Reactivity 3.1 Proton Transfer Reactions

IB DP CHEMISTRY Reactivity 3: What Are the Mechanisms of Chemical Change? Reactivity 3.1 Proton Transfer Reactions

Revision Notes · Standard and Higher Level Fahad H. Ahmad

+92 323 509 4443 | Megalecture.com

Original notes prepared for the IB Diploma Programme Chemistry course (2025 syllabus)

What the syllabus requires

Reactivity 3.1 treats acid-base chemistry as a family of proton transfer reactions. Use this checklist as a final revision map; items marked (HL) are assessed at Higher Level only.

Understanding You should be able to...

Brønsted–Lowry theory Define acids as proton (H+) donors and bases as proton acceptors; deduce conjugate acid–base pairs.

Amphiprotic species Recognise species such as water and HCO3

− that can donate or accept a proton.

Reactions of acids Write and balance equations for acids with metals, bases (oxides/hydroxides), carbonates and hydrogencarbonates; name the salts.

Strong and weak Distinguish strength (degree of dissociation) from concentration, and describe experiments that tell them apart.

The pH scale Use pH = −log[H+], the ion product Kw, and pH + pOH = 14 to find the pH of strong acids and bases.

(HL) Weak-acid equilibria Use Ka, Kb, pKa, pKb and KaKb = Kw to calculate the pH of weak acids.

(HL) Buffers Explain buffer action and calculate pH using the Henderson–Hasselbalch relationship.

(HL) pH curves Interpret titration curves, locate equivalence and half-equivalence points, and choose a suitable indicator.

Exam note: Sections 1–4 and 8 are common to SL and HL. Sections 5–7 (Ka/Kb calculations, buffers and titration curves) are HL only and are a rich source of multi-step exam questions.

1. Brønsted–Lowry acids and bases

The Brønsted–Lowry model defines acids and bases by what happens to a proton — a hydrogen ion, H+

— during reaction:

• An acid is a proton donor.
• A base is a proton acceptor.

Because a bare proton cannot exist in solution, an acid can only donate a proton if a base is present to accept it. Proton transfer is therefore always a competition between two bases for the proton. In water the accepted proton attaches to a water molecule to form the oxonium (hydronium) ion:

HCl + H2O → H3O+ + Cl−

The formula H+(aq) is a convenient shorthand for H3O+(aq); both are acceptable in the examination.

Conjugate acid–base pairs

When an acid donates its proton it becomes a species that can accept one back — its conjugate base. When a base accepts a proton it becomes its conjugate acid. The two members of a conjugate pair differ by exactly one H+. Consider ammonia dissolving in water:

NH3 + H2O ⇌ NH4

+ + OH−

Here NH3 accepts a proton (base) to give its conjugate acid NH4

+, while H2O donates a proton (acid) to give its conjugate base OH−. The two pairs are NH4

+/NH3 and H2O/OH−.

Acid Conjugate base Base Conjugate acid

HCl Cl− NH3 NH4

+

HNO3 NO3

− OH− H2O

H2SO4 HSO4

− CO3

2− HCO3

−

CH3COOH CH3COO− H2O H3O+

H3O+ H2O NH3 NH4

+

Spotting a pair: Two species form a conjugate pair only if you can turn one into the other by adding or removing a single H+. The charge always changes by +1 going from base to conjugate acid. NH3 and OH− are not a pair — they differ by more than one proton.

Amphiprotic and amphoteric species

A species that can either donate or accept a proton (depending on its reaction partner) is called amphiprotic. Water is the classic example: it acts as an acid towards ammonia but as a base towards hydrogen chloride. The hydrogencarbonate ion is another:

As an acid: HCO3

− + OH− → CO3

2− + H2O

As a base: HCO3

− + H3O+ → H2CO3 + H2O

Other amphiprotic species you should know include HSO4

−, H2PO4

− and HPO4

2−, and amino acids (which carry both –COOH and –NH2 groups).

Amphiprotic vs amphoteric: Amphoteric is the wider term: a substance that reacts with both acids and bases (e.g. Al2O3, ZnO). Amphiprotic is the narrower, proton-specific case — it must donate or accept H+. Every amphiprotic species is amphoteric, but not the reverse.

Worked example 1 — identifying conjugate pairs

In the reaction HCO3

− + H2O ⇌ H2CO3 + OH−, identify each acid, each base, and the two conjugate pairs.

