoms are in an alkane with 8 carbon atoms?
Enter an exact answer, without rounding.
Use 2n + 2.
✍️ Type your answer · 9 of 11
Hexane has the molecular formula C₆H₁₄.
How many structural isomers does C₆H₁₄ have (including hexane)?
Enter an exact answer, without rounding.
Shorten the main chain and move the branches.
✍️ Type your answer · 10 of 11
A molecule has 2 chiral centres and no internal symmetry.
What is the maximum number of stereoisomers it can have?
Enter an exact answer, without rounding.
The formula is 2ⁿ.
✏️ Quick check · 11 of 11
E/Z isomerism
Which of these can show E/Z isomerism?
Check both carbons of the C=C.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The name of CH₃CH₂CH₃ is:
Question 2 of 5
The functional group –COOH is found in:
Question 3 of 5
Which compound is saturated?
Question 4 of 5
Carbon always forms how many bonds in neutral organic compounds?
Question 5 of 5
Propanone has the functional group:
Lesson 6 · Organic
🔬 Organic Reactions
A handful of reaction types account for most of organic chemistry. Recognising the functional group tells you which one to expect.
🔁 Main reaction types
Alkanes undergo substitution, alkenes addition, and alcohols can undergo elimination or oxidation.
Addition joins atoms across a double bond, substitution swaps one atom for another, elimination removes a small molecule to form a double bond.
🍬 Esterification and yield
A carboxylic acid and an alcohol react (acid catalyst, reversible) to give an ester and water: RCOOH + R′OH ⇌ RCOOR′ + H₂O. Esters smell fruity and are used in flavourings.
ethanol + ethanoic acid → ethyl ethanoate + water
M(ethanol) = 46, M(ester C₄H₈O₂) = 88
4.6 g of ethanol = 0.1 mol → up to 0.1 mol of ester
theoretical mass = 0.1 · 88 = 8.8 g
6.6 g obtained → yield = 6.6 / 8.8 · 100 = 75%
Percentage yield = actual / theoretical × 100. It is below 100% for reversible reactions and when product is lost.
🧪 Worked examples: mechanisms, tests and yields
Electrophilic addition of HBr to an alkene. The π bond attacks the δ+ hydrogen, forming a carbocation. The more substituted carbocation is more stable, so it forms the major product.
Nucleophilic substitution of a halogenoalkane. The nucleophile (OH⁻, CN⁻ or NH₃) attacks the δ+ carbon, and the halide leaves. The weaker the C–X bond, the faster the reaction: C–I > C–Br > C–Cl.
CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻ (heat under reflux with aqueous NaOH)
Oxidation of alcohols with acidified K₂Cr₂O₇ (orange → green).
primary alcohol gently heat and DISTIL → aldehyde
heat under REFLUX → carboxylic acid
secondary alcohol heat under reflux → ketone
tertiary alcohol no reaction
Percentage yield.
23 g of ethanol (M = 46) = 0.50 mol → maximum 0.50 mol ethene = 14 g
obtained 8.4 g
yield = 8.4 ÷ 14 × 100 = 60%
🎯 Exam focus: reagents, conditions and curly arrows
Quote the reagent and conditions for each step. Curly arrows start at a lone pair or bond and finish at an atom, and they show the movement of a pair of electrons.
tests
C=C bromine water: orange → colourless
aldehyde Tollens’ reagent: silver mirror; Fehling’s/Benedict’s: red precipitate
ketone (no reaction) with Tollens’
acid effervesces with Na₂CO₃ (CO₂)
Common mistakes. Writing “heat” instead of “heat under reflux”. Drawing an arrow from an atom to an atom, instead of from a lone pair. Saying that tertiary alcohols are oxidised. Using the wrong carbocation, forgetting the major and minor products. Writing the yield without the 100% reference, and ignoring the stoichiometry.
Nucleophile or electrophile? A nucleophile donates a pair of electrons (needs a lone pair). An electrophile accepts a pair (δ+ carbon or a cation).
✏️ Quick check · 1 of 11
Alkene test
Bromine water is added to an unknown hydrocarbon and it turns from orange to colourless. The hydrocarbon is probably an:
Addition across the double bond.
✏️ Quick check · 2 of 11
Alkane reaction
Methane reacts with chlorine in UV light by:
Saturated molecules substitute.
✏️ Quick check · 3 of 11
Oxidation of alcohol
Oxidising a primary alcohol fully gives a:
Primary alcohol → aldehyde → acid.
✍️ Type your answer · 4 of 11
Ethene (C₂H₄) is hydrogenated to ethane.
How many moles of H₂ react with 1 mole of ethene?
Enter an exact answer, without rounding.
One double bond.
✍️ Type your answer · 5 of 11
C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O.
How many moles of CO₂ form from 0.5 mol of ethanol?
Enter an exact answer, without rounding.
Use the equation ratio.
✍️ Type your answer · 6 of 11
4.6 g of ethanol (M = 46) reacts with excess ethanoic acid. The ester has M = 88.
What is the theoretical mass of ester, in g?
Enter an exact answer, without rounding.
g
Convert to moles, then use 1 : 1.
