Acid-Base Reactions, Dissociation, and the pH Scale

Everything you need for Leaving Cert Higher Level Chemistry — syllabus-aligned explanations, key terms and self-check questions.

16 min readHigher LevelBy Studytok
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Acid–base chemistry runs through the whole Leaving Certificate course: it controls the pH of blood and natural water, forms the foundation of volumetric analysis in Question 1, and links directly to equilibrium, chemical kinetics, and thermochemistry. Understanding this area means moving comfortably between three theoretical frameworks, tracking proton transfers in solution, mastering logarithmic pH and dissociation constants, and handling quantitative laboratory procedures with precision.

Theories of Acids and Bases, Basicity, and Conjugate Pairs

Three complementary models explain acid-base behaviour, each broadening our view of how species interact in solution.

Three Theories of Acids and Bases

TheoryAcidBase
ArrheniusDissociates in water to produce H+\text{H}^+ ionsDissociates in water to produce OH−\text{OH}^- ions
Brønsted–LowryA proton donorA proton acceptor
Lewis (HL)An electron pair acceptorAn electron pair donor

The Arrhenius theory has clear limitations: it restricts acid-base reactions to aqueous solutions, and it fails to explain why substances lacking an OH−\text{OH}^- group, such as ammonia (NH3\text{NH}_3) or sodium carbonate (Na2CO3\text{Na}_2\text{CO}_3), act as bases in water.

Lewis theory widens the field to electron pairs. Consider the reaction between ammonia and boron trifluoride:

NH3+BF3→H3N–BF3\text{NH}_3 + \text{BF}_3 \rightarrow \text{H}_3\text{N–BF}_3

Here, ammonia donates a lone pair to form a dative covalent bond, acting as a Lewis base, while boron trifluoride accepts that pair, acting as a Lewis acid. Likewise, when a proton attaches to ammonia (H++:NH3→NH4+\text{H}^+ + :\text{NH}_3 \rightarrow \text{NH}_4^+), H+\text{H}^+ is the Lewis acid. Every Brønsted–Lowry base is also a Lewis base, because that unshared lone pair is precisely what binds the incoming proton.

Basicity of Acids

Basicity refers to the number of replaceable hydrogen ions per molecule of acid:

  • Monobasic (monoprotic): HCl\text{HCl}, HNO3\text{HNO}_3, CH3COOH\text{CH}_3\text{COOH} (releases 1 H+1\text{ H}^+).
  • Dibasic (diprotic): H2SO4\text{H}_2\text{SO}_4, H2CO3\text{H}_2\text{CO}_3 (releases 2 H+2\text{ H}^+).
  • Tribasic (triprotic): H3PO4\text{H}_3\text{PO}_4 (releases 3 H+3\text{ H}^+).

Notice that ethanoic acid (CH3COOH\text{CH}_3\text{COOH}) contains four hydrogens in total, yet it is strictly monobasic because only the polar hydrogen in the carboxyl group (−COOH-\text{COOH}) can dissociate.

Reading Conjugate Pairs from an Equation

A conjugate acid-base pair consists of two species that differ by one proton only. When writing them down, quote them as a pair with the acid first and the conjugate base second:

CH3COOH+H2O⇌CH3COO−+H3O+\text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+
  • Pair 1: CH3COOH / CH3COO−\text{CH}_3\text{COOH} \text{ / } \text{CH}_3\text{COO}^-
  • Pair 2: H3O+ / H2O\text{H}_3\text{O}^+ \text{ / } \text{H}_2\text{O}
NH3+H2O⇌NH4++OH−\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-
  • Pair 1: NH4+ / NH3\text{NH}_4^+ \text{ / } \text{NH}_3
  • Pair 2: H2O / OH−\text{H}_2\text{O} \text{ / } \text{OH}^-

Amphoteric (amphiprotic) species can donate or accept a proton depending on the reacting partner. Water is amphoteric: it acts as a base when accepting a proton from ethanoic acid, and as an acid when donating a proton to ammonia. Other common amphoteric species include HCO3−\text{HCO}_3^-, HSO4−\text{HSO}_4^-, and H2PO4−\text{H}_2\text{PO}_4^-.

