Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Chemistry guide (first assessment 2025).
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R3.1.1 Brønsted–Lowry acid
Brønsted–Lowry acid
A species that donates a proton (H⁺) to another species. In HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq), HCl is the Brønsted–Lowry acid. A species is classified by what it does in a particular reaction, not by its formula alone.
Brønsted–Lowry base
A species that accepts a proton (H⁺) from another species, using a lone pair of electrons to form a bond to the proton. In NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq), NH₃ is the Brønsted–Lowry base and H₂O is the acid. A base need not contain the hydroxide ion: NH₃, CH₃NH₂, O²⁻ and CO₃²⁻ are all bases.
H⁺(aq) and H₃O⁺(aq)
A proton in aqueous solution does not exist on its own; it is bonded to water molecules. It can be represented either as H⁺(aq) or as the oxonium (hydronium) ion H₃O⁺(aq). The two representations are equivalent: HCl(aq) → H⁺(aq) + Cl⁻(aq) and HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq) describe the same change.
Alkali
A base that is soluble in water and releases hydroxide ions, OH⁻(aq), into solution, for example NaOH(aq), KOH(aq) and NH₃(aq). Every alkali is a base, but not every base is an alkali: copper(II) oxide and magnesium oxide are bases that are insoluble in water, so they are not alkalis.
Students often think A Brønsted–Lowry acid is a proton acceptor and a base is a proton donor, so the species that gains H⁺ in a reaction is the acid. In fact It donates a proton (H⁺); a Brønsted–Lowry base accepts one.
Students often think 'Base' and 'alkali' are synonyms: every base is an alkali and every alkali is a base. In fact No. An alkali is a base that dissolves in water to release OH⁻(aq). Insoluble bases such as CuO and MgO are bases but not alkalis.
R3.1.2 Conjugate acid–base pair
Conjugate acid–base pair
Two species that differ by exactly one proton (H⁺). The acid of the pair has one more H⁺ than its conjugate base, so its charge is one unit more positive: H₂PO₄⁻/HPO₄²⁻, NH₄⁺/NH₃, H₃O⁺/H₂O, H₂O/OH⁻. In every Brønsted–Lowry reaction there are two conjugate pairs, one on each side of the equation.
Conjugate base and conjugate acid
The conjugate base of an acid is the species formed when the acid loses one H⁺ (remove one H and lower the charge by 1: HSO₄⁻ → SO₄²⁻). The conjugate acid of a base is the species formed when the base gains one H⁺ (add one H and raise the charge by 1: CO₃²⁻ → HCO₃⁻).
Students often think The species formed when a proton is lost is the acid (the acid is confused with its conjugate base), so the roles are read from the product side. In fact The acid is the reactant that loses H⁺. The species formed from it is its conjugate base.
Students often think Any two species related by the loss of protons form a conjugate pair, however many protons separate them. In fact No. A conjugate pair differs by exactly one H⁺. H₃PO₄ and HPO₄²⁻ differ by two; the conjugate base of H₃PO₄ is H₂PO₄⁻.
R3.1.3 Amphiprotic species
Amphiprotic species
A species that can act as a Brønsted–Lowry acid or as a Brønsted–Lowry base, depending on what it reacts with. It must contain an H that it can lose as H⁺ and be able to accept H⁺ (a lone pair). Examples: H₂O, HCO₃⁻, HSO₄⁻, H₂PO₄⁻. With OH⁻, HCO₃⁻ acts as an acid (HCO₃⁻ + OH⁻ ⇌ CO₃²⁻ + H₂O); with H₃O⁺ it acts as a base (HCO₃⁻ + H₃O⁺ ⇌ H₂CO₃ + H₂O). In any one reaction it plays only one of the two roles.
Students often think Water is neutral, so it cannot act as an acid or a base; it is only the solvent in which acid–base reactions happen. In fact Yes. Water is amphiprotic: it accepts H⁺ from HCl (base) and donates H⁺ to NH₃ (acid).
Students often think Negative ions are bases: being negatively charged, they can only attract and accept protons, never donate them. In fact No. An anion that contains an ionizable H, such as H₂PO₄⁻, HCO₃⁻ or HSO₄⁻, can donate H⁺ as well as accept it.
R3.1.4 pH
pH
pH = −log₁₀[H⁺], where [H⁺] is in mol dm⁻³; equivalently [H⁺] = 10⁻ᵖᴴ. pH has no unit. Because the scale is logarithmic, a change of one pH unit is a tenfold change in [H⁺] and a change of two units is a hundredfold change. A lower pH means a higher [H⁺]. pH can be below 0 or above 14 for very concentrated solutions.
Universal indicator and pH meter
Universal indicator gives a colour that is compared with a chart to estimate pH, usually to the nearest whole unit. A pH meter (pH probe), calibrated with buffer solutions of known pH, measures pH precisely, typically to 0.01 unit, and does not depend on judging a colour.
Students often think The pH of an acid solution is fixed by the acid used, so diluting it does not change [H⁺] or the pH. In fact Yes. [H⁺] falls in proportion to the dilution (for a strong acid), so the pH must be recalculated from the new concentration.
Students often think Any number produced on the way can be put into −log₁₀[H⁺]: a volume in cm³ can be used as if it were in dm³, or an amount in mol used as if it were a concentration. In fact The concentration of H⁺ in mol dm⁻³, found by converting any volume in cm³ to dm³ (÷ 1000) and dividing an amount by the solution's volume.
R3.1.5 Ion product constant of water, Kw
Ion product constant of water, Kw
Water ionizes slightly: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq). Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ at 298 K. Because the product is constant at a given temperature, [H⁺] and [OH⁻] are inversely related: if [OH⁻] increases tenfold, [H⁺] falls tenfold. Every aqueous solution contains both ions.
Acidic, neutral and basic solutions
An aqueous solution is acidic when [H⁺] > [OH⁻], neutral when [H⁺] = [OH⁻], and basic when [H⁺] < [OH⁻]. At 298 K a neutral solution has [H⁺] = [OH⁻] = 1.00 × 10⁻⁷ mol dm⁻³ and pH 7.00; the criterion itself is the relative size of [H⁺] and [OH⁻].
Students often think In any aqueous solution [H⁺] and [OH⁻] are equal, because water always ionizes into one H⁺ and one OH⁻. In fact No. [H⁺] = [OH⁻] only in a neutral solution. Kw fixes the product [H⁺][OH⁻], not the individual values: in a basic solution [OH⁻] > [H⁺].
Students often think [H⁺] and [OH⁻] are independent of each other, so each is judged on its own against its value in pure water at 298 K (1.0 × 10⁻⁷ mol dm⁻³): adding an acid increases [H⁺] but leaves [OH⁻] at its pure-water value. In fact Yes. [OH⁻] = Kw/[H⁺], so when [H⁺] rises, [OH⁻] falls in inverse proportion: a hundredfold rise in [H⁺] gives a hundredfold fall in [OH⁻].
