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Chemistry of Aqueous Solutions

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Chemistry of Aqueous Solutions

Calculate acidity, follow titrations and buffers, and predict whether an ionic solid dissolves or precipitates.

A-Level 9476 (2026-2027)

01

Strength is the extent of ionisation, not the amount dissolved

A dilute strong acid and a concentrated weak acid are different ideas.

A strong acid is essentially fully ionised in dilute aqueous solution; a weak acid establishes an equilibrium with only a fraction ionised. For a weak monoprotic acid, HA(aq) ⇌ H+(aq) + A-(aq). A strong base such as dissolved NaOH provides hydroxide essentially completely; a weak base such as NH3 reacts only partly with water: NH3 + H2O ⇌ NH4+ + OH-.

Keep three independent descriptions separate
DescriptionMeaningExample distinction
Strong or weakExtent of acid/base ionisation or reaction with water.HCl is strong; ethanoic acid is weak.
Concentrated or diluteAmount of dissolved substance per volume.A weak acid can be concentrated.
Soluble or sparingly solubleHow much substance can dissolve under the conditions.A sparingly soluble hydroxide is not automatically a weak base.

At the same analytical concentration, a weak monoprotic acid usually gives a lower hydrogen-ion concentration than a strong monoprotic acid. Dilution can increase the fraction of weak acid ionised even while the absolute hydrogen-ion concentration decreases. Degree of ionisation and [H+] must not be treated as the same quantity.

02

Constants connect an acid to its conjugate base

Ka and Kb quantify equilibrium; p values are their negative base-10 logarithms.

The definitions used in calculations
QuantityDefinition and use
pH-log10[H+], using the usual dilute-solution concentration convention.
Ka for HA[H+][A-]/[HA] for HA ⇌ H+ + A-. A larger Ka means a stronger acid.
pKa-log10 Ka. A smaller pKa means a stronger acid.
Kb for B[BH+][OH-]/[B] for B + H2O ⇌ BH+ + OH-. A larger Kb means a stronger base.
pKb-log10 Kb. A smaller pKb means a stronger base.
Kw[H+][OH-], the ionic product of water at the stated temperature.

At 298 K, Kw is approximately 1.00 × 10-14 mol2 dm-6, so pH + pOH = 14.00 under the usual concentration convention. Kw changes with temperature. A neutral solution always has [H+] = [OH-], but neutral pH is exactly 7.00 only when pKw = 14.00.

For a conjugate pair HA/A-, Ka(HA)Kb(A-) = Kw, hence pKa + pKb = pKw. The species must be a conjugate pair. Multiplying the acid constant of ethanoic acid by the base constant of ammonia does not give Kw.

Worked example

Find the strength of a conjugate base

At 298 K, ethanoic acid has Ka = 1.80 × 10-5. Find Kb for ethanoate.

  1. Ethanoate is the conjugate base of ethanoic acid.
  2. Kb = Kw/Ka = (1.00 × 10-14)/(1.80 × 10-5).
Answer

Kb = 5.56 × 10-10. Ethanoate is a weak base and its aqueous salt solution is alkaline.

03

Choose the chemical model before taking a logarithm

Strong substances use stoichiometry; weak substances use an equilibrium approximation.

Useful dilute-solution models
SolutionFirst findThen calculate
Strong monoprotic acid at concentration c[H+] approximately c.pH = -log10[H+].
Strong base[OH-] from complete dissociation and the formula stoichiometry.pOH = -log10[OH-]; pH = pKw - pOH.
Weak monoprotic acid HA at concentration c[H+] approximately sqrt(Ka c), if ionisation is small and water contribution negligible.pH from [H+]; check [H+]/c is small.
Weak monoacidic base B at concentration c[OH-] approximately sqrt(Kb c), under corresponding assumptions.pOH, then pH using pKw.

The weak-acid approximation comes from Ka = x2/(c - x), with x = [H+] = [A-]. If x is small compared with c, replace c - x with c. A small computed ionised fraction supports the approximation. If it is not small, the approximation is not justified; the required syllabus problems will not demand solving a quadratic.

Worked example

Compare four solutions at 298 K

Find pH for 0.0200 mol dm-3 HNO3; 0.00500 mol dm-3 dissolved Ca(OH)2; 0.100 mol dm-3 HA with Ka = 1.80 × 10-5; and 0.100 mol dm-3 NH3 with Kb = 1.80 × 10-5.

