What do the Grade 12 Chemistry cards cover?

22 cards across 5 chapters of the national textbook: Acid-base equilibria, Electrochemistry, Industrial chemistry, Polymers and Introduction to environmental chemistry. You can take any chapter one card at a time on the page itself.

Do these help with the Grade 12 national exam?

Yes. Grade 12 sits a national exam, and these are the Chemistry rules and formulas it expects you to know. Past papers with a worked answer behind every question are on the same site, also free.

Where do these cards come from?

They are drafted from Grade 12 Chemistry, the Ministry of Education textbook for this grade. A second pass that cannot see the chapter then re-derives every formula, constant and table row, and anything it cannot confirm is held back instead of published.

Yes. Every card here is free to read and the printable sheet is free to download. Neither needs an account.

When was this last checked?

6 September 2026. Cards arrive chapter by chapter, and the line under each one says when that card was last read through.

Free short notes and revision cards

Grade 12 Chemistry

Grade 12 Chemistry on Temari has 22 revision cards, arranged by the chapters of the Ethiopian national curriculum. Every card says when the rule applies, what each symbol in it stands for, and the mistake students most often make with it. They are free to read and need no account.

22
Cards: 5
Chapters: 6
Formulas: 9
Reference tables: Grade 12 Chemistry
Textbook: Free
Price: 6 September 2026
Last checked

Everything22Reference table9Formula6Common mistake3Rule3Method1

Acid-base equilibria

Henderson Hasselbalch EquationFormula

\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)

\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)

$\text{pH}$: pH of the buffer solution $dimensionless$
$\text{p}K_a$: negative logarithm of acid dissociation constant $dimensionless$
$[\text{A}^-]$: molar concentration of conjugate base $\text{mol}\,\text{L}^{-1}$
$[\text{HA}]$: molar concentration of weak acid $\text{mol}\,\text{L}^{-1}$

When you use it

Use when finding the pH of a buffer solution composed of a weak acid and its conjugate base.

Watch out

The ratio inside the logarithm must place the conjugate base concentration in the numerator and the weak acid in the denominator.

Drafted from Grade 12 Chemistry, pages 3-56, then checked twice before it went up

Ion Product of Water and pH RelationFormula

K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0 \times 10^{-14}

K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0 \times 10^{-14}

$K_w$: ion product constant for water at 25 °C $\text{mol}^2\,\text{L}^{-2}$
$[\text{H}_3\text{O}^+]$: hydronium ion concentration $\text{mol}\,\text{L}^{-1}$
$[\text{OH}^-]$: hydroxide ion concentration $\text{mol}\,\text{L}^{-1}$

When you use it

Use when converting between hydronium and hydroxide ion concentrations or calculating pH and pOH in aqueous solutions at 25 °C.

Watch out

The value 1.0 times 10^-14 is strictly at 25 °C, and changes when temperature varies.

Drafted from Grade 12 Chemistry, pages 3-56, then checked twice before it went up

Neglecting the Five Percent Rule in ApproximationsCommon mistake

When you use it

Reach for this when simplifying equilibrium expressions for weak acids and weak bases by dropping x from initial concentration.

Watch out

If the ionized concentration x exceeds 5 percent of the initial acid concentration, the approximation fails and the quadratic formula must be used.

Drafted from Grade 12 Chemistry, pages 3-56, then checked twice before it went up

Salt Hydrolysis and Solution pHReference tableStrong · Weak

Parent Acid	Parent Base	Solution Nature	pH Range at 25 °C
Strong	Strong	Neutral	$\text{pH} = 7$
Weak	Strong	Basic	$\text{pH} > 7$
Strong	Weak	Acidic	$\text{pH} < 7$
Weak	Weak	Depends on $K_a$ and $K_b$	$K_a > K_b \Rightarrow \text{pH} < 7$

When you use it

Use to predict whether a dissolved salt yields an acidic, basic, or neutral aqueous solution.

Watch out

For salts of weak acid and weak base, compare Ka of the cation and Kb of the anion to decide acidity.

Drafted from Grade 12 Chemistry, pages 3-56, then checked twice before it went up

Electrochemistry

Balancing Redox Reactions in Basic MediumMethod

When you use it

Use when balancing oxidation-reduction equations occurring in basic or alkaline solution using the half-reaction method.

Watch out

First balance fully as if in acidic medium using H+ and H2O, then add equal numbers of OH- to BOTH sides for every H+, and combine H+ and OH- into H2O.