Primary Acid Reactions and Evidence for Acid Strength

Reactions involving acids in the laboratory follow three predictable stoichiometric pathways:

  1. Neutralisation (Acid + Base):
Acid+Base→Salt+Water\text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water}HCl+NaOH→NaCl+H2O\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}

Writing out dissolved ions produces the complete ionic equation:

H++Cl−+Na++OH−→Na++Cl−+H2O\text{H}^+ + \text{Cl}^- + \text{Na}^+ + \text{OH}^- \rightarrow \text{Na}^+ + \text{Cl}^- + \text{H}_2\text{O}

Sodium (Na+\text{Na}^+) and chloride (Cl−\text{Cl}^-) are spectator ions because they remain unaltered in solution. Cancelling them leaves the net ionic equation:

H++OH−→H2OorH3O++OH−→2H2O\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O} \quad \text{or} \quad \text{H}_3\text{O}^+ + \text{OH}^- \rightarrow 2\text{H}_2\text{O}

In medicine, antacids containing true bases like magnesium hydroxide neutralise excess stomach acid:

Mg(OH)2+2HCl→MgCl2+2H2O\text{Mg(OH)}_2 + 2\text{HCl} \rightarrow \text{MgCl}_2 + 2\text{H}_2\text{O}

In agriculture, slaked lime (calcium hydroxide, Ca(OH)2\text{Ca(OH)}_2) neutralises soil acidity.

  1. Acid + Carbonate (or Hydrogencarbonate):
Acid+Carbonate→Salt+Water+Carbon Dioxide\text{Acid} + \text{Carbonate} \rightarrow \text{Salt} + \text{Water} + \text{Carbon Dioxide}2HCl+CaCO3→CaCl2+H2O+CO22\text{HCl} + \text{CaCO}_3 \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2

Carbonate ions (CO32−\text{CO}_3^{2-}) encounter hydronium ions, forming carbonic acid (H2CO3\text{H}_2\text{CO}_3), which immediately decomposes: H2CO3→H2O+CO2\text{H}_2\text{CO}_3 \rightarrow \text{H}_2\text{O} + \text{CO}_2. Escaping CO2\text{CO}_2 gas produces rapid effervescence.

  1. Acid + Reactive Metal:
Acid+Metal→Salt+Hydrogen Gas\text{Acid} + \text{Metal} \rightarrow \text{Salt} + \text{Hydrogen Gas}Mg+2HCl→MgCl2+H2\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2

Diatomic hydrogen gas ignites with a characteristic sharp sound during the pop test. Unreactive metals like gold and copper do not displace hydrogen from dilute acids.

Strength Versus Concentration

Strength depends solely on the extent of dissociation in aqueous solution, whereas concentration is the number of moles of solute dissolved per litre of solution (mol l−1\text{mol l}^{-1}).

Strong (fully dissociated)Weak (partially dissociated)
Concentrated11 mol l−1 HCl11\text{ mol l}^{-1}\text{ HCl}Glacial ethanoic acid ( 17 mol l−1~17\text{ mol l}^{-1})
Dilute0.001 mol l−1 HCl0.001\text{ mol l}^{-1}\text{ HCl}Diluted vinegar ( 0.08 mol l−1~0.08\text{ mol l}^{-1})

Experimental Comparison: Strong vs Weak Acid (0.1 mol l−1 HCl0.1\text{ mol l}^{-1}\text{ HCl} vs 0.1 mol l−1 CH3COOH0.1\text{ mol l}^{-1}\text{ CH}_3\text{COOH})

Test0.1 mol l−1 HCl0.1\text{ mol l}^{-1}\text{ HCl}0.1 mol l−1 CH3COOH0.1\text{ mol l}^{-1}\text{ CH}_3\text{COOH}Reason
pH meter reading≈1.0\approx 1.0≈2.9\approx 2.9Lower [H3O+][\text{H}_3\text{O}^+] in weak acid
Electrical conductivityHighLowFewer mobile ions to carry current
Rate of reaction with Mg ribbonVigorous fizzing, rapidSlow bubblingLower [H3O+][\text{H}_3\text{O}^+] gives fewer effective collisions

Crucially, equal volumes of both acids require the exact same volume of sodium hydroxide solution for complete neutralisation. Both solutions contain identical total quantities of available acidic protons; strength dictates how many ions exist at equilibrium, not the total neutralising capacity.