  1. HNO3: [H+] = 0.0200; pH = 1.70.
  2. Ca(OH)2: [OH-] = 2(0.00500) = 0.0100; pOH = 2.00 and pH = 12.00.
  3. HA: [H+] = sqrt(1.80 × 10-5 × 0.100) = 1.34 × 10-3; pH = 2.87. The ionised fraction is 1.34%, supporting the approximation.
  4. NH3: [OH-] = 1.34 × 10-3; pOH = 2.87, so pH = 11.13.
Answer

The pH values are 1.70, 12.00, 2.87 and 11.13 respectively. The calcium hydroxide calculation concerns the stated dissolved concentration, not an unlimited solubility assumption.

04

Different regions of a titration are controlled by different species

Equivalence means stoichiometric reaction, not necessarily pH 7.

Strong and weak acids titrated with strong base

Both curves reach equivalence at 25 cubic centimetres of sodium hydroxide. Strong acid begins near pH 1 and has equivalence pH 7. Weak acid begins near pH 2.87, has a buffer region and has an alkaline equivalence near pH 8.72. Beyond equivalence the curves approach one another because excess hydroxide controls pH.

Calculated ideal dilute-solution curves at 298 K: 25.0 cm3 of 0.100 mol dm-3 monoprotic acid, titrated with 0.100 mol dm-3 NaOH. Weak-acid Ka = 1.80 × 10-5. Curves use charge balance; students do not need to solve the numerical model to interpret them.
Follow a weak acid with strong base through four regions
RegionMain chemistryWhat controls pH
Before any baseWeak acid partly ionises.Ka and initial acid concentration.
Before equivalenceAdded OH- converts HA into A-. Both HA and A- remain.The acid/conjugate-base buffer ratio. At half-equivalence, pH = pKa.
At equivalenceAll initial HA has been converted stoichiometrically into A-.Hydrolysis of the weak conjugate base: A- + H2O ⇌ HA + OH-. pH is above 7 at 298 K.
After equivalenceStrong base is in excess.Excess OH- divided by total solution volume.
The four acid-base strength combinations
Titration pairEquivalence at 298 KCurve feature
Strong acid + strong baseApproximately pH 7.Large, steep pH change around equivalence.
Weak acid + strong baseAbove pH 7, from conjugate-base hydrolysis.A buffer region before equivalence; lower-pH part of the jump is shortened.
Strong acid + weak baseBelow pH 7, from conjugate-acid dissociation.When acid is added to weak base, pH falls through a base/conjugate-acid buffer region.
Weak acid + weak baseDepends on relative Ka and Kb, not automatically 7.Usually no large sharp jump, making a visual indicator endpoint unreliable.

Worked example

After equivalence, count the excess

30.0 cm3 of 0.100 mol dm-3 NaOH is added to 25.0 cm3 of 0.100 mol dm-3 monoprotic acid. Find pH at 298 K when strong-base excess controls the result.

  1. Initial acid = 0.00250 mol; added OH- = 0.00300 mol.
  2. Excess OH- = 0.000500 mol in 0.0550 dm3.
  3. [OH-] = 0.00909 mol dm-3; pOH = 2.04.
Answer

pH = 11.96. Using only the base volume in the denominator would overestimate [OH-].

05

Choose an indicator whose transition sits inside the steep change

An endpoint should closely match equivalence for the actual titration curve.

An acid-base indicator is usually a weak acid or base whose two forms have different colours. Its visible colour change occurs over a pH interval, commonly around pKa ± 1. The endpoint is the observed colour change; the equivalence point is the stoichiometric point. They should be close, but they are not definitions of the same event.

Use the supplied transition interval and curve
SituationSuitable reasoning
Strong acid/strong base with a broad steep jumpSeveral indicators may work if their entire useful transition interval lies in the steep region.
Weak acid/strong baseAn alkaline-range indicator such as phenolphthalein, roughly pH 8.2-10.0, commonly matches the steep region; methyl orange, roughly pH 3.1-4.4, changes in the buffer region too early.
Strong acid/weak baseChoose an indicator in the acidic steep region using the actual curve; an alkaline-range indicator usually changes too early when acid is added.
Weak acid/weak baseA broad gradual transition makes visual endpoint selection poor; a pH meter and suitable analysis may be preferable.