Drafted from Grade 12 Chemistry, pages 57-135, then checked twice before it went up

Electrolysis Products of Aqueous SolutionsReference tableDilute NaCl (inert) · Concentrated NaCl / Brine (inert) · Dilute H_2SO_4 (inert) · +2

Electrolyte and Electrodes	Cathode Product	Anode Product	Net Result
Dilute $\text{NaCl}$ (inert)	$\text{H}_2\text{ (g)}$	$\text{O}_2\text{ (g)}$	Decomposition of water
Concentrated $\text{NaCl}$ / Brine (inert)	$\text{H}_2\text{ (g)}$	$\text{Cl}_2\text{ (g)}$	$\text{NaOH}$ remains in solution
Dilute $\text{H}_2\text{SO}_4$ (inert)	$\text{H}_2\text{ (g)}$	$\text{O}_2\text{ (g)}$	Decomposition of water
$\text{CuSO}_4$ with inert electrodes (Pt)	$\text{Cu (s)}$	$\text{O}_2\text{ (g)}$	Solution becomes acidic ( $\text{H}_2\text{SO}_4$ )
$\text{CuSO}_4$ with active $\text{Cu}$ electrodes	$\text{Cu (s)}$ deposited	$\text{Cu}^{2+}\text{ (aq)}$ dissolved	$\text{CuSO}_4$ concentration unchanged

When you use it

Use to determine the products liberated at the cathode and anode based on ion concentrations and electrode activity.

Watch out

Active copper anodes dissolve into Cu2+ ions instead of oxidizing water or hydroxide ions to oxygen gas.

Drafted from Grade 12 Chemistry, pages 57-135, then checked twice before it went up

Faraday First Law of ElectrolysisFormula

m = \frac{M I t}{n F}

m = \frac{M I t}{n F}

$m$: mass of substance deposited or liberated $\text{g}$
$M$: molar mass of the substance $\text{g\,mol}^{-1}$
$I$: electric current $\text{A}$
$t$: time of electrolysis $\text{s}$
$n$: valence (number of electrons transferred per ion) $1$
$F$: Faraday constant (96,500 C/mol) $\text{C\,mol}^{-1}$

When you use it

Use to calculate the mass of a substance deposited or liberated at an electrode during electrolysis given current and time.

Watch out

Time t must be converted to seconds (s) before calculating, not left in minutes or hours.

Drafted from Grade 12 Chemistry, pages 57-135, then checked twice before it went up

Faraday Second Law of ElectrolysisFormula

\frac{m_1}{E_1} = \frac{m_2}{E_2}

\frac{m_1}{E_1} = \frac{m_2}{E_2}

$m_1$: mass of first substance $\text{g}$
$E_1$: equivalent weight of first substance (M/n) $\text{g\,eq}^{-1}$
$m_2$: mass of second substance $\text{g}$
$E_2$: equivalent weight of second substance (M/n) $\text{g\,eq}^{-1}$

When you use it

Use when the same quantity of electric current passes through two or more electrolytic cells connected in series.

Watch out

Equivalent weight E equals molar mass divided by valence (n). Do not confuse valence with the formula subscript of the salt.

Drafted from Grade 12 Chemistry, pages 57-135, then checked twice before it went up

Metallic versus Electrolytic ConductionReference tableCharge carriers · Matter transfer · Chemical change · +1

Property	Metallic Conduction	Electrolytic Conduction
Charge carriers	Delocalized mobile electrons	Mobile cations and anions
Matter transfer	No transfer of matter	Actual transfer of matter
Chemical change	No chemical decomposition	Accompanied by chemical decomposition
Effect of temperature	Conductivity decreases as temperature rises	Conductivity increases as temperature rises

When you use it

Use to distinguish how electrical charge moves through metals compared to molten or aqueous electrolytes.

Watch out

Remember that raising temperature increases electrolytic conductivity due to higher ion mobility, but decreases metallic conductivity due to lattice vibrations.

Drafted from Grade 12 Chemistry, pages 57-135, then checked twice before it went up

Industrial chemistry

Four Steps of the Contact ProcessRule

When you use it

Use these sequential steps to write the chemical equations for the industrial production of sulfuric acid.

Watch out

SO3 is dissolved into concentrated H2SO4 to make oleum (H2S2O7), never directly into water.