Self-Ionisation of Water and the Ionic Product (Kw)

Pure deionised water conducts an extremely small, detectable electric current because water molecules spontaneously dissociate to establish a reversible equilibrium:

H2O+H2O⇌H3O++OH−(ΔH=+55.8 kJ mol−1)\text{H}_2\text{O} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^- \quad (\Delta H = +55.8\text{ kJ mol}^{-1})

Because dissociation occurs to such a tiny extent, the concentration of undissociated water remains virtually constant. We simplify the equilibrium expression to the ionic product of water (KwK_w):

Kw=[H3O+][OH−]K_w = [\text{H}_3\text{O}^+][\text{OH}^-]

At 25 °C, Kw=1.0×10−14 mol2 l−2K_w = 1.0 \times 10^{-14}\text{ mol}^2\text{ l}^{-2}. In pure water, each dissociation produces an equal number of hydronium and hydroxide ions, so [H3O+]=[OH−]=1.0×10−14=1.0×10−7 mol l−1[\text{H}_3\text{O}^+] = [\text{OH}^-] = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7}\text{ mol l}^{-1}.

Temperature Dependence of KwK_w

Self-ionisation is endothermic (ΔH=+55.8 kJ mol−1\Delta H = +55.8\text{ kJ mol}^{-1}). Applying thermal energy shifts the position of equilibrium to the right by Le Chatelier's principle, causing KwK_w to rise systematically as temperature increases:

Temperature (°C)KwK_w (mol2 l−2\text{mol}^2\text{ l}^{-2})pH of pure water
00.11×10−140.11 \times 10^{-14}7.47
251.0×10−141.0 \times 10^{-14}7.00
10051.3×10−1451.3 \times 10^{-14}6.14

The exam point here is vital: chemical neutrality requires [H3O+]=[OH−][\text{H}_3\text{O}^+] = [\text{OH}^-]. Pure boiling water at 100 °C gives a pH reading of 6.14, but it is still completely neutral because hydronium and hydroxide ions remain present in equal numbers.

The pH Scale and Quantitative Calculations

The pH scale was devised by Sørensen in 1909 so that very small hydrogen ion concentrations could be expressed as manageable numbers between 0 and 14.

pH=−log⁡10[H3O+][H3O+]=10−pH\text{pH} = -\log_{10}[\text{H}_3\text{O}^+] \qquad [\text{H}_3\text{O}^+] = 10^{-\text{pH}}

Because the scale is logarithmic, each step of 1 pH unit represents a 10-fold change in [H3O+][\text{H}_3\text{O}^+]. A difference of 2 pH units means a 100-fold change, and shifting by nn pH units alters the concentration by a factor of 10n10^n.

Significant Figures Rule for pH: Only digits after the decimal point are significant. A concentration stated to two significant figures, such as 2.5×10−4 mol l−12.5 \times 10^{-4}\text{ mol l}^{-1}, yields a pH expressed to two decimal places (3.60).

Dilution and the pH Scale

Diluting a strong acid by a factor of 10 raises its pH by 1 unit. Diluting a strong base by a factor of 10 lowers its pH by 1 unit. However, progressive dilution cannot push an acid above pH 7 or a base below pH 7; as the solution becomes infinitely dilute, water's own self-ionisation dominates.

Calculations for Strong Species

  • Strong Monobasic Acid: [H3O+]=c[\text{H}_3\text{O}^+] = c. For 0.02 mol l−1 HCl0.02\text{ mol l}^{-1}\text{ HCl}, pH=−log⁡10(0.02)=1.70\text{pH} = -\log_{10}(0.02) = 1.70.
  • Strong Dibasic Acid: [H3O+]=2×c[\text{H}_3\text{O}^+] = 2 \times c. For 0.05 mol l−1 H2SO40.05\text{ mol l}^{-1}\text{ H}_2\text{SO}_4, [H3O+]=2×0.05=0.10 mol l−1[\text{H}_3\text{O}^+] = 2 \times 0.05 = 0.10\text{ mol l}^{-1}, so pH=−log⁡10(0.10)=1.00\text{pH} = -\log_{10}(0.10) = 1.00.
  • Strong Base: Use Kw=[H3O+][OH−]=1.0×10−14K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1.0 \times 10^{-14} or pH=14−pOH\text{pH} = 14 - \text{pOH}.