R3.1.6 Strong acid and strong base
Strong acid and strong base
An acid or base that is (effectively) completely ionized in aqueous solution. The strong acids to know are HCl, HBr, HI, HNO₃ and H₂SO₄: HCl(aq) → H⁺(aq) + Cl⁻(aq). The group 1 hydroxides (LiOH, NaOH, KOH) are strong bases. For a strong monoprotic acid, [H⁺] equals the concentration of the acid.
Weak acid and weak base
An acid or base that is only partially ionized in aqueous solution, so an equilibrium lies well to the left: CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq); NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq). Carboxylic acids, carbonic acid, ammonia and amines are weak. At equal concentration a weak acid has a lower [H⁺] (higher pH) and lower electrical conductivity than a strong acid, but it neutralizes the same amount of base.
Strong/weak versus concentrated/dilute
Strength describes the extent of ionization of the acid or base; concentration describes the amount of acid or base dissolved per unit volume (mol dm⁻³). The two are independent: 0.0010 mol dm⁻³ HNO₃ is a dilute solution of a strong acid, and 10 mol dm⁻³ CH₃COOH is a concentrated solution of a weak acid.
Position of acid–base equilibria
An acid–base equilibrium lies on the side of the weaker acid and weaker base (the weaker conjugates). HCl + H₂O → H₃O⁺ + Cl⁻ lies far to the right because H₃O⁺ is a weaker acid than HCl and Cl⁻ is a weaker base than H₂O. The stronger an acid, the weaker its conjugate base.
Students often think Strength and concentration are the same thing: a strong acid is a concentrated one and a weak acid a dilute one (so diluting a strong acid makes it weak), and an extreme pH shows a strong acid or base whatever the conce… In fact No. Strong/weak describes the extent of ionization; concentrated/dilute describes the amount of acid or base per dm³. pH depends on both, so strength can be judged from pH only when the concentration is known.
Students often think All acids, strong ones included, ionize only partly, so a solution of HNO₃ contains mostly HNO₃ molecules in equilibrium with a few ions. In fact No. A strong acid such as HNO₃ is effectively completely ionized in water, so there are essentially no HNO₃ molecules in the solution.
R3.1.7 Neutralization
Neutralization
The reaction of an acid with a base. Acid + metal oxide or metal hydroxide → salt + water (CuO + H₂SO₄ → CuSO₄ + H₂O). Acid + carbonate or hydrogencarbonate → salt + water + carbon dioxide (Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂; CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂). Acid + ammonia or an amine → salt only (NH₃ + HCl → NH₄Cl; CH₃NH₂ + HCl → CH₃NH₃Cl). Weak organic acids such as ethanoic acid undergo all of these reactions.
Salt and its parent acid and base
A salt is an ionic compound formed when the H⁺ of an acid is replaced by a metal ion or an ammonium-type ion. Its parent acid supplies the anion and its parent base supplies the cation: CH₃COONa comes from CH₃COOH and NaOH; NH₄NO₃ from HNO₃ and NH₃; CH₃NH₃Cl from HCl and CH₃NH₂.
Students often think Every neutralization follows 'acid + base → salt + water', so the products are always a salt and water only. In fact No. Acids with carbonates and hydrogencarbonates also give carbon dioxide, and acids with ammonia or amines give a salt only, with no water.
Students often think Acids release hydrogen gas in all their reactions, so the gas formed when an acid reacts with a carbonate is hydrogen (or includes hydrogen). In fact No. Hydrogen is produced when an acid reacts with a reactive metal. Acid–base reactions transfer H⁺ to the base; carbonates give carbon dioxide, not hydrogen.
R3.1.8 pH curve for a strong acid–strong base titration
pH curve for a strong acid–strong base titration
A plot of pH against volume of titrant added. For 25.0 cm³ of 0.100 mol dm⁻³ HCl titrated with 0.100 mol dm⁻³ NaOH, the curve intercepts the pH axis at pH 1.0, rises only slowly, then rises almost vertically from about pH 3 to pH 11 around 25.0 cm³, passes through the equivalence point at pH 7, and then levels off, approaching (but not reaching) the pH of the titrant, about 12.5 after 50 cm³.
Equivalence point
The point in a titration at which the amounts of acid and base added are in the stoichiometric ratio of the equation, so neither is in excess. For a strong monoprotic acid and a strong base the solution then contains only a neutral salt and water, so the pH at the equivalence point is 7 at 298 K.
Students often think pH changes linearly with the volume of alkali added, so the pH curve is a straight line from the starting pH to the final pH. In fact No. The pH changes only slowly at first, then very sharply around the equivalence point, then slowly again, because pH is logarithmic.
Students often think Once the acid has been neutralized the pH stays at 7, because the solution is now neutral and more alkali cannot change it. In fact No. Beyond the equivalence point there is excess NaOH, so the pH continues to rise and levels off approaching the pH of the NaOH solution.
R3.1.9 pOH HL
pOH
pOH = −log₁₀[OH⁻]; [OH⁻] = 10⁻ᵖᴼᴴ. At 298 K, pH + pOH = 14.00, which follows from Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴. For example, a solution of pOH 4.30 has pH 9.70 and [H⁺] = 10⁻⁹·⁷⁰ = 2.0 × 10⁻¹⁰ mol dm⁻³.
Students often think pH and pOH (and likewise pKa and pKb) are interchangeable, so a pOH value can be read as a pH, or a pH read at the half-equivalence point of a weak base titration can be taken as its pKb. In fact No. pH refers to [H⁺] and pOH to [OH⁻]; pKa refers to an acid and pKb to a base. At 298 K they are linked by pH + pOH = 14.00 and pKa + pKb = 14.00, and the right one must be chosen.
Students often think Because Kw = [H⁺][OH⁻] (and Ka × Kb = Kw) is a product, the p-values also multiply: pH × pOH = 14.00 and pKa × pKb = 14.00. In fact No. Taking −log₁₀ of a product gives a sum: pH + pOH = 14.00 at 298 K (and likewise pKa + pKb = 14.00 for a conjugate pair).
R3.1.10 Acid dissociation constant, Ka, and pKa HL
Acid dissociation constant, Ka, and pKa
For HA(aq) ⇌ H⁺(aq) + A⁻(aq), Ka = [H⁺][A⁻]/[HA] at equilibrium (unit mol dm⁻³, usually omitted); pKa = −log₁₀Ka. A larger Ka, and therefore a smaller pKa, means a stronger acid. Ethanoic acid: Ka = 1.74 × 10⁻⁵, pKa = 4.76; methanoic acid: pKa = 3.75, so methanoic acid is the stronger acid.
Base dissociation constant, Kb, and pKb
For B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq), Kb = [BH⁺][OH⁻]/[B]; pKb = −log₁₀Kb. A larger Kb, and therefore a smaller pKb, means a stronger base. Ammonia: pKb = 4.75.
Students often think A larger pKa value means a stronger acid (and a larger pKb a stronger base), because a bigger number indicates more strength. In fact No. pKa = −log₁₀Ka, so a larger pKa means a smaller Ka and a weaker acid.