Drafted from Grade 12 Chemistry, pages 136-213, then checked twice before it went up

Solvay Process ReactionsRule

When you use it

Use this reaction sequence to describe the production of sodium carbonate (Na2CO3) and identify its major by-product.

Watch out

The overall raw materials are CaCO3 and NaCl, while CaCl2 is the main by-product and NH3 is recycled.

Drafted from Grade 12 Chemistry, pages 136-213, then checked twice before it went up

Sulfuric Acid Dilution SafetyCommon mistake

When you use it

Use this rule when describing the laboratory or industrial dilution and handling of concentrated sulfuric acid.

Watch out

Never add water to concentrated sulfuric acid. Always add acid slowly to water because the dissolution is violently exothermic.

Drafted from Grade 12 Chemistry, pages 136-213, then checked twice before it went up

Synthesis of Urea and DAPReference tableUrea · DAP

Fertilizer	Raw Materials	Synthesis Reactions
Urea	$\text{NH}_3$ , $\text{CO}_2$	$2\text{NH}_3 + \text{CO}_2 \rightarrow \text{NH}_2\text{COONH}_4 \rightarrow (\text{NH}_2)_2\text{CO} + \text{H}_2\text{O}$
DAP	$\text{NH}_3$ , $\text{H}_3\text{PO}_4$	$3\text{NH}_3 + 2\text{H}_3\text{PO}_4 \rightarrow \text{NH}_4\text{H}_2\text{PO}_4 + (\text{NH}_4)_2\text{HPO}_4$

When you use it

Use this table when writing industrial preparation reactions for nitrogen-based fertilizers like urea and diammonium hydrogen phosphate.

Watch out

Urea synthesis forms ammonium carbamate as an intermediate before dehydrating to urea.

Drafted from Grade 12 Chemistry, pages 136-213, then checked twice before it went up

Polymers

Addition versus Condensation PolymerizationReference tableMonomer requirement · By-product formation · Growth mechanism · +2

Feature	Addition Polymerization	Condensation Polymerization
Monomer requirement	Unsaturated molecules with $C=C$ or $C\equiv C$	Molecules with two or more functional groups
By-product formation	No small molecules eliminated	Eliminates small molecules like $H_2O$ or $CH_3OH$
Growth mechanism	Chain-growth (initiation, propagation, termination)	Step-growth reaction between functional groups
Polymer molar mass	Exact multiple of monomer molar mass	Less than sum of individual monomer masses
Examples	Polyethylene, PVC, Polystyrene, Teflon	Nylon 66, Dacron, Starch, Proteins

When you use it

Use this comparison to distinguish the mechanism, monomer requirements, and products of the two main polymerization types.

Watch out

Condensation polymerization requires polyfunctional monomers and always splits off a small molecule such as water.

Drafted from Grade 12 Chemistry, pages 214-240, then checked twice before it went up

Common Synthetic Addition PolymersReference tablePolyethylene (PE) · Polypropylene (PP) · Polyvinyl chloride (PVC) · +2

Polymer Name	Monomer Name	Monomer Structure	Key Application
Polyethylene (PE)	Ethylene (Ethene)	$CH_2=CH_2$	Plastic bags, squeeze bottles, trash bags
Polypropylene (PP)	Propylene (Propene)	$CH_2=CH-CH_3$	Ropes, carpets, dishwasher safe containers
Polyvinyl chloride (PVC)	Vinyl chloride	$CH_2=CHCl$	Pipes, floor tiles, raincoats
Polymethyl methacrylate (PMMA)	Methyl methacrylate	$CH_2=C(CH_3)COOCH_3$	Airplane windows, streetlights (Lucite)
Polytetrafluoroethylene (PTFE)	Tetrafluoroethylene	$CF_2=CF_2$	Non-stick cooking pans, electrical insulation

When you use it

Use this reference to match synthetic addition polymers to their monomer formulas and commercial applications.

Watch out

Do not confuse the monomer name propylene with polymer polypropylene when identifying starting materials.

Drafted from Grade 12 Chemistry, pages 214-240, then checked twice before it went up

Degree of PolymerizationFormula

n = \frac{M_{\text{polymer}}}{M_{\text{monomer}}}

n = \frac{M_{\text{polymer}}}{M_{\text{monomer}}}

$n$: degree of polymerization (number of monomer units) $1$
$M_{\text{polymer}}$: molar mass of the polymer $\text{g mol}^{-1}$
$M_{\text{monomer}}$: molar mass of the repeating monomer unit $\text{g mol}^{-1}$

When you use it

Use this formula to calculate the number of monomer units present in a polymer molecule from molar mass data.