Calculations for Weak Acids (KaK_a)

For the equilibrium HA+H2O⇌H3O++A−\text{HA} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{A}^-:

Ka=[H3O+][A−][HA]K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}

Two working assumptions carry marks when stated explicitly:

  1. [H3O+]=[A−][\text{H}_3\text{O}^+] = [\text{A}^-], because both arise in equal amounts from the dissociation.
  2. Equilibrium [HA]≈c[\text{HA}] \approx c (initial concentration), because weak acids dissociate only slightly.

This gives the working expression:

[H3O+]=Ka×c[\text{H}_3\text{O}^+] = \sqrt{K_a \times c}

KaK_a values quantify weak acid strength: ethanoic acid (1.8×10−51.8 \times 10^{-5}), methanoic acid (1.8×10−41.8 \times 10^{-4}), and hydrofluoric acid (6.8×10−46.8 \times 10^{-4}). The larger the KaK_a, the stronger the weak acid. Strong acids are fully dissociated, so equilibrium constants are not quoted for them in aqueous solution.

The degree of dissociation is given by:

Degree of dissociation=[H3O+]c×100\text{Degree of dissociation} = \frac{[\text{H}_3\text{O}^+]}{c} \times 100

Indicators, Titration Curves, and Salt Hydrolysis

The definition examiners look for is straightforward: an indicator is a substance that changes colour according to the pH of the solution it is in. In chemical terms, an indicator is itself a weak acid (or weak base) where the undissociated molecule and its conjugate base display distinctly different colours:

HIn⇌H++In−orHIn+H2O⇌H3O++In−\text{HIn} \rightleftharpoons \text{H}^+ + \text{In}^- \quad \text{or} \quad \text{HIn} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{In}^- (Colour A)(Colour B)\text{(Colour A)} \qquad\qquad\qquad\qquad\qquad\qquad \text{(Colour B)}

Adding acid increases [H3O+][\text{H}_3\text{O}^+], driving the equilibrium to the left by Le Chatelier's principle so Colour A dominates. Adding base removes hydronium ions, pulling the equilibrium to the right so Colour B dominates.

IndicatorpH rangeColour in acid →\rightarrow colour in base
Methyl orange3.1 – 4.4Red →\rightarrow Yellow
Litmus5.0 – 8.0Red →\rightarrow Blue
Phenolphthalein8.3 – 10.0Colourless →\rightarrow Pink

Choosing an Indicator from Titration Curves

The equivalence point occurs where acid and base have reacted in exact stoichiometric proportions. The end point is where the indicator changes colour. A suitable indicator has its entire pH transition range falling within the steep, vertical section of the titration curve.

CombinationpH at equivalenceSteep range (approx.)Suitable indicator
Strong acid + strong base (HCl/NaOH\text{HCl} / \text{NaOH})73 – 11Methyl orange or phenolphthalein
Weak acid + strong base (CH3COOH/NaOH\text{CH}_3\text{COOH} / \text{NaOH})≈9\approx 9 (basic)7 – 11Phenolphthalein only
Strong acid + weak base (HCl/NH3\text{HCl} / \text{NH}_3)≈5\approx 5 (acidic)3 – 7Methyl orange only
Weak acid + weak base≈7\approx 7No steep sectionNone suitable (use a calibrated pH meter)

Salt Hydrolysis Explains Equivalence pH

Why is the equivalence point for ethanoic acid with sodium hydroxide basic rather than neutral? The salt produced, sodium ethanoate, separates into ions. The ethanoate ion hydrolyses with water molecules:

CH3COO−+H2O⇌CH3COOH+OH−\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-

Generating hydroxide ions makes the final solution slightly alkaline (pH≈9\text{pH} \approx 9). Conversely, ammonium chloride (from strong acid + weak base) hydrolyses to release hydronium ions, leaving the solution acidic (pH≈5\text{pH} \approx 5):

NH4++H2O⇌NH3+H3O+\text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}_3\text{O}^+

Visualising the Titration Curve: Sketch an xyxy-graph with pH (0 to 14) on the vertical axis and volume of base added (cm3\text{cm}^3) on the horizontal axis. For a strong acid with a strong base, the curve starts low at pH 1, rises very gradually, shoots almost vertically between pH 3 and pH 11, and flattens out around pH 13. The equivalence point sits at the midpoint of that vertical rise at pH 7.