Students often think All weak acids are equally weak, so their strengths cannot be compared; only the strong/weak classification matters. In fact No. Weak acids have a wide range of Ka values, and their relative strengths are compared using Ka or pKa.
R3.1.11 Ka × Kb = Kw HL
Ka × Kb = Kw
For a conjugate acid–base pair HA/A⁻ (or BH⁺/B), multiplying the two expressions gives Ka × Kb = [H⁺][OH⁻] = Kw, so at 298 K Ka × Kb = 1.00 × 10⁻¹⁴ and pKa + pKb = 14.00. NH₄⁺ has pKa 9.25, so NH₃ has pKb 4.75. The stronger an acid, the weaker its conjugate base.
pH of a weak acid (approximation)
Because a weak acid is only slightly ionized, [HA]equilibrium ≈ [HA]initial and [H⁺] ≈ [A⁻], so Ka ≈ [H⁺]²/[HA]initial and [H⁺] = √(Ka × [HA]initial). For 0.100 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵): [H⁺] = 1.32 × 10⁻³ mol dm⁻³, pH = 2.88. The same method with Kb gives [OH⁻] for a weak base.
Students often think The acid and base of a conjugate pair share the same K value, so the Ka (or pKa) of an acid can be used as the Kb (or pKb) of its conjugate base, or a species' pKa read as its strength as a base. In fact No. For a conjugate pair Ka × Kb = Kw, so pKa + pKb = 14.00 at 298 K. A pKa value describes the acid of the pair; it is not a measure of the base's strength.
Students often think For a weak acid, [H⁺] = Ka × [HA]initial, with no square root. In fact [H⁺] = √(Ka × [HA]), because Ka ≈ [H⁺]²/[HA] when [H⁺] = [A⁻].
R3.1.12 Salt hydrolysis HL
Salt hydrolysis
The reaction of an ion from a salt with water, which changes [H⁺] and so the pH. The conjugate acid of a weak base makes the solution acidic: NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq). The conjugate base of a weak acid makes the solution basic: RCOO⁻(aq) + H₂O(l) ⇌ RCOOH(aq) + OH⁻(aq); CO₃²⁻(aq) + H₂O(l) ⇌ HCO₃⁻(aq) + OH⁻(aq); HCO₃⁻(aq) + H₂O(l) ⇌ H₂CO₃(aq) + OH⁻(aq). Ions from strong acids and strong bases (Cl⁻, NO₃⁻, Na⁺, K⁺) do not hydrolyse measurably.
Students often think Salts are the products of neutralization, so every salt solution is neutral and every titration has its equivalence point at pH 7. In fact No. A salt of a weak acid and a strong base is basic (CH₃COONa), a salt of a strong acid and a weak base is acidic (NH₄Cl), so equivalence points can lie above or below 7.
Students often think Hydrolysis is the reverse of neutralization: the salt reacts with water to re-form its parent acid and parent base as molecules, e.g. NH₄Cl + H₂O ⇌ NH₄OH + HCl. In fact No. Hydrolysis is an equilibrium in which one ion (not the whole salt) transfers a proton to or from water, usually to a small extent; e.g. NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺.
R3.1.13 pH curves for strong and weak acid–base combinations HL
pH curves for strong and weak acid–base combinations
Intercept: a weak acid starts at a higher pH than a strong acid of equal concentration (0.100 mol dm⁻³ CH₃COOH: pH 2.88; HCl: pH 1.00). Equivalence point: pH 7 for strong acid–strong base; above 7 for weak acid–strong base (about 8.7 for 0.100 mol dm⁻³ CH₃COOH with NaOH); below 7 for strong acid–weak base (about 5.3 for 0.100 mol dm⁻³ NH₃ with HCl); no sharp vertical section for weak acid–weak base. A weak acid or weak base gives a buffer region before the equivalence point.
Half-equivalence point
The point at which half the volume of titrant needed for equivalence has been added. For a weak acid titrated with a strong base, [HA] = [A⁻] there, so [H⁺] = Ka and pH = pKa. For a weak base titrated with a strong acid, [B] = [BH⁺], so pOH = pKb, and pH = pKa of the conjugate acid BH⁺.
Students often think The pH of any solution of a weak acid, or of any buffer made from it, equals the pKa of the acid. In fact No. pH = pKa only when [HA] = [A⁻], as at the half-equivalence point or in a buffer with equal concentrations of acid and conjugate base.
Students often think The half-equivalence point and the equivalence point are the same point, so the pH at half the equivalence volume is the pH at equivalence. In fact No. The half-equivalence point is where half the titrant needed for equivalence has been added; there [HA] = [A⁻] and pH = pKa. The equivalence point is where all of the acid has reacted.
R3.1.14 Acid–base indicator HL
Acid–base indicator
A weak acid, HInd, whose undissociated form and conjugate base have different colours: HInd(aq) ⇌ H⁺(aq) + Ind⁻(aq). Adding acid shifts the equilibrium to the left (colour of HInd); adding alkali removes H⁺ and shifts it to the right (colour of Ind⁻). The colour changes over a range of about pKa ± 1, and the end point is at pH ≈ pKa, where [HInd] = [Ind⁻]. Examples: methyl orange, range about 3.1–4.4; phenolphthalein, about 8.3–10.0.
Universal indicator
A mixture of several indicators whose colour-change ranges lie at different pH values, so that the mixture shows a different colour across a wide pH range (about 1 to 14).
Students often think At equilibrium the two sides are present in equal amounts, so an indicator at equilibrium always contains equal amounts of HInd and Ind⁻. In fact No. At equilibrium the rates of the forward and backward reactions are equal; the concentrations are constant but usually not equal.
Students often think Every indicator changes colour at pH 7, showing its acid colour below 7 and its alkali colour above 7. In fact No. Each indicator changes colour over a range around its own pKa (about pKa ± 1): methyl orange about 3.1–4.4, phenolphthalein about 8.3–10.0.
R3.1.15 End point HL
End point
The point in a titration at which the indicator changes colour. An indicator is appropriate when its colour-change range lies within the steep (near-vertical) section of the pH curve, so that the end point coincides with the equivalence point. For a weak acid–strong base titration (equivalence pH above 7, salt such as CH₃COONa) phenolphthalein is suitable; for a strong acid–weak base titration (equivalence pH below 7, salt such as NH₄Cl) methyl orange is suitable.
Students often think An indicator changes colour at the pKa of the acid being tested, so the best indicator is one whose range includes the analyte's pKa. In fact No. The colour-change pH is set by the indicator's own pKa. The indicator is chosen so that this range falls within the steep section around the equivalence point.
Students often think Any indicator gives the correct end point in any titration, because the pH changes through every indicator's range near the equivalence point. In fact Yes. The indicator's colour-change range must fall within the steep section of the pH curve; otherwise the end point is not at the equivalence point.