HCO3

− gains a proton, so it acts as a base; its conjugate acid is H2CO3.

H2O loses a proton, so it acts as an acid; its conjugate base is OH−.

Conjugate pairs: H2CO3 / HCO3

− and H2O / OH−. Because HCO3

− behaves as a base here but as an acid in the previous equation, it is amphiprotic.

2. Reactions of acids

Acids undergo four characteristic reactions that all produce a salt — an ionic compound in which the acid's replaceable hydrogen has been swapped for a metal (or ammonium) ion. Learn the general pattern first, then the examples.

Reactant General equation Products

Reactive metal acid + metal → salt + hydrogen salt + H2

Metal oxide (base) acid + metal oxide → salt + water salt + H2O

Metal hydroxide (base/alkali) acid + hydroxide → salt + water salt + H2O

Carbonate acid + carbonate → salt + water + carbon dioxide salt + H2O + CO2

Hydrogencarbonate acid + hydrogencarbonate → salt + water + carbon dioxide salt + H2O + CO2

Worked balanced examples

• Metal: Mg + 2HCl → MgCl2 + H2
• Metal oxide: CuO + H2SO4 → CuSO4 + H2O
• Hydroxide (alkali): NaOH + HCl → NaCl + H2O
• Carbonate: CaCO3 + 2HCl → CaCl2 + H2O + CO2
• Hydrogencarbonate: NaHCO3 + HCl → NaCl + H2O + CO2

Neutralisation

When an acid reacts with a hydroxide, the reaction that actually occurs is proton transfer from H3O+ to OH−. Cancelling the spectator ions leaves the same net ionic equation every time — this is the essence of neutralisation:

H+(aq) + OH−(aq) → H2O(l) ∆H ≈ −57 kJ mol−1

The enthalpy of neutralisation is almost constant (≈ −57 kJ mol−1) for any strong acid with any strong base, precisely because the underlying reaction is always H+ + OH− → H2O. With weak acids or bases it is a little less exothermic, because some energy is used to complete their dissociation.

Salt naming: The salt takes its name from the metal and the acid: HCl gives chlorides, HNO3 gives nitrates, H2SO4 gives sulfates, and CH3COOH gives ethanoates.

Titration stoichiometry

Because neutralisation reacts in fixed mole ratios, a titration lets you find an unknown concentration. Work in moles: n = c × V, match the ratio from the balanced equation, then convert back.

Worked example — finding an unknown concentration

In a titration, 25.0 cm3 of sodium hydroxide is exactly neutralised by 20.0 cm3 of 0.100 mol dm−3 sulfuric acid. Find the concentration of the NaOH.

Equation: 2NaOH + H2SO4 → Na2SO4 + 2H2O (ratio NaOH : H2SO4 = 2 : 1).

n(H2SO4) = 0.100 × (20.0/1000) = 2.00 × 10−3 mol. n(NaOH) = 2 × 2.00 × 10−3 = 4.00 × 10−3 mol.

c(NaOH) = n / V = (4.00 × 10−3) / (25.0/1000) = 0.160 mol dm−3.

3. Strong and weak acids and bases

Strength measures the degree of dissociation (ionisation) of an acid or base in water — the fraction of molecules that actually transfer a proton. It is a completely separate idea from concentration, which measures how much solute is dissolved per unit volume.

Strong Weak

Dissociation essentially complete (~100%) only partial (often <5%)

Equation one-way arrow, → equilibrium arrow, ⇌

Example acids HCl, HNO3, H2SO4 CH3COOH (ethanoic), carbonic, most organic acids

Example bases NaOH, KOH, Ba(OH)2 NH3 (ammonia), amines

Compare the two representative equilibria at equal concentration:

Strong: HCl(aq) → H+(aq) + Cl−(aq) Weak: CH3COOH(aq) ⇌ H+(aq) + CH3COO−(aq)

In the weak acid, the position of equilibrium lies well to the left, so [H+] is far lower than the acid concentration. A 0.10 mol dm−3 solution of HCl has [H+] = 0.10, but the same concentration of ethanoic acid has [H+] ≈ 0.0013 mol dm−3.