✍️ Type your answer · 7 of 11
Only 6.6 g of ester is collected, from a theoretical 8.8 g.
What is the percentage yield?
Enter an exact answer, without rounding.
%
actual / theoretical.
✍️ Type your answer · 8 of 11
23 g of ethanol (M = 46) is dehydrated to ethene (M = 28), and 8.4 g of ethene is collected.
What is the percentage yield?
Enter an exact answer, without rounding.
%
Calculate the theoretical mass first.
✍️ Type your answer · 9 of 11
5.6 g of ethene (M = 28) reacts completely with excess HBr to form bromoethane (M = 109).
What mass of bromoethane is formed, in g?
Enter an exact answer, without rounding.
g
The ratio is 1 : 1.
✏️ Quick check · 10 of 11
Oxidising a primary alcohol
A primary alcohol is heated under reflux with acidified potassium dichromate(VI). What is the main organic product?
Reflux or distil?
✏️ Quick check · 11 of 11
Halogenoalkane reactivity
Which halogenoalkane reacts fastest with aqueous sodium hydroxide?
Compare bond strengths.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
Elimination of water from ethanol produces:
Question 2 of 5
Esterification is catalysed by:
Question 3 of 5
Which reaction has 100% atom economy?
Question 4 of 5
What is released when a halogenoalkane is hydrolysed by NaOH(aq)?
Question 5 of 5
Complete combustion of a hydrocarbon gives:
Lesson 7 · Organic
🧵 Polymers & Biomolecules
Polymers are long chains made of many repeating units. Plastics, fabrics, proteins and starch are all built this way.
🛍️ Addition polymers
In addition polymerisation many alkene monomers join with no atoms lost, giving polymers like poly(ethene), poly(chloroethene) (PVC) and polystyrene.
n CH₂=CH₂ → –(CH₂–CH₂)ₙ–
monomer 28, polymer M = 28000 → n = 1000
PVC repeat unit C₂H₃Cl: 24 + 3 + 35.5 = 62.5
polymer 31250 → n = 500
Addition polymers do not biodegrade because the carbon–carbon backbone is unreactive.
🖼️ See it
Proteins are condensation polymers of amino acids, joined by peptide bonds.
🧶 Condensation polymers
In condensation polymerisation two functional groups react and release a small molecule such as water. Polyesters (diacid + diol) and polyamides like nylon (diacid + diamine) form this way, and so do proteins from amino acids.
100 monomers joined end to end → 99 links
→ 99 water molecules released
glycine H₂N–CH₂–COOH M = 75
dipeptide Gly–Gly: 2·75 − 18 = 132
starch repeat C₆H₁₀O₅ = 162
1000 units → M = 162000
🧪 Worked examples: repeat units and masses
Addition polymer from the monomer. Open the C=C and show the repeat unit with the extension bonds.
propene CH₂=CH(CH₃) → –[CH₂–CH(CH₃)]ₙ–
M(repeat unit) = 42 polymer M = 42 000 → n = 1000
125 g of chloroethene at 100% yield gives 125 g of PVC (no atoms are lost)
Condensation polymer from two monomers. Each link loses one small molecule, usually water.
Peptides. A tripeptide of glycine (M = 75): 3 × 75 − 2 × 18 = 189. Hydrolysis needs the same water back, so a tripeptide needs 2 molecules of water to break into 3 amino acids.
Triglycerides. A fat is made of one glycerol and three fatty acids, joined by three ester links. Hydrolysis with NaOH gives glycerol and the sodium salts of the acids (soap).
Biodegradability. Polyesters and polyamides can be hydrolysed, so they are potentially biodegradable. Addition polymers have an unreactive C–C backbone, so they are not.
🎯 Exam focus: drawing repeat units and linking to properties
For an addition polymer, the repeat unit has two carbons in the chain with bonds going out at both ends. For a condensation polymer, include the ester or amide link and show at least one complete link.
nylon-6,6 –NH–(CH₂)₆–NH–CO–(CH₂)₄–CO–
strong because of hydrogen bonds between the chains
PET –O–CH₂CH₂–O–CO–C₆H₄–CO–
Common mistakes. Drawing the repeat unit of an addition polymer with a C=C still in it. Leaving out the water molecule in the balanced equation for condensation. Counting n monomers as making n links (a chain of n monomers has n − 1 links). Confusing the type of bond: ester (–COO–), amide or peptide (–CONH–).
Amino acid behaviour. Amino acids are amphoteric: –NH₂ accepts H⁺ and –COOH donates H⁺. They exist as zwitterions near neutral pH.
✏️ Quick check · 1 of 11
Addition polymer
Poly(ethene) is made from:
The name gives the monomer.
✏️ Quick check · 2 of 11
Condensation
Condensation polymerisation releases:
Think of peptide bonds.
✏️ Quick check · 3 of 11
Nylon
Nylon is a:
It has the same link as proteins.
✍️ Type your answer · 4 of 11
Ethene has M = 28. A poly(ethene) molecule has M = 28000.
How many monomer units are in the chain?