R3.1.16 Buffer solution HL
Buffer solution
A solution that resists a change in pH when small amounts of acid or alkali are added. An acidic buffer contains a weak acid and its conjugate base in similar amounts (CH₃COOH and CH₃COONa), or is made by adding a strong base to an excess of weak acid. A basic buffer contains a weak base and its conjugate acid (NH₃ and NH₄Cl), or is made by adding a strong acid to an excess of weak base.
Buffer action
In CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) with large amounts of both CH₃COOH and CH₃COO⁻: added H⁺ is removed by CH₃COO⁻ (CH₃COO⁻ + H⁺ → CH₃COOH); added OH⁻ is removed by CH₃COOH (CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O). In an NH₃/NH₄⁺ buffer, NH₃ removes added H⁺ and NH₄⁺ removes added OH⁻. The ratio of the two components changes only slightly, so the pH changes only slightly; it is not held exactly constant.
Students often think Any acid mixed with a salt of that acid forms a buffer, so HCl with NaCl is a buffer just as CH₃COOH with CH₃COONa is. In fact No. A buffer needs a weak acid and its conjugate base (or a weak base and its conjugate acid) in similar amounts. HCl with NaCl is not a buffer.
Students often think Any mixture of a weak acid and a strong base is a buffer, whatever the amounts used. In fact No. The weak acid must be in excess so that both HA and A⁻ remain. Equal amounts of CH₃COOH and NaOH react completely to give sodium ethanoate only.
R3.1.17 pH of a buffer HL
pH of a buffer
From Ka = [H⁺][A⁻]/[HA]: [H⁺] = Ka × [HA]/[A⁻], or pH = pKa + log₁₀([A⁻]/[HA]). The pH depends on the pKa of the acid and the ratio [A⁻]/[HA]; when [HA] = [A⁻], pH = pKa. For a basic buffer, pOH = pKb + log₁₀([BH⁺]/[B]). Diluting a buffer changes both concentrations by the same factor, so the ratio, and to a first approximation the pH, stays the same; dilution does reduce the buffer's capacity to absorb added acid or alkali.
Students often think [H⁺] = Ka × [A⁻]/[HA], so the ratio is salt over acid in the [H⁺] expression. In fact The acid, [HA]. More acid relative to conjugate base gives a higher [H⁺] and a lower pH.
Students often think Diluting a buffer with an equal volume of water halves [H⁺], just as it would for a solution of a strong acid, so the pH rises by 0.30. In fact No. [HA] and [A⁻] fall by the same factor, so [H⁺] = Ka[HA]/[A⁻] and the pH stay approximately unchanged.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 Ammonia reacts with water:
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)
Which species acts as a Brønsted–Lowry acid in the forward reaction?
Answer and reasoning
NH₃, as it gains a proton to form NH₄⁺ — Gaining a proton is what a Brønsted–Lowry base does. NH₃ accepts H⁺, so it is the base; the acid is H₂O, which donates the proton.
OH⁻, as it forms when H₂O loses a proton — OH⁻ is what is left after the acid has donated its proton: it is the conjugate base of H₂O. The acid is the reactant that loses H⁺, which is H₂O itself.
None, as water is neutral and cannot donate — Water is amphiprotic. Here it donates H⁺ to NH₃, becoming OH⁻, so H₂O is the Brønsted–Lowry acid.
H₂O, as it donates a proton to NH₃ — In the forward reaction H₂O loses H⁺ (becoming OH⁻) and NH₃ gains it (becoming NH₄⁺). The proton donor, H₂O, is the Brønsted–Lowry acid; NH₃ is the base.
2 Which pair of species is a conjugate acid–base pair?
Answer and reasoning
H₃PO₄ and HPO₄²⁻ — H₃PO₄ and HPO₄²⁻ differ by two protons. A conjugate pair differs by one: the conjugate base of H₃PO₄ is H₂PO₄⁻.
CH₃COOH and NH₃ — CH₃COOH and NH₃ react with each other, but that makes them the acid and base of one reaction, not a conjugate pair. They do not differ by one proton; the pairs are CH₃COOH/CH₃COO⁻ and NH₄⁺/NH₃.
H₂PO₄⁻ and HPO₄²⁻ — The two species differ by exactly one proton: H₂PO₄⁻ − H⁺ = HPO₄²⁻. One H fewer and a charge one unit more negative is the signature of a conjugate pair.
HSO₄⁻ and SO₄⁻ — Losing a proton (H⁺) lowers the charge by one. The conjugate base of HSO₄⁻ is SO₄²⁻; 'SO₄⁻' results from removing a hydrogen atom without changing the charge.
3 Which statement about measuring the pH of an aqueous solution is correct?
Answer and reasoning
Universal indicator is as precise as a pH meter. — Universal indicator is read by matching a colour to a chart; each colour covers a range of pH and judging colour is subjective. It gives an estimate, typically to the nearest whole unit, not a precise value.
A new pH meter reads accurately before it is calibrated. — A pH meter must be calibrated with buffers of known pH before use, because the probe's response drifts with age and temperature. Without calibration its precise-looking readings can be inaccurate.
A calibrated pH meter can give the pH to about 0.01 unit. — A pH meter (probe), calibrated with buffer solutions of known pH, gives a numerical reading, typically to 0.01 pH unit. Universal indicator only estimates pH, usually to the nearest whole unit.
A reading of 7.00 shows a neutral solution at 323 K. — At 323 K Kw is larger, so a neutral solution has pH below 7 (about 6.6); a reading of 7.00 at 323 K is slightly basic. A solution is neutral when [H⁺] = [OH⁻], which corresponds to pH 7.00 only at 298 K.
4 A solution of nitric acid, HNO₃(aq), has a concentration of 0.0010 mol dm⁻³. Which statement about this solution is correct?
Answer and reasoning
It is a weak acid, as its concentration is low. — Weak and dilute are different ideas. A low concentration makes the solution dilute; HNO₃ is still a strong acid because it is fully ionized.
It contains mainly HNO₃ molecules with a few ions. — That describes a weak acid. HNO₃ is strong: it is effectively completely ionized, so the solution contains H⁺(aq) and NO₃⁻(aq) with essentially no HNO₃ molecules.
It is a dilute solution of an acid that is fully ionized. — HNO₃ is a strong acid, so it is fully ionized at this concentration: HNO₃ → H⁺ + NO₃⁻, and [H⁺] = 0.0010 mol dm⁻³. At 0.0010 mol dm⁻³ the solution is dilute. Strength (extent of ionization) and concentration are independent.
It is neutral, as this much water neutralizes the acid. — Dilution spreads the H⁺ through more water but does not remove it. [H⁺] = 1.0 × 10⁻³ mol dm⁻³ (pH 3), far greater than [OH⁻], so the solution is acidic.
5 Aqueous ethanoic acid is added to solid sodium carbonate until the reaction stops. What are the products of the reaction?