Distinguishing strong from weak experimentally

Given two acids of the same concentration (this control is essential), a stronger acid gives a higher [H+] and therefore:

Measurement Strong acid Weak acid Why pH lower higher more dissociation → higher [H+]

Electrical conductivity higher lower more ions carry more current

Rate with Mg / carbonate faster fizzing slower fizzing higher [H+] → faster reaction

Vol. of alkali to neutralise same same depends on amount (moles), not strength

Key distinction: The final volume of alkali needed to neutralise a fixed amount of acid is the same for a strong and a weak monoprotic acid — as the weak acid is neutralised, its equilibrium shifts right and eventually all the acidic hydrogen reacts. Strength changes the pH and rate, not the stoichiometry.

Monoprotic, diprotic and polyprotic acids

Acids differ in how many protons each molecule can donate. This affects the stoichiometry of neutralisation and the [H+] produced.

Type Protons donated Examples Neutralisation with NaOH

Monoprotic HCl, HNO3, CH3COOH 1 : 1

Diprotic H2SO4, H2CO3 1 : 2

Triprotic H3PO4 1 : 3

Polyprotic acids lose their protons in successive steps, each with its own Ka; the first proton is always the most readily lost (Ka1 > Ka2 > Ka3).

Worked example — percentage dissociation

A 0.20 mol dm−3 solution of a weak acid has [H+] = 2.0 × 10−3 mol dm−3. Find the percentage dissociation and comment on the acid's strength.

% dissociation = ([H+] dissociated / initial concentration) × 100

= (2.0 × 10−3 / 0.20) × 100 = 1.0%.

Only 1% of molecules have ionised, confirming this is a weak acid. A strong acid at the same concentration would be ~100% dissociated, giving [H+] = 0.20.

4. The pH scale

The concentration of hydrogen ions in aqueous solution spans many orders of magnitude, so it is compressed onto a logarithmic scale:

pH = −log10[H+] [H+] = 10−pH

A fall of one pH unit corresponds to a ten-fold rise in [H+]. Most familiar solutions lie between pH 0 and 14, but the scale is not formally bounded; concentrated strong acids can give negative pH values.

Figure 4.1. The pH scale (0−14). pH = −log[H+], so a lower pH means a higher [H+]; each unit is a ten−fold change. For example pH 2 gives [H+] = 0.01 mol dm−3 and pH 11 gives 1×10−11 mol dm−3.

The ionic product of water

Water itself is very slightly ionised (self-ionisation):

2H2O(l) ⇌ H3O+(aq) + OH−(aq)

The equilibrium is described by the ionic product of water:

Kw = [H+][OH−] = 1.0 × 10−14 (at 298 K)

Taking negative logs of the whole expression gives the vital relationship between pH and pOH (where pOH = −log[OH−]):

pH + pOH = 14.00 (at 298 K)

In pure water [H+] = [OH−] = 1.0 × 10−7 mol dm−3, giving pH = 7.00 (neutral). Acidic solutions have [H+] > [OH−] and pH < 7; basic solutions the reverse.

Kw depends on temperature: Self-ionisation is endothermic, so Kw increases as temperature rises. At 50 °C, Kw ≈ 5.5 × 10−14 and neutral pH falls to about 6.6. The water is still neutral — [H+] = [OH−] — even though pH is below 7. Neutral does not always mean pH 7.

pH of strong acids

A strong monoprotic acid is fully dissociated, so [H+] equals the acid concentration directly. For a diprotic strong acid such as H2SO4, each mole releases two moles of H+.

Worked example 2 — pH of strong acids

Find the pH of (a) 0.010 mol dm−3 HCl and (b) 0.0050 mol dm−3 H2SO4 (assume complete dissociation of both protons).

• [H+] = 0.010 mol dm−3. pH = −log(0.010) = 2.00.
• Each H2SO4 gives 2H+, so [H+] = 2 × 0.0050 = 0.010 mol dm−3. pH = −log(0.010) = 2.00.

Both give the same pH: the diprotic acid is half the concentration but releases twice the protons.

Worked example — finding [H+] from pH, and dilution

• A solution has pH 3.40. Find [H+]. (b) 25.0 cm3 of this solution is diluted to 250 cm3. Find the new pH (assume a strong acid).
• [H+] = 10−pH = 10−3.40 = 4.0 × 10−4 mol dm−3.
• Dilution by a factor of 10 divides [H+] by 10: new [H+] = 4.0 × 10−5 mol dm−3.