A transition interval does not have to be centred at pH 7. The useful question is how much titrant volume changes during the indicator transition. In a steep region this volume is small, so the endpoint error is small. If given unfamiliar indicators, compare their ranges directly with the curve rather than choosing a familiar name.

06

A buffer contains a reservoir for both added acid and added base

A weak acid/conjugate-base pair or weak base/conjugate-acid pair limits small pH changes.

An acidic buffer contains appreciable amounts of a weak acid HA and its conjugate base A-, often made by mixing the acid with a soluble salt or partially neutralising the acid. Added H+ is consumed by A-; added OH- is consumed by HA. An alkaline buffer similarly contains B and BH+, such as NH3/NH4+.

Follow the added species, not just an equilibrium arrow
AdditionAcidic buffer reactionAlkaline ammonia buffer reaction
A small amount of acidA- + H+ → HANH3 + H+ → NH4+
A small amount of baseHA + OH- → A- + H2ONH4+ + OH- → NH3 + H2O

These reactions prevent the added strong acid or base from remaining freely in solution. The component ratio changes only modestly when the addition is small relative to both reservoirs, so pH changes modestly. A buffer does not keep pH perfectly fixed, and it fails when one component is substantially exhausted.

Buffers are used to maintain suitable pH for enzyme reactions, chemical measurements and calibration. Diluting both components by the same factor leaves their ratio approximately unchanged, so ideal buffer pH changes little, but the smaller amounts per volume give lower buffer capacity. At extreme dilution, the simple approximation also breaks down.

07

Neutralise the added reagent first, then calculate the new ratio

The equilibrium formula uses the remaining buffer components.

Rearrange Ka = [H+][A-]/[HA]: pH = pKa + log10([A-]/[HA]). When both components share a solution volume, their mole ratio gives the same ratio. This buffer approximation uses the analytical component amounts after any strong acid/base reaction, provided equilibrium ionisation changes are small relative to those amounts.

Worked example

A buffer before and after added acid

100.0 cm3 of buffer contains 0.0200 mol HA and 0.0300 mol A-. Ka = 1.80 × 10-5. Find its initial pH and its pH after adding 0.00500 mol HCl, with volume effects common to both components.

  1. pKa = -log10(1.80 × 10-5) = 4.7447.
  2. Initial pH = 4.7447 + log10(0.0300/0.0200) = 4.92.
  3. Added H+ reacts with A-: new A- amount = 0.0250 mol and new HA amount = 0.0250 mol.
  4. The new ratio is 1, so pH = pKa.
Answer

pH changes from 4.92 to 4.74. Substituting the added HCl concentration directly into -log[H+] would ignore its reaction with the buffer.

For a weak base B and conjugate acid BH+, either use the acid constant of BH+ in the same equation or use pOH = pKb + log10([BH+]/[B]), followed by pH = pKw - pOH. Equal amounts of NH3 and NH4+ with Kb = 1.80 × 10-5 give pH about 9.26 at 298 K.

Check your understandingCan the buffer equation be used after adding 0.0400 mol HCl to the original buffer above?Think it through, then reveal the answer
Not with the original buffer model. Only 0.0300 mol A- is available, so it is exhausted and 0.0100 mol strong acid remains in excess. First determine the final volume and excess acid concentration; there is no substantial A- reservoir left.
08

Carbonate buffering limits acidification but cannot stop unlimited carbon dioxide input

More dissolved CO2 consumes carbonate and changes the balance of the ocean carbon system.

Follow added carbon dioxide into the carbonate system
  1. CO2 enters the water

    CO2(g) ⇌ CO2(aq). Dissolved CO2 participates in acid-base equilibria; the simplified combined step is CO2(aq) + H2O(l) ⇌ H+(aq) + HCO3-(aq).

  2. Carbonate consumes some added acid

    CO32-(aq) + H+(aq) → HCO3-(aq). This limits the rise in free hydrogen-ion concentration.

  3. The buffer balance changes

    The carbonate/hydrogencarbonate ratio falls, so the buffered pH falls. Overall: CO2 + CO32- + H2O → 2HCO3-.