Mandatory Experiment: Volumetric Analysis and Primary Standards

Volumetric analysis determines unknown concentrations by measuring the precise volume of one solution required to react with a known volume of another.

Primary Standards

A primary standard is a substance available in a stable, highly pure form from which a standard solution of known concentration can be made up directly by dissolving a known mass in deionised water and making up to a known volume.

Four essential requirements:

  1. Highly pure state.
  2. Stable in air and solution (neither hygroscopic, deliquescent, nor efflorescent).
  3. High relative molecular mass (MrM_r), minimising proportional weighing errors.
  4. Readily soluble in water and reacts completely in a single, predictable reaction.

Examples include anhydrous sodium carbonate (Na2CO3\text{Na}_2\text{CO}_3) and ammonium iron(II) sulfate. Solid sodium hydroxide cannot serve as a primary standard because it is deliquescent and absorbs atmospheric CO2\text{CO}_2. Concentrated hydrochloric acid cannot serve because it is a volatile solution of uncertain concentration. Both must be standardised against a primary standard.

Mandatory Experiment Block: Acid-Base Titration

  • Purpose: We determine the concentration of an unknown solution, such as ethanoic acid in commercial vinegar, using a standardised solution of sodium hydroxide.
  • Apparatus: Retort stand, burette (50 cm350\text{ cm}^3), pipette (25.0 cm325.0\text{ cm}^3), pipette filler, conical flask, white tile, wash bottle with deionised water, volumetric flask (250 cm3250\text{ cm}^3 or 500 cm3500\text{ cm}^3), dropper, filter funnel.
  • Described Diagram: Retort stand clamped vertically to a burette filled with sodium hydroxide solution. The burette tap and delivery jet are fully filled with solution, free from air bubbles. Directly below rests a conical flask sitting on a white tile to ensure sharp visual detection of the end point. The flask contains 25.0 cm325.0\text{ cm}^3 of acid transferred via pipette, with 2 to 3 drops of indicator.
  • Rinsing Rules:
  • Burette: Rinse with deionised water, then with the solution it will contain.
  • Pipette: Rinse with deionised water, then with the solution it will contain.
  • Conical flask: Rinse with deionised water only (rinsing with acid would add extra moles).
  • Volumetric flask: Rinse with deionised water only.
  • Procedure:
  1. Pipette 25.0 cm325.0\text{ cm}^3 of acid into the conical flask; add 2–3 drops of phenolphthalein.
  2. Fill the burette using a funnel, remove the funnel, and open the tap to ensure the jet is filled with no trapped air bubble. Adjust the liquid so the bottom of the meniscus rests on zero at eye level.
  3. Titrate on a white tile with continuous swirling. Wash down flask walls with deionised water near the end point.
  4. Add base dropwise until a single drop causes a permanent colour transition (colourless to faint pink).
  5. Complete one rough titration followed by accurate repeats until two titres agree within 0.1 cm30.1\text{ cm}^3. Average the concordant titres.
  • Sources of Error: Trapped air bubble in burette jet, parallax error when reading the meniscus, overshooting the end point, adding excessive indicator, failure to swirl continuously.
  • Advantage of Diluting Concentrated Samples: Diluting vinegar ensures the required titre falls within the accurate working range of the burette (15–35 cm315\text{–}35\text{ cm}^3), keeping percentage reading errors small.

Mandatory Experiments: Heat of Neutralisation and Measuring pH

Heat of Neutralisation

The heat of neutralisation is the heat change when one mole of H+\text{H}^+ ions from an acid reacts with one mole of OH−\text{OH}^- ions from a base to form one mole of water, under standard conditions. For any strong acid reacting with any strong base, ΔH≈−57.1 kJ mol−1\Delta H \approx -57.1\text{ kJ mol}^{-1}.

Why is this value constant? Because a strong acid and a strong base both dissociate fully in aqueous solution, the actual chemical process taking place in every case is simply H++OH−→H2O\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}. The spectator ions undergo no change at all. When a weak acid is used, the enthalpy change is less exothermic because a portion of the heat released must be absorbed to drive the complete dissociation of the weak acid molecules.