New pH = −log(4.0 × 10−5) = 4.40. A ten-fold dilution raises the pH of a strong acid by exactly 1 unit.

pH of strong bases

For a strong base, first find [OH−], then convert. Two routes: either use pOH and pH + pOH = 14, or find [H+] from Kw = [H+][OH−].

Worked example 3 — pH of a strong base

Find the pH of 0.050 mol dm−3 NaOH at 298 K.

NaOH is a strong base, fully dissociated: [OH−] = 0.050 mol dm−3.

Route 1: pOH = −log(0.050) = 1.30, so pH = 14.00 − 1.30 = 12.70.

Route 2: [H+] = Kw / [OH−] = (1.0 × 10−14) / 0.050 = 2.0 × 10−13; pH = −log(2.0 × 10−13) = 12.70. Both agree.

5. (HL) Weak-acid and weak-base equilibria

Because a weak acid only partly dissociates, [H+] cannot be read off the concentration; it must be found from an equilibrium constant. For a weak acid HA:

HA(aq) ⇌ H+(aq) + A−(aq) Ka = [H+][A−] / [HA]

For a weak base B, the base-dissociation constant is defined analogously:

B(aq) + H2O(l) ⇌ BH+(aq) + OH−(aq) Kb = [BH+][OH−] / [B]

A larger Ka means a stronger acid (equilibrium further right). As with pH, these constants are usually quoted as negative logarithms:

pKa = −log Ka pKb = −log Kb

The logarithm reverses the ranking: a smaller pKa means a stronger acid. A change of one pKa unit is a ten-fold change in Ka.

For any conjugate acid–base pair, Ka and Kb are linked through the ionic product of water:

Ka × Kb = Kw pKa + pKb = 14.00 (at 298 K)

So the stronger an acid, the weaker its conjugate base, and vice versa.

Acid Ka / mol dm−3 pKa Relative strength

Chloric(VII) (perchloric) very large < 0 strong

Methanoic, HCOOH 1.8 × 10−4 3.75 weak (stronger)

Ethanoic, CH3COOH 1.8 × 10−5 4.76 weak

Carbonic (1st), H2CO3 4.3 × 10−7 6.37 weak (weaker)

Ammonium, NH4

+ 5.6 × 10−10 9.25 very weak

Calculating the pH of a weak acid

For a weak acid of concentration c where dissociation is small, two approximations make the algebra manageable: (1) [H+] = [A−] (both come from the same dissociation), and (2) the equilibrium [HA] ≈ c (little dissociates). Substituting into Ka:

Ka ≈ [H+]2 / c so [H+] ≈ (Ka × c)1/2

Worked example 4 — pH of a weak acid from Ka

Calculate the pH of 0.10 mol dm−3 ethanoic acid, Ka = 1.8 × 10−5 mol dm−3.

[H+] ≈ (Ka × c)1/2 = (1.8 × 10−5 × 0.10)1/2 = (1.8 × 10−6)1/2 = 1.34 × 10−3 mol dm−3.

pH = −log(1.34 × 10−3) = 2.87.

Check: only 1.3% of the acid dissociated, so the approximations are safe. Compare with 0.10 mol dm−3 HCl, which would give pH 1.00 — the weak acid is far less acidic at the same concentration.

When the approximation fails: The formula [H+] = (Kac)1/2 assumes dissociation below about 5%. For very dilute or relatively strong weak acids you must solve the full quadratic — but IB questions are almost always designed so the approximation holds.

6. (HL) Buffer solutions

A buffer resists changes in pH when small amounts of acid or base are added, or on dilution. It contains appreciable amounts of both members of a conjugate pair, so it has a reservoir to neutralise added acid and added base.

• An acidic buffer (pH < 7) is a weak acid plus its salt, e.g. CH3COOH + CH3COONa.
• A basic buffer (pH > 7) is a weak base plus its salt, e.g. NH3 + NH4Cl.

How it works

Take the ethanoic acid / ethanoate buffer, which contains a large store of both CH3COOH and CH3COO−:

Add acid (H+): CH3COO− + H+ → CH3COOH (the conjugate base mops up added H+)

Add base (OH−): CH3COOH + OH− → CH3COO− + H2O (the acid mops up added OH−)

Because both components are present in large amounts, the ratio [base]/[acid] barely changes, so the pH barely changes.