The pair HCO3-/CO32- is an acid/conjugate-base buffer: carbonate accepts H+, while hydrogencarbonate can consume OH- to form carbonate and water. Ocean water contains other buffering species too, but this pair links increasing atmospheric CO2 directly to carbonate availability.

The rapid atmospheric CO2 increase drives additional uptake by surface waters. Buffering consumes carbonate as it resists the added acidity, so it has finite capacity and the equilibrium pH decreases. Ocean acidification means a fall in pH; it does not require seawater to become acidic below pH 7. Reduced carbonate availability also affects the equilibria involved in forming calcium carbonate shells and skeletons.

09

A saturated solution has a fixed product of free-ion concentrations

Ksp describes an equilibrium with an undissolved solid at a specified temperature.

For AgCl(s) ⇌ Ag+(aq) + Cl-(aq), Ksp = [Ag+][Cl-]. The pure solid is omitted. For CaF2(s) ⇌ Ca2+(aq) + 2F-(aq), Ksp = [Ca2+][F-]2. The exponent follows the dissolution coefficient.

Worked example

Convert Ksp into molar solubility

Find the molar solubility of AgCl with Ksp = 1.80 × 10-10, and CaF2 with Ksp = 3.20 × 10-11, in pure water under the simple dissolution model.

  1. For AgCl, let solubility be s. Both ion concentrations are s, so s2 = 1.80 × 10-10 and s = 1.34 × 10-5 mol dm-3.
  2. For CaF2, [Ca2+] = s and [F-] = 2s, so Ksp = s(2s)2 = 4s3.
  3. s = cube root(3.20 × 10-11 / 4) = 2.00 × 10-4 mol dm-3.
Answer

AgCl: 1.34 × 10-5 mol dm-3; CaF2: 2.00 × 10-4 mol dm-3. Comparing raw Ksp values alone is not a valid solubility ranking for different stoichiometries.

The ionic product Q has the same form as Ksp but uses the current free-ion concentrations. Q < Ksp means unsaturated; Q = Ksp means equilibrium saturation; Q > Ksp makes precipitation thermodynamically favoured. When mixing solutions, first calculate concentrations in the combined volume.

Worked example

Predict precipitation after mixing

Mix 10.0 cm3 of 0.00200 mol dm-3 AgNO3 with 10.0 cm3 of 0.00100 mol dm-3 NaCl. Does AgCl precipitate if Ksp = 1.80 × 10-10?

  1. Before precipitation, dilution gives [Ag+] = 0.00100 and [Cl-] = 0.000500 mol dm-3.
  2. Q = 0.00100 × 0.000500 = 5.00 × 10-7.
  3. This exceeds Ksp by a large factor; ions are removed into the solid until the equilibrium condition is approached.
Answer

AgCl precipitates. These diluted concentrations test the initial tendency; they are not the final dissolved concentrations after precipitation.

10

A common ion and a ligand pull the dissolution equilibrium in opposite directions

Ksp stays fixed at fixed temperature, but the free-ion concentrations respond.

Adding Cl- to saturated AgCl solution increases a product concentration and favours precipitation, reducing the salt's solubility. If a 0.0100 mol dm-3 chloride background dominates the tiny dissolved contribution, [Ag+] ≈ Ksp/[Cl-] = 1.80 × 10-8 mol dm-3. The approximation must be checked when the background common-ion amount is not large.

Ammonia binds free silver ions: Ag+(aq) + 2NH3(aq) ⇌ [Ag(NH3)2]+(aq). Removing free Ag+ lowers [Ag+][Cl-] below Ksp, so more AgCl dissolves. The total dissolved silver can increase while the free Ag+ concentration remains small. Ksp uses the free ion, not the sum of free and complexed silver.

The silver-ion and ammonia tests for halides
HalideWith aqueous AgNO3 after acidifying with dilute HNO3With aqueous ammonia
Cl-White AgCl precipitate.Dissolves in dilute ammonia.
Br-Cream AgBr precipitate.Does not readily dissolve in dilute ammonia; dissolves in concentrated ammonia.
I-Yellow AgI precipitate.Remains insoluble in concentrated ammonia under the test conditions.