Mandatory Experiment Block: Heat of Neutralisation

  • Purpose: Measure the heat of neutralisation of HCl\text{HCl} by NaOH\text{NaOH}.
  • Apparatus: Expanded polystyrene cup with lid, thermometer (0.1 ∘C0.1\text{ }^\circ\text{C} graduations) or temperature probe, two measuring cylinders (50 cm350\text{ cm}^3 or 100 cm3100\text{ cm}^3).
  • Procedure:
  1. Measure 50 cm350\text{ cm}^3 of 1.0 mol l−1 HCl1.0\text{ mol l}^{-1}\text{ HCl} into an expanded polystyrene cup; record its temperature.
  2. Measure 50 cm350\text{ cm}^3 of 1.0 mol l−1 NaOH1.0\text{ mol l}^{-1}\text{ NaOH}; ensure its temperature matches the acid (or record the average initial temperature).
  3. Add the base quickly to the acid in the cup, replace the lid, stir continuously, and record the maximum temperature achieved.
  4. Calculate ΔT=Tfinal−Tinitial\Delta T = T_{\text{final}} - T_{\text{initial}}.
  • Why Polystyrene: Expanded polystyrene provides excellent thermal insulation and has negligible heat capacity, minimising heat transfer to the surroundings compared to glass beakers.
  • Calculation Formula: q=mcΔTq = mc\Delta T, where mm is the combined mass of solution (taking density as 1.0 g cm−31.0\text{ g cm}^{-3}), and c=4.18 kJ kg−1 K−1c = 4.18\text{ kJ kg}^{-1}\text{ K}^{-1}. Divide qq by the moles of water formed to determine ΔH\Delta H in kJ mol−1\text{kJ mol}^{-1} (assigning a negative sign for exothermic change).
  • Sources of Error: Heat loss during mixing, slow thermometer response, inaccurate volume measurement.

Measuring pH of Household Substances

Universal indicator and pH papers supply quick, visual estimates by matching colours to a chart, but cannot give readings to decimal places and fail in bleaching or intensely coloured solutions (such as bleach). A digital pH meter provides continuous readings to two decimal places. Before use, the meter must be calibrated against standard buffer solutions of known pH (e.g. pH 4 and pH 7). Between readings, the glass electrode must be thoroughly rinsed with deionised water and blotted dry to avoid cross-contamination.

Environmental Chemistry: Acid Rain and Natural Water

Even clean, unpolluted rainwater is slightly acidic, showing a pH of around 5.6. This happens because carbon dioxide in the air dissolves in rain droplets to produce weak carbonic acid:

CO2+H2O⇌H2CO3\text{CO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{CO}_3

Rainwater with a pH below 5.6 is classified as acid rain, produced by atmospheric pollutants:

  • Sulfur Dioxide (SO2\text{SO}_2): When power stations and oil refineries burn fuels carrying sulfur impurities, they pump sulfur dioxide gas straight into the atmosphere—this is the specific pollutant examiners usually ask you to name. It dissolves in moisture to form sulfurous acid, which then oxidises to sulfuric acid:
SO2+H2O→H2SO32H2SO3+O2→2H2SO4\text{SO}_2 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_3 \qquad 2\text{H}_2\text{SO}_3 + \text{O}_2 \rightarrow 2\text{H}_2\text{SO}_4
  • Oxides of Nitrogen (NOx\text{NO}_x): Formed in internal combustion engines when high spark temperatures cause nitrogen and oxygen from the air to combine:
N2+O2→2NO2NO+O2→2NO2\text{N}_2 + \text{O}_2 \rightarrow 2\text{NO} \qquad 2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2

Nitrogen dioxide dissolves in rain droplets to produce nitric acid (HNO3\text{HNO}_3).

Environmental Impact and Remediation

Acid rain leaches vital plant nutrients (calcium, magnesium) from soil and releases dissolved aluminium ions (Al3+\text{Al}^{3+}) into waterways, which are toxic to aquatic organisms. It also attacks limestone buildings and monuments through acid-carbonate decomposition:

CaCO3+H2SO4→CaSO4+H2O+CO2\text{CaCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + \text{H}_2\text{O} + \text{CO}_2

Acidity in damaged lakes and farmland is treated by spreading agricultural lime (CaCO3\text{CaCO}_3 or Ca(OH)2\text{Ca(OH)}_2) to restore neutral conditions. Vehicle emissions are limited by fitting catalytic converters to convert nitrogen monoxide into harmless nitrogen gas.