Watch out

Both molar masses must be expressed in the exact same units, typically g/mol, before dividing.

Drafted from Grade 12 Chemistry, pages 214-240, then checked twice before it went up

Thermoplastics versus ThermosetsReference tableMolecular architecture · Response to heat · Recyclability · +2

Property	Thermoplastics	Thermosets
Molecular architecture	Linear or lightly branched chains	Extensive three-dimensional cross-linked network
Response to heat	Melt when heated, resolidify when cooled	Do not melt; decompose irreversibly at high heat
Recyclability	Recyclable (can be remelted and reshaped)	Non-recyclable
Mechanical properties	Flexible and remoldable	Hard, rigid, heat resistant, corrosion resistant
Examples	Polyethylene, PVC, Nylon, Teflon	Bakelite, Epoxy resin, Vulcanized rubber

When you use it

Use this table to classify polymers according to their molecular structure, thermal behavior, and recyclability.

Watch out

Nylon and Dacron are condensation polymers but act as thermoplastics because they lack extensive covalent cross-linking.

Drafted from Grade 12 Chemistry, pages 214-240, then checked twice before it went up

Introduction to environmental chemistry

Atmospheric Abundance vs Greenhouse WarmingCommon mistake

When you use it

Use when identifying which atmospheric gases contribute to the greenhouse effect.

Watch out

Nitrogen and oxygen make up roughly 99 percent of the atmosphere but do not absorb infrared radiation and do not cause the greenhouse effect.

Drafted from Grade 12 Chemistry, page 241 onwards, then checked twice before it went up

Atom EconomyFormula

\text{Atom Economy} = \frac{m_{\text{desired}}}{m_{\text{total}}} \times 100

\text{Atom Economy} = \frac{m_{\text{desired}}}{m_{\text{total}}} \times 100

$m_{\text{desired}}$: Mass of the desired product $\text{kg}$
$m_{\text{total}}$: Total mass of all products $\text{kg}$

When you use it

Use this formula to evaluate the efficiency of a chemical reaction according to green chemistry principles.

Watch out

Divide by the total mass of all products formed, not just the mass of reactants or byproducts alone.

Drafted from Grade 12 Chemistry, page 241 onwards, then checked twice before it went up

Environmental Quality IndicatorsReference tableDissolved Oxygen (DO) · Biological Oxygen Demand (BOD) · Threshold Limit Value (TLV)

Parameter	Definition	Standard value or condition
Dissolved Oxygen (DO)	Oxygen dissolved in water vital for aquatic life	$4 - 8\text{ mg L}^{-1}$ is good quality, $< 4\text{ mg L}^{-1}$ is polluted
Biological Oxygen Demand (BOD)	Capacity of dissolved organic matter to consume oxygen	Measured experimentally over a 5 day period in a sealed sample
Threshold Limit Value (TLV)	Permissible level of toxic pollutant in the atmosphere	Safe for exposure over an 8 hour day without adverse effects

When you use it

Use when assessing water quality levels or safe atmospheric exposure limits for chemical species.

Watch out

A higher BOD value indicates more organic pollution and lower water quality, whereas high DO indicates healthy water.

Drafted from Grade 12 Chemistry, page 241 onwards, then checked twice before it went up

Greenhouse Gas ContributionsReference tableCO_2 · CH_4 · Halogenated compounds (CFCs and HCFCs) · +2

Greenhouse gas	Atmospheric lifetime	Contribution to global warming
$\text{CO}_2$	80% lasts 200 years, 20% up to 30000 years	52.92%
$\text{CH}_4$	12 years	14.88%
Halogenated compounds (CFCs and HCFCs)	Months to tens of thousands of years	10.78%
Tropospheric ozone ( $\text{O}_3$ )	Few months	10.72%
Nitrous oxide ( $\text{N}_2\text{O}$ )	114 years	10.70%

When you use it

Use when comparing the sources, persistence, and warming impact of major anthropogenic greenhouse gases.

Watch out

Carbon dioxide contributes the largest percentage to global warming even though other gases like CFCs may persist longer or have higher intrinsic warming potential.