Answer and reasoning
sodium ethanoate and water, with no gas — 'Acid + base → salt + water' is incomplete for carbonates. CO₃²⁻ + 2H⁺ → H₂O + CO₂, so carbon dioxide is also produced (the mixture fizzes).
sodium ethanoate, hydrogen and carbon dioxide — Hydrogen is produced when an acid reacts with a reactive metal, not with a carbonate. The protons are accepted by CO₃²⁻, forming water and carbon dioxide.
none, as weak acids do not react with carbonates — Weak acids undergo all the typical acid reactions. As the carbonate removes H⁺, more ethanoic acid ionizes, so the reaction continues until one reactant is used up; vinegar fizzes with carbonates.
sodium ethanoate, water and carbon dioxide — Acid + carbonate → salt + water + carbon dioxide: 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂. The carbonate ion accepts two protons, forming H₂O and CO₂.
7 At 298 K the pKa of the ammonium ion, NH₄⁺, is 9.25 and the pKa of the methylammonium ion, CH₃NH₃⁺, is 10.66. Which statement about the bases NH₃ and CH₃NH₂ is correct? HL
Answer and reasoning
CH₃NH₂ is the stronger base, and its pKb is 1.31. — 1.31 is 14.00 ÷ 10.66, treating pKa × pKb = 14.00. Taking logarithms of Ka × Kb = Kw gives a sum: pKb = 14.00 − 10.66 = 3.34.
CH₃NH₂ is the stronger base, since its pKb is 3.34. — pKa + pKb = 14.00 for a conjugate pair at 298 K. pKb(CH₃NH₂) = 14.00 − 10.66 = 3.34 and pKb(NH₃) = 14.00 − 9.25 = 4.75. The smaller pKb means the stronger base, so CH₃NH₂ is stronger.
NH₃ is the stronger base, since its pKb is 9.25. — 9.25 is the pKa of NH₄⁺, the conjugate acid. The base has pKb = 14.00 − 9.25 = 4.75, and CH₃NH₂ (pKb 3.34) is the stronger base.
NH₃ is the stronger base, as its pKb of 4.75 is larger. — The pKb of 4.75 is right, but a larger pKb means a smaller Kb and a weaker base. CH₃NH₂, with pKb 3.34, is the stronger base.
Working pKb = 14.00 − pKa(conjugate acid). NH₃: 14.00 − 9.25 = 4.75. CH₃NH₂: 14.00 − 10.66 = 3.34. Smaller pKb, larger Kb (10⁻³·³⁴ = 4.6 × 10⁻⁴ against 10⁻⁴·⁷⁵ = 1.8 × 10⁻⁵), so CH₃NH₂ is the stronger base.
8 Which equation represents the hydrolysis reaction that determines the pH of aqueous ammonium chloride, NH₄Cl(aq)? HL
Answer and reasoning
Cl⁻(aq) + H₂O(l) ⇌ HCl(aq) + OH⁻(aq) — Cl⁻ is the conjugate base of a strong acid, so it is an extremely weak base and does not take H⁺ from water to any measurable extent. The hydrolysing ion is NH₄⁺.
NH₄Cl(s) + aq → NH₄⁺(aq) + Cl⁻(aq) — This is the salt dissolving. It produces no H₃O⁺ or OH⁻, so it cannot explain the pH. Hydrolysis is the next step: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺.
NH₄Cl(aq) + H₂O(l) ⇌ NH₄OH(aq) + HCl(aq) — Hydrolysis is not neutralization in reverse. HCl is strong and does not exist as molecules in water; only the NH₄⁺ ion reacts, donating H⁺ to water: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺.
NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq) — NH₄⁺ is the conjugate acid of the weak base NH₃, so it donates H⁺ to water, forming H₃O⁺. This makes NH₄Cl(aq) acidic. Cl⁻, the conjugate base of the strong acid HCl, does not hydrolyse measurably.
9 Universal indicator shows a different colour at almost every whole-number pH from about 1 to 14. What is the reason for this? HL
Answer and reasoning
It is a single weak acid whose one equilibrium shifts gradually across that range. — A single indicator has only two coloured forms (HInd and Ind⁻) and changes colour over about pKa ± 1. Covering 1 to 14 needs a mixture of indicators with different pKa values.
It reacts with the acid or alkali present, forming a differently coloured salt. — An indicator does not neutralize the solution; it is present in tiny amounts. Its colour depends on the position of its own equilibrium HInd ⇌ H⁺ + Ind⁻, which shifts with pH.
It is a mixture of indicators whose colour changes occur at different pH values. — Each component indicator changes colour over about 2 pH units around its own pKa. Mixing several indicators with different pKa values gives a sequence of colours across the whole range.
It changes colour at the pKa of whichever acid is being tested in the sample. — An indicator responds to [H⁺], and its colour change is set by its own pKa, not by the identity of the acid tested. Universal indicator is a mixture of indicators with a spread of pKa values.
10 A buffer solution contains 0.10 mol dm⁻³ CH₃COOH(aq) and 0.10 mol dm⁻³ CH₃COONa(aq). A small amount of HCl(aq) is added. Which statement about the buffer's response is correct? HL
Answer and reasoning
CH₃COO⁻ ions react with the added H⁺, forming CH₃COOH. — CH₃COO⁻ + H⁺ → CH₃COOH removes almost all the added H⁺. The ratio [CH₃COOH]/[CH₃COO⁻] rises only a little, so the pH falls only slightly.
CH₃COOH molecules react with the added H⁺, removing it. — CH₃COOH is an acid and does not take up H⁺. The base of the pair, CH₃COO⁻, removes added H⁺; CH₃COOH removes added OH⁻.
Na⁺ ions combine with the added Cl⁻, forming NaCl. — Na⁺ and Cl⁻ are spectator ions; they stay as separate hydrated ions and remove no H⁺. Buffering is done by the conjugate pair CH₃COOH/CH₃COO⁻.
The buffer holds [H⁺] fixed, so the pH cannot change at all. — A buffer reduces a change in pH; it does not prevent it. Converting some CH₃COO⁻ into CH₃COOH changes the ratio [CH₃COOH]/[CH₃COO⁻] slightly, so the pH falls slightly.
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40 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 Copper(II) oxide, CuO, is insoluble in water. It reacts with dilute sulfuric acid to form copper(II) sulfate and water. How should copper(II) oxide be classified?
Answer and reasoning
As an alkali, as it neutralizes an acid — Neutralizing an acid shows that CuO is a base, but 'alkali' means a base that is soluble in water and releases OH⁻(aq). Insoluble CuO is a base but not an alkali.
As a base, but it is not an alkali — CuO accepts protons from the acid (O²⁻ + 2H⁺ → H₂O), so it is a base. An alkali is a base that dissolves in water to release OH⁻(aq); CuO is insoluble, so it is not an alkali.
As neither, as it contains no OH⁻ ions — A Brønsted–Lowry base need not contain OH⁻; it must accept protons. The oxide ion in CuO accepts H⁺ to form water, so CuO is a base (though not an alkali).
As neither, since it is insoluble in water — Proton transfer does not require the base to dissolve in water first. CuO accepts protons from sulfuric acid, so it is a base; its insolubility is only why it is not an alkali.