Cross-Curricular Synthesis

How do these separate acid-base ideas tie into the rest of your exam questions? The constants KwK_w and KaK_a are dynamic equilibrium constants governed by Le Chatelier's principle, while the degree of acid dissociation directly dictates reaction rates with metals by setting the concentration of reacting ions. Meanwhile, neutralisation gives off heat of reaction, and every volumetric calculation you do comes right back to the mole concept.

Key terms

Acid (Arrhenius)
A substance that dissociates in water to produce hydrogen ions (H⁺).
Base (Arrhenius)
A substance that dissociates in water to produce hydroxide ions (OH⁻).
Acid (Brønsted–Lowry)
A substance that acts as a proton donor.
Base (Brønsted–Lowry)
A substance that acts as a proton acceptor.
Lewis Acid (HL)
A substance that acts as an electron pair acceptor.
Lewis Base (HL)
A substance that acts as an electron pair donor.
Conjugate acid-base pair
Two species that differ by a single proton.
Amphoteric substance
A substance that can act as either an acid or a base.
Strong acid
An acid that is completely (fully) dissociated into ions in aqueous solution.
Weak acid
An acid that is only partially dissociated into ions in aqueous solution.
Ionic product of water (Kw)
The equilibrium constant for the self-ionisation of water, Kw = [H₃O⁺][OH⁻], equal to 1.0 × 10⁻¹⁴ mol² l⁻² at 25 °C.
pH
The negative logarithm to the base 10 of the hydronium ion concentration in moles per litre: pH = -log₁₀[H₃O⁺].
Primary standard
A substance available in a pure, stable state from which a standard solution can be made up directly by dissolving a known mass in deionised water to a known volume.
Standard solution
A solution whose concentration is accurately known.
Equivalence point
The point in a titration at which the acid and base have reacted in exact stoichiometric proportions according to the balanced equation.
End point
The point in a titration where the indicator changes colour.
Heat of neutralisation
The heat change when one mole of H⁺ ions from an acid reacts with one mole of OH⁻ ions from a base to form one mole of water, under standard conditions.
% (w/v)
The mass of solute in grams dissolved per 100 cm³ of solution.

Check yourself

  1. What is the Arrhenius definition of a base?

    A substance that dissociates in water to produce hydroxide ions (OH⁻).

  2. State one limitation of the Arrhenius theory.

    It applies only to aqueous solutions and fails to explain the basic properties of substances without OH groups, such as NH₃.

  3. Identify both conjugate acid-base pairs in: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻.

    Pair 1: NH₄⁺ / NH₃; Pair 2: H₂O / OH⁻.

  4. Why is the conical flask rinsed only with deionised water during a titration?

    Rinsing with the solution it will contain leaves extra droplets of solute on the walls, introducing extra moles and ruining accuracy.

  5. Name a suitable indicator for titrating ethanoic acid against sodium hydroxide, and state its end-point colour change.

    Phenolphthalein; colourless to permanent faint pink.

  6. What is the pH range and colour change of methyl orange?

    pH 3.1 to 4.4; changes from red in acid to yellow in base.

  7. List two requirements of a primary standard.

    Must be available in a highly pure state, stable in air and solution, have a high relative molecular mass, and be water-soluble.

  8. Why is the heat of neutralisation constant (-57.1 kJ mol⁻¹) for any strong acid with any strong base?

    Both are fully dissociated, so the only reaction is H⁺ + OH⁻ → H₂O; spectator ions do not participate.

  9. Calculate the pH of a 0.05 mol l⁻¹ solution of sulfuric acid (H₂SO₄).

    pH = 1.00 (since [H₃O⁺] = 2 × 0.05 = 0.10 mol l⁻¹, and -log₁₀(0.10) = 1.00).

  10. What is the pH of unpolluted rainwater, and what chemical causes its acidity?

    pH ≈ 5.6; caused by dissolved carbon dioxide forming carbonic acid (H₂CO₃).

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