The Henderson–Hasselbalch relationship

Rearranging Ka = [H+][A−]/[HA] and taking negative logs gives a formula for buffer pH:

pH = pKa + log ( [A−] / [HA] ) = pKa + log ( [salt] / [acid] )

Two consequences are worth memorising: the pH of a buffer is close to the pKa of its weak acid, and when [salt] = [acid] the log term is zero, so pH = pKa. This is why a buffer is chosen so that the required pH is near the pKa of the acid used.

Figure 6.1. Henderson−Hasselbalch line: pH = pKa + log([salt]/[acid]). When [salt] = [acid] the log term is zero, so pH = pKa =

4.74; the 0.30 / 0.20 mol dm−3 mixture of Worked example 5 gives pH 4.92.

Worked example 5 — buffer pH

A buffer is made by mixing ethanoic acid (final concentration 0.20 mol dm−3) with sodium ethanoate (0.30 mol dm−3). Ka(ethanoic) = 1.8 × 10−5, so pKa = 4.74. Find the pH.

pH = pKa + log([salt]/[acid]) = 4.74 + log(0.30 / 0.20)

= 4.74 + log(1.50) = 4.74 + 0.18 = 4.92.

The pH sits just above pKa because there is more conjugate base than acid.

Worked example 6 — buffer action on adding acid

To 1.00 dm3 of the buffer above (0.20 mol CH3COOH and 0.30 mol CH3COO−) is added 0.050 mol of HCl. Estimate the new pH, and compare with adding the same acid to pure water.

Added H+ converts ethanoate to ethanoic acid: acid becomes 0.20 + 0.050 = 0.25 mol; salt becomes 0.30 − 0.050 = 0.25 mol.

pH = 4.74 + log(0.25 / 0.25) = 4.74 + log(1) = 4.74.

The pH fell by only 0.18 units. Adding 0.050 mol HCl to 1 dm3 of pure water would give [H+] = 0.050 and pH = 1.30 — a dramatic change. That is buffer action.

Where buffers matter

Buffers are central to any system that must hold a steady pH. Human blood is buffered near pH 7.4 mainly by the carbonic acid / hydrogencarbonate pair (H2CO3 / HCO3

−); a shift of even a few tenths of a pH unit is dangerous. Buffers also control the pH of shampoos, fermentation vats, and the standard solutions used to calibrate pH meters.

Worked example — designing a buffer

You need a buffer of pH 5.0. Choose a suitable weak acid from the list in Section 5 and state the ratio [salt] : [acid] required.

Choose an acid with pKa close to 5.0. Ethanoic acid (pKa = 4.76) is ideal — the target is within one unit of its pKa.

pH = pKa + log([salt]/[acid]) → 5.0 = 4.76 + log(ratio) log(ratio) = 0.24, so [salt]/[acid] = 100.24 = 1.7. Use ethanoic acid with ethanoate ions in a ratio of about 1.7 : 1.

Buffer requirements: A working buffer needs (i) a weak acid or base with (ii) a comparable amount of its conjugate partner. A strong acid has no conjugate base reservoir, so it cannot buffer. Best buffering occurs within about ±1 pH unit of the pKa.

Worked example 6b — using KaKb = Kw

Ethanoic acid has Ka = 1.8 × 10−5. Find Kb for its conjugate base, the ethanoate ion, and hence pKb.

Kb = Kw / Ka = (1.0 × 10−14) / (1.8 × 10−5) = 5.6 × 10−10.

pKb = −log(5.6 × 10−10) = 9.25. Check: pKa + pKb = 4.74 + 9.25 ≈ 14. The very small Kb confirms ethanoate is only a weak base.

7. (HL) pH (titration) curves

A pH curve plots the pH of the solution in the flask against the volume of titrant added. Its shape depends on whether the acid and base are strong or weak, and it lets you locate the equivalence point and choose an indicator.