2 Dihydrogenphosphate ions react with ammonia:
H₂PO₄⁻(aq) + NH₃(aq) ⇌ HPO₄²⁻(aq) + NH₄⁺(aq)
Which statement describes the role of H₂PO₄⁻ in the forward reaction?
Answer and reasoning
It is an acid, as it gives up a proton to NH₃ — H₂PO₄⁻ becomes HPO₄²⁻: it loses H⁺, which NH₃ accepts. So in this reaction H₂PO₄⁻ is a Brønsted–Lowry acid. With a stronger acid such as H₃O⁺, the same ion would act as a base.
It is a base, as it loses a proton to NH₃ — Losing a proton is the behaviour of an acid, not a base. H₂PO₄⁻ donates H⁺ to NH₃, so here it acts as a Brønsted–Lowry acid.
It is a base, since negative ions only accept H⁺ — Some anions contain an ionizable H and can donate it. H₂PO₄⁻ loses H⁺ in this reaction, so it acts as an acid, whatever its charge.
It is both an acid and a base in this reaction — H₂PO₄⁻ is amphiprotic, meaning it can act as either, but in one reaction it plays one role. Here it only loses H⁺ (to NH₃), so it acts as an acid.
3 10.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid is diluted with distilled water to a total volume of 1.00 dm³. What is the pH of the diluted solution?
Answer and reasoning
1.00 — 1.00 is the pH of the original 0.100 mol dm⁻³ acid. After a 100-fold dilution [H⁺] is 100 times smaller, so the pH is 2 units higher: 3.00.
3.00 — Amount of HCl = 0.100 mol dm⁻³ × 0.0100 dm³ = 1.00 × 10⁻³ mol. In 1.00 dm³, [H⁺] = 1.00 × 10⁻³ mol dm⁻³, so pH = −log₁₀(1.00 × 10⁻³) = 3.00. A 100-fold dilution raises the pH by 2 units.
0.00 — This comes from 0.100 × 10.0 / 1.00 = 1.00 mol dm⁻³, using 10.0 cm³ as if it were 10.0 dm³. Converting gives 0.0100 dm³, [H⁺] = 1.00 × 10⁻³ mol dm⁻³ and pH 3.00.
6.91 — 6.91 is −ln(1.00 × 10⁻³), the natural logarithm. pH uses log₁₀: −log₁₀(1.00 × 10⁻³) = 3.00.
4 At 298 K, solution P has [H⁺] = 1.0 × 10⁻³ mol dm⁻³ and solution Q has [H⁺] = 1.0 × 10⁻⁵ mol dm⁻³. Kw = 1.00 × 10⁻¹⁴ at 298 K. How does [OH⁻] in P compare with [OH⁻] in Q?
Answer and reasoning
It is 100 times larger in P than in Q. — This assumes [OH⁻] = [H⁺] in every solution. That is true only in a neutral solution. Because [H⁺][OH⁻] = Kw is constant, the solution with 100 times more H⁺ has 100 times less OH⁻.
It is the same, as acid has no effect on [OH⁻]. — [H⁺] and [OH⁻] are linked by Kw = [H⁺][OH⁻]. Adding acid shifts H₂O ⇌ H⁺ + OH⁻ to the left, lowering [OH⁻]: P has 1.0 × 10⁻¹¹ and Q 1.0 × 10⁻⁹ mol dm⁻³.
Neither contains OH⁻ ions, as both are acidic. — Every aqueous solution contains both H⁺ and OH⁻. Acidic solutions simply have [H⁺] > [OH⁻]: here [OH⁻] = Kw/[H⁺] = 1.0 × 10⁻¹¹ mol dm⁻³ in P and 1.0 × 10⁻⁹ mol dm⁻³ in Q.
It is 100 times smaller in P than in Q. — [OH⁻] = Kw/[H⁺]. P: 1.00 × 10⁻¹⁴ / 1.0 × 10⁻³ = 1.0 × 10⁻¹¹ mol dm⁻³; Q: 1.00 × 10⁻¹⁴ / 1.0 × 10⁻⁵ = 1.0 × 10⁻⁹ mol dm⁻³. P has 100 times the [H⁺] of Q, so it has 1/100 of the [OH⁻].
5 A student's data show that an aqueous solution at 323 K has [H⁺] = [OH⁻] = 2.34 × 10⁻⁷ mol dm⁻³, so its pH is 6.63. Which statement about this solution is correct?
Answer and reasoning
It is neutral, as its [H⁺] and [OH⁻] are equal. — A solution is neutral when [H⁺] = [OH⁻], at any temperature. pH 7.00 marks neutrality only at 298 K; at 323 K a neutral solution has a lower pH because water is more ionized.
It is acidic, as its pH is lower than 7.00. — pH 7.00 is the neutral point only at 298 K. The test is the relative size of [H⁺] and [OH⁻]; here they are equal, so the solution is neutral even though its pH is 6.63.
It is acidic, as it contains H⁺(aq) ions. — Every aqueous solution contains both H⁺ and OH⁻ ions. A solution is acidic only when [H⁺] > [OH⁻]; here they are equal, so it is neutral.
Both acidic and basic: each ion exceeds 1.0 × 10⁻⁷. — Acidic and basic are judged by comparing [H⁺] with [OH⁻], not by comparing each with its 298 K value. The two are equal here, so the solution is neutral; 1.0 × 10⁻⁷ mol dm⁻³ is the neutral value only at 298 K.
6 Hydrogen chloride reacts with water:
HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq)
The position of this equilibrium lies so far to the right that the reaction is effectively complete. Which statement about this reaction is correct?
Answer and reasoning
Cl⁻ is the acid, since HCl forms it. — Cl⁻ is what remains after HCl donates its proton: it is the conjugate base of HCl, not an acid. The acid in the forward reaction is HCl, the reactant that loses H⁺.
Cl⁻ is a weaker base than H₂O. — Acid–base equilibria lie on the side of the weaker acid and weaker base. The products, H₃O⁺ and Cl⁻, are favoured, so Cl⁻ is a weaker base than H₂O (and H₃O⁺ a weaker acid than HCl).
H₃O⁺ is a stronger acid than HCl. — An equilibrium lies on the side of the weaker acid. H₃O⁺ is favoured, so it is the weaker acid; HCl, which gives up its proton almost completely, is the stronger.
H₂O is the acid, as it gains a proton. — Gaining a proton makes H₂O the Brønsted–Lowry base here; HCl donates the proton and is the acid. Because the equilibrium lies to the right, the products are the weaker acid and base, so Cl⁻ is a weaker base than H₂O.
7 Separate 25.0 cm³ samples of 0.100 mol dm⁻³ HCl(aq) and 0.100 mol dm⁻³ CH₃COOH(aq) are each titrated with 0.100 mol dm⁻³ NaOH(aq). Which statement is correct?