Enter an exact answer, without rounding.
Divide the polymer mass by the monomer mass.
✍️ Type your answer · 5 of 11
The repeat unit of PVC (C₂H₃Cl) has M = 62.5. A chain has M = 31250.
How many repeat units does it have?
Enter an exact answer, without rounding.
Divide the chain mass by the unit mass.
✍️ Type your answer · 6 of 11
A condensation polymer chain is made from 100 monomers joined end to end.
How many water molecules are released?
Enter an exact answer, without rounding.
One fewer than the monomers.
✍️ Type your answer · 7 of 11
Glycine has M = 75. Two glycine molecules join to make a dipeptide, losing one water (M = 18).
What is the dipeptide’s molar mass, in g/mol?
Enter an exact answer, without rounding.
g/mol
Subtract the lost water.
✍️ Type your answer · 8 of 11
The repeat unit of poly(propene) has M = 42, and a sample has a relative molecular mass of 42 000.
How many repeat units does the polymer chain have?
Enter an exact answer, without rounding.
Divide the polymer mass by the repeat unit mass.
✍️ Type your answer · 9 of 11
A polyester is made from 50 diacid molecules and 50 diol molecules, all joined in a single chain.
How many molecules of water are released?
Enter an exact answer, without rounding.
Links = monomers − 1.
✍️ Type your answer · 10 of 11
Glycine (M = 75) forms a tripeptide by condensation.
What is the relative molecular mass of the tripeptide?
Enter an exact answer, without rounding.
Two water molecules are lost.
✏️ Quick check · 11 of 11
Ester links
How many ester links are in one molecule of a triglyceride?
Count the –OH groups.
📝 Lesson quiz · 5 questions
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Re-read the lesson, then give it another go.
Question 1 of 5
A monomer is:
Question 2 of 5
Protein chains are condensation polymers of:
Question 3 of 5
Why is plastic waste a problem?
Question 4 of 5
A polyester is made from:
Question 5 of 5
Starch is a polymer of:
Lesson 8 · Energetics
🌡️ Thermodynamics: Enthalpy, Entropy & Gibbs
Why do some reactions happen on their own and others not? The answer balances the heat released, the disorder created and the temperature.
🔥 Enthalpy and Hess’s law
The enthalpy change ΔH is the heat at constant pressure (negative means exothermic). Hess’s law: the total ΔH is the same whichever route is taken. Bond enthalpies: ΔH = energy to break bonds − energy released forming bonds.
A reaction is spontaneous when ΔG is negative. Here entropy wins at high temperature, so the reaction only goes above 500 K.
🎲 Entropy and Gibbs free energy
Entropy S measures the number of ways energy can be spread out. Gas > liquid > solid. The Gibbs free energy combines both factors: ΔG = ΔH − TΔS. A reaction is spontaneous when ΔG < 0.
ΔH = +60 kJ, ΔS = +0.2 kJ/K, T = 200 K
ΔG = 60 − 200·0.2 = 20 kJ → not spontaneous
becomes spontaneous when TΔS > ΔH
T > 60 / 0.2 = 300 K
ΔH < 0 and ΔS > 0: spontaneous at all temperatures
Spontaneous does not mean fast. ΔG tells you whether a reaction can happen, and the activation energy tells you how quickly.
🧪 Worked examples: formation, Born–Haber, entropy and Gibbs
From enthalpies of formation: ΔH = Σ ΔHf(products) − Σ ΔHf(reactants).
The lattice enthalpy is exothermic (formation of the lattice from gaseous ions). It is larger for ions with higher charge and a smaller radius, because the electrostatic attraction is stronger.
MgO (Mg²⁺, O²⁻) ≈ −3791 kJ mol⁻¹
NaCl (Na⁺, Cl⁻) ≈ −787 kJ mol⁻¹
calorimetry: q = m c ΔT, ΔH = − q ÷ n (kJ per mole)
Common mistakes. Mixing J and kJ in ΔG = ΔH − TΔS. Forgetting that the first electron affinity is exothermic but the second is endothermic. Using the wrong sign for ΔH when the temperature rises (exothermic: ΔH is negative). Using a mass of the substance instead of the mass of water in q = m c ΔT. Saying that a feasible reaction is fast.
Combustion calorimetry. 0.0100 mol of fuel heats 200 g of water by 12.5 K: q = 200 × 4.2 × 12.5 = 10 500 J = 10.5 kJ, so ΔHc = −10.5 ÷ 0.0100 = −1050 kJ mol⁻¹.
✏️ Quick check · 1 of 11
Entropy order
Which state has the highest entropy for the same substance?
Most disorder.
✏️ Quick check · 2 of 11
Spontaneous
A reaction is thermodynamically spontaneous when:
Look at the sign of ΔG.
✏️ Quick check · 3 of 11
Hess
Hess’s law is a consequence of:
Energy is conserved.
✍️ Type your answer · 4 of 11
C + O₂ → CO₂ has ΔH = −394 kJ and CO + ½O₂ → CO₂ has ΔH = −283 kJ.
How much heat is released when 1 mol of C burns to CO, in kJ?