Titration Start pH Equivalence pH Curve feature

Strong acid – strong base very low (~1) 7 (neutral) long, near-vertical jump (~pH 3–11)

Weak acid – strong base higher (~3) > 7 (basic) buffer region; shorter jump (~pH 7–11)

Strong acid – weak base very low (~1) < 7 (acidic) shorter jump (~pH 3–7)

Weak acid – weak base higher (~3) ~7 no sharp vertical jump — no clear end point

Figure 7.1. Titration of 25.0 cm3 of 0.100 mol dm−3 acid with 0.100 mol dm−3 NaOH. Both curves reach equivalence at 25.0 cm3; the strong−acid curve passes through pH 7.00, while the weak−acid curve is basic at equivalence (pH ≈ 8.72) and shows a buffer region where pH = pKa at half−equivalence.

Equivalence and half-equivalence

The equivalence point is where the acid and base have been mixed in exactly stoichiometric amounts — the steep middle of the jump. Its pH is not always 7: for a weak acid with a strong base the salt is basic, so equivalence is above 7; for a strong acid with a weak base it is below 7.

The half-equivalence point (halfway to equivalence, in a weak acid titration) is where exactly half the weak acid has been neutralised. There [salt] = [acid], so the Henderson–Hasselbalch equation gives a memorable result:

at half-equivalence: pH = pKa

This is the standard experimental method for finding the pKa of a weak acid — read the pH at the half-equivalence volume straight off the curve.

Choosing an indicator

An acid–base indicator is itself a weak acid (HIn) whose conjugate base (In−) has a different colour. It changes colour over a range of roughly pKa(In) ± 1. For a sharp end point the indicator's colour-change range must fall within the near-vertical section of the pH curve — i.e. its pKa should be close to the equivalence pH.

Indicator Range (pH) Acid colour → base colour Best for

Methyl orange 3.1 – 4.4 red → yellow strong acid – weak base

Bromothymol blue 6.0 – 7.6 yellow → blue strong acid – strong base

Phenolphthalein 8.3 – 10.0 colourless → pink weak acid – strong base

Worked example 7 — choosing an indicator

Which indicator suits the titration of 0.10 mol dm−3 ethanoic acid (weak) with 0.10 mol dm−3

NaOH (strong)?

This is a weak acid – strong base titration, so the equivalence point is basic (pH ≈ 8.7). The near-vertical section lies roughly pH 7–11.

Phenolphthalein (range 8.3–10.0) changes colour inside this jump, so it gives a sharp end point. Methyl orange (3.1–4.4) would change far too early, in the buffer region, and is unsuitable.

Weak–weak titrations: When both partners are weak, there is no sharp vertical section, so no indicator changes colour cleanly. Such titrations are not used for accurate analysis — a pH meter would be needed.

8. Salt hydrolysis (qualitative)

Dissolving a salt does not always give a neutral solution. The ions of a salt can react with water (“hydrolyse”) if they are the conjugate of a weak acid or base. The rule follows from which parent was weak.

Salt from... Example Ion that hydrolyses Resulting pH strong acid + strong base NaCl, KNO3 neither neutral (≈ 7) strong acid + weak base NH4Cl, NH4NO3 cation (NH4

+ donates H+) acidic (< 7) weak acid + strong base CH3COONa, Na2CO3 anion (accepts H+) basic (> 7) weak acid + weak base CH3COONH4 both ions depends on Ka, Kb For example, the ethanoate ion is the conjugate base of a weak acid, so it removes protons from water and leaves excess OH−:

CH3COO− + H2O ⇌ CH3COOH + OH− (solution basic)

The ammonium ion is the conjugate acid of a weak base, so it donates protons to water and leaves excess H3O+:

NH4

+ + H2O ⇌ NH3 + H3O+ (solution acidic)

Quick rule: “Strong wins.” The salt takes on the character of the stronger parent: strong-acid + weak-base → acidic; weak-acid + strong-base → basic; strong + strong → neutral.

9. Common pitfalls

• Strong is not the same as concentrated. Strength is degree of dissociation; concentration is amount per volume. A dilute strong acid can have a higher pH than a concentrated weak one.
• Forgetting the minus sign in pH = −log[H+], or taking log of a negative exponent incorrectly.
• Treating Kw = 1.0 × 10−14 as fixed. It is only that value at 298 K; it rises with temperature, so neutral pH is not always 7.
• Reading [H+] straight off the concentration of a weak acid — you must use Ka.
• Assuming every equivalence point is at pH 7. Only strong–strong titrations are neutral at equivalence.
• Calling a mixture a buffer when one component is a strong acid or base, or when the conjugate partner is absent.
• Confusing amphiprotic (donates/accepts H+) with amphoteric (reacts with acids and bases) — use the narrower term when a proton is transferred.