Answer and reasoning
Both need the same volume of NaOH(aq), as the acid amounts are equal. — Both samples contain 2.50 × 10⁻³ mol of a monoprotic acid, so both need 2.50 × 10⁻³ mol of NaOH (25.0 cm³). As OH⁻ removes H⁺, the ethanoic acid equilibrium shifts to the right until all the acid has reacted.
The HCl needs more NaOH(aq), as it releases more H⁺ ions in water. — HCl does have the higher [H⁺] at the start, but the amount of base needed depends on the amount of acid. Ethanoic acid keeps ionizing as its H⁺ is removed, so both need 25.0 cm³ of NaOH(aq).
Both acids start at the same pH, as their acid concentrations are equal. — Only a strong acid has [H⁺] equal to its concentration. HCl is fully ionized (pH 1.0); ethanoic acid is only slightly ionized, so its [H⁺] is lower and its starting pH higher.
The CH₃COOH starts at the lower pH, as each molecule has four H atoms. — Only the H of the –COOH group in ethanoic acid can ionize, and only partly. CH₃COOH is a weak monoprotic acid, so its [H⁺] is lower than that of HCl and its pH is higher.
8 Methylammonium chloride, CH₃NH₃Cl, is a salt. Which statement about this salt is correct?
Answer and reasoning
It forms from HCl and CH₃NH₂, with water as a second product. — An amine, like ammonia, simply gains H⁺: CH₃NH₂ + H⁺ → CH₃NH₃⁺. No OH⁻ or O²⁻ is involved, so no water forms; the salt is the only product.
It forms from HCl and CH₃NH₂, with no water produced. — CH₃NH₂ + HCl → CH₃NH₃Cl. The amine accepts a proton on the lone pair of its nitrogen atom, so the only product is the salt. Its parent acid is HCl and its parent base is the amine CH₃NH₂.
Its parent base is CH₃NH₃OH, since a base must contain OH. — CH₃NH₃OH, written by analogy with 'NH₄OH', is not the parent base. An amine is a base because the lone pair on its nitrogen atom accepts H⁺, and it contains no OH: CH₃NH₂ + HCl → CH₃NH₃Cl. The parent base is CH₃NH₂.
Its parent base is CH₃NH₃⁺, the cation present in the salt. — CH₃NH₃⁺ is the ion in the salt, formed when the base gained a proton. Removing that proton gives the parent base, CH₃NH₂.
9 25.0 cm³ of 0.100 mol dm⁻³ HCl(aq) is placed in a flask and 25.0 cm³ of distilled water is added. The mixture is then titrated with 0.100 mol dm⁻³ NaOH(aq). At what pH does the pH curve intercept the pH axis (0.00 cm³ of NaOH(aq) added)?
Answer and reasoning
1.30 — n(HCl) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol in 0.0500 dm³, so [H⁺] = 0.0500 mol dm⁻³ and pH = −log₁₀(0.0500) = 1.30.
1.00 — 1.00 is the pH of the undiluted 0.100 mol dm⁻³ acid. The added water doubles the volume, halving [H⁺] to 0.0500 mol dm⁻³, so the pH is 1.30.
3.00 — 3.00 is −ln(0.0500), the natural logarithm. pH uses log₁₀: −log₁₀(0.0500) = 1.30.
2.60 — 2.60 is −log₁₀(2.50 × 10⁻³), using the amount of HCl in mol as if it were a concentration. Dividing by the volume, 0.0500 dm³, gives [H⁺] = 0.0500 mol dm⁻³ and pH 1.30.
10 25.0 cm³ of 0.100 mol dm⁻³ HCl(aq) is titrated with 0.100 mol dm⁻³ NaOH(aq) at 298 K until 50.0 cm³ of NaOH(aq) has been added. Which statement about the pH curve is correct?
Answer and reasoning
After 25.0 cm³ the pH stays at 7, as all the acid is neutralized. — Beyond 25.0 cm³ the added NaOH has no acid left to react with, so OH⁻ accumulates and the pH keeps rising, levelling off at about 12.5 by 50.0 cm³.
The pH rises by an equal amount for each 1.00 cm³ of NaOH(aq) added. — pH is logarithmic. Early additions remove only a small fraction of the H⁺, so the pH hardly changes; near 25.0 cm³ it jumps from about 3 to about 11 within 1 cm³.
The curve starts near pH 13, the pH of the NaOH(aq) added. — The intercept is the pH of the solution in the flask before any titrant is added: 0.100 mol dm⁻³ HCl, pH 1.0. The curve rises towards, but does not reach, the pH of the NaOH(aq).
At 25.0 cm³ the pH is 7, as only NaCl(aq) and water are present. — 25.0 cm³ of NaOH(aq) contains the same amount as the HCl, so this is the equivalence point. The solution contains only NaCl and water, and a strong acid–strong base salt gives pH 7 at 298 K.
11 At 298 K the pKa of methanoic acid, HCOOH, is 3.75 and the pKa of ethanoic acid, CH₃COOH, is 4.76. Which statement about 0.10 mol dm⁻³ solutions of the two acids is correct? HL
Answer and reasoning
Ethanoic acid is the stronger acid, as its pKa is the larger. — pKa = −log₁₀Ka, so a larger pKa means a smaller Ka and a weaker acid. Ethanoic acid (pKa 4.76) is weaker than methanoic acid (pKa 3.75).
Both are ionized to the same extent, as both are weak acids. — 'Weak' covers a wide range of strengths. The pKa values differ by about 1, so methanoic acid has a Ka about ten times larger and is ionized to a greater extent.
Methanoic acid needs more NaOH(aq) to neutralize the same volume of it. — Equal volumes of the two 0.10 mol dm⁻³ monoprotic acids contain equal amounts of acid, so they need the same amount of NaOH, whatever their strengths.
Methanoic acid is ionized to a greater extent than ethanoic acid. — The smaller pKa means the larger Ka (1.8 × 10⁻⁴ against 1.7 × 10⁻⁵), so methanoic acid is the stronger acid and is ionized to a greater extent at the same concentration.
12 What is the pH of 0.100 mol dm⁻³ ethanoic acid, CH₃COOH(aq), at 298 K? Ka = 1.74 × 10⁻⁵. Assume that the equilibrium concentration of CH₃COOH equals its initial concentration. HL
Answer and reasoning
2.88 — Ka = [H⁺]²/[CH₃COOH], so [H⁺] = √(1.74 × 10⁻⁵ × 0.100) = 1.32 × 10⁻³ mol dm⁻³ and pH = 2.88.
1.00 — 1.00 is −log₁₀(0.100), which treats ethanoic acid as fully ionized. It is weak, so [H⁺] = √(Ka × [HA]) = 1.32 × 10⁻³ mol dm⁻³ and the pH is 2.88.
5.76 — 5.76 is −log₁₀(Ka × [HA]), without the square root. Ka × [HA] equals [H⁺]², so [H⁺] = √(1.74 × 10⁻⁶) = 1.32 × 10⁻³ mol dm⁻³ and pH = 2.88.