Drafted from Grade 12 Chemistry, page 241 onwards, then checked twice before it went up

Four air pollution problems and their controlsRuleCut the emission at source: efficient furnaces and engines, and electricity from hydro, wind, solar and geothermal instead of coal and oil. Afforestation is a supporting sink that stores part of the carbon. · Electrostatic precipitators and bag filters take out the particles, wet scrubbers take out the acidic gases, catalytic converters treat vehicle exhaust, and the authority enforces emission limits at the chimney. · Burn low sulfur fuel, strip SO_2 out of the flue gas with lime or limestone, which fixes it as calcium sulfite and then calcium sulfate, and fit catalytic converters to cut nitrogen oxides. · +1

Control measure	The problem it works on	The pollutant and the harm it does
Cut the emission at source: efficient furnaces and engines, and electricity from hydro, wind, solar and geothermal instead of coal and oil. Afforestation is a supporting sink that stores part of the carbon.	Global warming	Burning fossil fuels releases $\text{CO}_2$ , which absorbs the infrared radiation leaving the Earth's surface and warms the lower atmosphere.
Electrostatic precipitators and bag filters take out the particles, wet scrubbers take out the acidic gases, catalytic converters treat vehicle exhaust, and the authority enforces emission limits at the chimney.	Respiratory illness	Industrial and vehicle emissions carry particulate matter, $\text{SO}_2$ , $\text{NO}_x$ and $\text{CO}$ . Fine particles and the acidic gases inflame the airways and bring on bronchitis and asthma attacks, while carbon monoxide binds to haemoglobin and blocks oxygen transport in the blood.
Burn low sulfur fuel, strip $\text{SO}_2$ out of the flue gas with lime or limestone, which fixes it as calcium sulfite and then calcium sulfate, and fit catalytic converters to cut nitrogen oxides.	Acid rain	$2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3$ , then $\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4$ , and $3\text{NO}_2 + \text{H}_2\text{O} \rightarrow 2\text{HNO}_3 + \text{NO}$ . The acid acidifies soil and lakes, scorches leaves and lowers crop yield, corrodes metal and eats limestone and marble.
The Montreal Protocol of 1987 phased out CFCs worldwide, and industry moved to substitutes that carry no chlorine.	Ozone layer depletion	CFCs are unreactive in the troposphere, so they survive to drift up. Ultraviolet light splits a chlorine atom off, and it destroys ozone catalytically: $\text{Cl} + \text{O}_3 \rightarrow \text{ClO} + \text{O}_2$ , then $\text{ClO} + \text{O} \rightarrow \text{Cl} + \text{O}_2$ . The chlorine comes back, so one atom destroys many ozone molecules and more UV-B reaches the ground, raising rates of skin cancer and cataracts and lowering crop yield.

When you use it

Use for the environmental chemistry question that asks you to name an air pollution problem, the pollutant behind it and one control measure. Each row pairs the control with the chemistry it is fixing, so your answer names the equipment or the treaty instead of saying that pollution should be reduced.

Watch out

Students merge the two big problems and write that the ozone hole lets more heat in, or that carbon dioxide destroys the ozone layer. Keep them apart: ozone depletion is chlorine chemistry high in the stratosphere and what comes through is ultraviolet light, while global warming is carbon dioxide and methane low in the atmosphere holding back the infrared going out. The same slip turns up in acid rain answers, where carbon dioxide gets blamed. Acid rain is sulfur dioxide and nitrogen oxides, and carbon dioxide only accounts for the pH 5.6 that clean rain already has.

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Questions students ask

What do the Grade 12 Chemistry cards cover?: 22 cards across 5 chapters of the national textbook: Acid-base equilibria, Electrochemistry, Industrial chemistry, Polymers and Introduction to environmental chemistry. You can take any chapter one card at a time on the page itself.
Do these help with the Grade 12 national exam?: Yes. Grade 12 sits a national exam, and these are the Chemistry rules and formulas it expects you to know. Past papers with a worked answer behind every question are on the same site, also free.
Where do these cards come from?: They are drafted from Grade 12 Chemistry, the Ministry of Education textbook for this grade. A second pass that cannot see the chapter then re-derives every formula, constant and table row, and anything it cannot confirm is held back instead of published.
Is this free?: Yes. Every card here is free to read and the printable sheet is free to download. Neither needs an account.
When was this last checked?: 6 September 2026. Cards arrive chapter by chapter, and the line under each one says when that card was last read through.

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