Quick-reference summary

Quantity / rule Expression

Acid / base (Brønsted–Lowry) acid = H+ donor; base = H+ acceptor

Conjugate pair differ by one H+ (charge changes by 1)

Neutralisation H+ + OH− → H2O pH / [H+] pH = −log[H+]; [H+] = 10−pH

Ionic product Kw = [H+][OH−] = 1.0 × 10−14 (298 K) pH and pOH pH + pOH = 14 (298 K)

(HL) Weak acid Ka = [H+][A−]/[HA]; [H+] ≈ (Kac)1/2

(HL) Conjugate constants KaKb = Kw; pKa + pKb = 14

(HL) Buffer pH = pKa + log([salt]/[acid])

(HL) Half-equivalence pH = pKa

10. Test yourself

Attempt all ten without notes; full worked answers follow. Questions marked (HL) are Higher Level only.

• Give the conjugate base of each: H2SO4, HNO3, NH4 +. Give the conjugate acid of each: NH3, CO3

2−, OH−.

• Write balanced equations for dilute hydrochloric acid reacting with (a) magnesium, (b) copper(II) oxide,
• calcium carbonate.
• State two experimental observations that would distinguish equal-concentration solutions of a strong and a weak acid.
• Calculate the pH of 0.025 mol dm−3 nitric acid.
• Calculate the pH of 0.040 mol dm−3 sodium hydroxide at 298 K.
• (HL) A weak acid of concentration 0.20 mol dm−3 has Ka = 1.0 × 10−5. Find [H+] and the pH.
• (HL) A 0.10 mol dm−3 solution of a weak acid HA has pH 3.00. Find Ka and pKa.
• (HL) Calculate the pH of a buffer that is 0.15 mol dm−3 in propanoic acid (pKa = 4.87) and 0.10 mol dm−3 in sodium propanoate.
• (HL) State, with a reason, which indicator you would use to titrate ethanoic acid against potassium hydroxide.
• Predict whether each salt solution is acidic, neutral or basic: (a) NH4NO3, (b) KNO3, (c) sodium ethanoate, (d) sodium carbonate.

Worked answers

• Conjugate bases: HSO4 −, NO3

−, NH3. Conjugate acids: NH4

+, HCO3

−, H2O. (Each differs from its partner by one H+.)

• (a) Mg + 2HCl → MgCl2 + H2. (b) CuO + 2HCl → CuCl2 + H2O. (c) CaCO3 + 2HCl → CaCl2 + H2O + CO2.
• Any two: the strong acid has the lower pH; the strong acid has higher electrical conductivity; the strong acid reacts faster with magnesium or a carbonate (more vigorous fizzing). All comparisons require equal concentration.
• HNO3 is a strong monoprotic acid: [H+] = 0.025 mol dm−3. pH = −log(0.025) = 1.60.
• [OH−] = 0.040 mol dm−3; pOH = −log(0.040) = 1.40; pH = 14.00 − 1.40 = 12.60.
• (HL) [H+] ≈ (Kac)1/2 = (1.0 × 10−5 × 0.20)1/2 = (2.0 × 10−6)1/2 = 1.41 × 10−3 mol dm−3; pH = −log(1.41 × 10−3) = 2.85.
• (HL) [H+] = 10−3.00 = 1.0 × 10−3 mol dm−3. Ka ≈ [H+]2/c = (1.0 × 10−3)2 / 0.10 = 1.0 × 10−5; pKa = −log(1.0 × 10−5) = 5.00.
• (HL) pH = pKa + log([salt]/[acid]) = 4.87 + log(0.10/0.15) = 4.87 + log(0.667) = 4.87 − 0.18 = 4.69.
• (HL) Ethanoic acid (weak) with KOH (strong) gives a basic equivalence point (pH ≈ 8–9), so use phenolphthalein (range 8.3–10.0): its colour change falls inside the near-vertical section. Methyl orange would change too early.
• (a) acidic (NH4 + is the conjugate acid of weak NH3); (b) neutral (both parents strong); (c) basic (ethanoate is the conjugate base of a weak acid); (d) basic (carbonate is the conjugate base of weak carbonic acid).