Nitric acid removes interfering carbonate or hydroxide without introducing a halide. Hydrochloric acid would introduce chloride and give a false positive. The ammonia trend reflects the balance between the very small solubility products and complex formation, rather than a claim that ammonia breaks every silver-halide lattice equally easily.

Check your understandingWhy can acidifying a clear silver-ammonia solution make AgCl reappear when chloride is still present?Think it through, then reveal the answer
H+ converts NH3 to NH4+, reducing the free ligand concentration. The silver-ammonia complex dissociates, free Ag+ rises and the ionic product can exceed Ksp, so AgCl precipitates again.

Quick revision

Revisit the essentials, then return to an explanation when you need it.

Choose the pH calculation
  1. Has acid reacted with base?

    Do the mole balance and dilution first.

  2. Identify what remains

    Strong excess: concentration directly. Weak acid/base alone: small-dissociation model. Conjugate pair: buffer ratio.

  3. Check conditions

    Use the stated temperature, correct Kw, justified approximation and relevant indicator range.

Buffers resist small changes by consuming the addition; they have finite capacity. Ocean CO2 uptake consumes carbonate and lowers pH. For solubility, compare the free-ion product with Ksp: a common ion suppresses dissolution, while complexing a dissolved ion can promote it.

Scope and references

Learning outcomes and sources

10. Chemistry of Aqueous Solutions. Use the outcome map to find the explanation for a particular syllabus requirement.

See the learning outcome map
  1. 10.1(a) Explain strong/weak acid and base behaviour.

    • Extent of dissociation or reaction with water
    • Strength distinguished from concentration and solubility

    Strength is the extent of ionisation, not the amount dissolved

  2. 10.1(b) Define and calculate with acid-base equilibrium quantities.

    • pH, Ka, pKa, Kb, pKb, Kw
    • Kw = Ka Kb for a conjugate pair
    • Temperature dependence of Kw

    Constants connect an acid to its conjugate baseChoose the chemical model before taking a logarithm

  3. 10.1(c) Calculate hydrogen-ion concentration and pH.

    • Strong acids
    • Weak monobasic/monoprotic acids
    • Strong bases
    • Weak monoacidic bases
    • No quadratic solving required

    Choose the chemical model before taking a logarithm

  4. 10.1(d) Explain acid-base titration pH changes.

    • Strong and weak acid/base combinations
    • Buffer, equivalence and excess regions
    • Hydrolysis at equivalence

    Different regions of a titration are controlled by different species

  5. 10.1(e) Choose indicators from supplied titration data.

    • Transition interval within the steep pH change
    • Endpoint versus equivalence
    • Unfamiliar indicator ranges

    Choose an indicator whose transition sits inside the steep change

  6. 10.1(f) Explain buffer action and applications.

    • (i) Control of pH after small acid/base additions
    • (ii) Uses including ocean CO3^2-/HCO3- buffering
    • Rapid atmospheric CO2 increase and ocean acidification
    • Finite buffer capacity

    A buffer contains a reservoir for both added acid and added baseCarbonate buffering limits acidification but cannot stop unlimited carbon dioxide input

  7. 10.1(g) Calculate buffer pH.

    • Acidic and alkaline buffer ratios
    • Strong acid/base neutralisation before equilibrium calculation

    Neutralise the added reagent first, then calculate the new ratio

  8. 10.2(a) Understand and apply the solubility-product concept.

    • Saturated equilibrium with solid
    • Free-ion product and precipitation criterion

    A saturated solution has a fixed product of free-ion concentrations

  9. 10.2(b) Calculate Ksp from concentrations and vice versa.

    • Dissolution stoichiometry
    • Molar solubility
    • Dilution on mixing

    A saturated solution has a fixed product of free-ion concentrations

  10. 10.2(c) Explain changes in ionic-salt solubility.

    • (i) Common-ion effect
    • (ii) Complex formation
    • Halide ions with aqueous silver ions then aqueous ammonia
    • Free versus total dissolved ion

    A common ion and a ligand pull the dissolution equilibrium in opposite directions

  • SEAB H2 Chemistry 9476, examination 2026

    Topic 10, printed pages 22-23. Seven acid-base and three solubility outcomes, including the current ocean-buffer requirement and all silver-halide/ammonia observations, inspected.