4.76 — 4.76 is the pKa. pH = pKa only when [CH₃COOH] = [CH₃COO⁻], as at half-equivalence. In the acid alone [CH₃COO⁻] is far smaller, and pH = 2.88.
13 Aqueous solutions of sodium carbonate, Na₂CO₃, and sodium hydrogencarbonate, NaHCO₃, each of concentration 0.10 mol dm⁻³, are prepared at 298 K. pKa(H₂CO₃) = 6.35 and pKa(HCO₃⁻) = 10.33. Which statement is correct? HL
Answer and reasoning
NaHCO₃(aq) has the higher pH, since HCO₃⁻ has the larger pKa (10.33). — 10.33 is the pKa of HCO₃⁻ acting as an acid; it describes CO₃²⁻ as the base of that pair (pKb 3.67). HCO₃⁻ acting as a base has pKb = 14.00 − 6.35 = 7.65, so it is the weaker base and NaHCO₃(aq) has the lower pH.
NaHCO₃(aq) is acidic, as HCO₃⁻ contains an H⁺ that it can donate. — HCO₃⁻ is amphiprotic, but its strength as a base (pKb 7.65) exceeds its strength as an acid (pKa 10.33), so its hydrolysis to OH⁻ dominates and NaHCO₃(aq) is weakly basic.
Na₂CO₃(aq) has the higher pH, as CO₃²⁻ has the smaller pKb (3.67). — CO₃²⁻ is the conjugate base of HCO₃⁻: pKb = 14.00 − 10.33 = 3.67. HCO₃⁻ as a base is the conjugate base of H₂CO₃: pKb = 14.00 − 6.35 = 7.65. CO₃²⁻ is the much stronger base, so CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻ produces more OH⁻.
Both solutions are at pH 7, as each salt is a product of neutralization. — Both anions come from a weak acid and react with water to form OH⁻: CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻ and HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻. Both solutions are basic.
14 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid (pKa = 4.76) is titrated with 0.100 mol dm⁻³ NaOH(aq) at 298 K. What is the pH after 12.5 cm³ of NaOH(aq) has been added? HL
Answer and reasoning
9.24 — 9.24 is 14.00 − 4.76, the pKb of CH₃COO⁻. At the half-equivalence point of a weak acid it is the pH that equals pKa: pH = 4.76.
8.73 — 8.73 is the pH at the equivalence point (25.0 cm³), where only CH₃COO⁻ remains. At 12.5 cm³, half the acid remains, [CH₃COOH] = [CH₃COO⁻] and pH = pKa = 4.76.
4.76 — 12.5 cm³ is half the equivalence volume (25.0 cm³). Half of the CH₃COOH has been converted to CH₃COO⁻, so [CH₃COOH] = [CH₃COO⁻], [H⁺] = Ka and pH = pKa = 4.76.
1.48 — 1.48 treats the remaining 0.0333 mol dm⁻³ CH₃COOH as fully ionized. It is weak, and with an equal concentration of CH₃COO⁻ present, pH = pKa = 4.76.
Working n(CH₃COOH) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol; equivalence at 25.0 cm³ NaOH. After 12.5 cm³: n(NaOH) = 1.25 × 10⁻³ mol, leaving 1.25 × 10⁻³ mol CH₃COOH and forming 1.25 × 10⁻³ mol CH₃COO⁻. [CH₃COOH] = [CH₃COO⁻], so [H⁺] = Ka and pH = pKa = 4.76.
15 25.0 cm³ of a 0.100 mol dm⁻³ aqueous solution of a base, B, is titrated with 0.100 mol dm⁻³ HCl(aq) at 298 K. The pH curve starts at pH 11.1 and falls gently through a region in which the pH is 9.25 at 12.5 cm³. It then falls steeply near 25.0 cm³, through an equivalence point at pH 5.3, and levels off near pH 1.5 by 50.0 cm³. Which statement about B is correct? HL
Answer and reasoning
B is a weak base, and the pKb of B itself is 9.25. — At half-equivalence it is the pOH that equals pKb. The pH there, 9.25, equals the pKa of the conjugate acid, so pKb = 14.00 − 9.25 = 4.75.
B is a weak base, and the pKa of its conjugate acid is 9.25. — A 0.100 mol dm⁻³ strong base would start at pH 13.0, and the buffer region and acidic equivalence point (5.3) also show a weak base. At half-equivalence (12.5 cm³) [B] = [BH⁺], so pOH = pKb and pH = pKa(BH⁺) = 9.25. (These data fit ammonia.)
B is a strong base, as the curve has a steep section. — The steep section appears because the titrant, HCl, is strong. The starting pH of 11.1 (not 13.0), the buffer region and the equivalence point below 7 all show that B is weak.
B is a strong base, as the curve starts above pH 11. — pH depends on concentration as well as strength. A 0.100 mol dm⁻³ strong base would give pH 13.0; pH 11.1 at this concentration shows that B is only partly ionized, i.e. weak.
16 An indicator, HInd, has pKa = 3.7:
HInd(aq) ⇌ H⁺(aq) + Ind⁻(aq)
HInd is red and Ind⁻ is yellow. What colour does the indicator show in a solution of pH 6.0, and why? HL
Answer and reasoning
Yellow, as Ind⁻ predominates when the pH is above the pKa. — At pH 6.0, [H⁺] = 1 × 10⁻⁶ mol dm⁻³, far below Ka = 2 × 10⁻⁴, so the equilibrium lies to the right: [Ind⁻]/[HInd] = Ka/[H⁺] ≈ 200. The solution shows the colour of Ind⁻, yellow.
Red, as the solution is acidic, with its pH below 7. — An indicator switches colour at pH ≈ its own pKa (3.7 here), not at 7. At pH 6.0 the solution is acidic, but [H⁺] is too low to keep this indicator in its HInd form, so it is yellow.
Red, as HInd predominates when the pH is above the pKa. — Above the pKa, [H⁺] < Ka, so [Ind⁻]/[HInd] = Ka/[H⁺] > 1: Ind⁻ predominates. Raising the pH removes H⁺ and shifts the equilibrium to the right, giving yellow.
Orange, as HInd and Ind⁻ are equal at equilibrium. — Equilibrium means equal rates, not equal amounts. [HInd] = [Ind⁻] only when pH = pKa (3.7). At pH 6.0, Ind⁻ is about 200 times more concentrated, so the colour is yellow.
17 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid (pKa = 4.76) is titrated with 0.100 mol dm⁻³ NaOH(aq) at 298 K. Three indicators are available, with these pH ranges: methyl orange 3.1–4.4; bromothymol blue 6.0–7.6; phenolphthalein 8.3–10.0. Which choice of indicator, with its reason, is correct? HL
Answer and reasoning
Phenolphthalein, as the salt CH₃COONa makes the equivalence pH above 7. — At equivalence the flask contains sodium ethanoate; CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻ makes the solution basic (about pH 8.7, with [CH₃COO⁻] = 0.0500 mol dm⁻³). Phenolphthalein's range, 8.3–10.0, lies within the steep section.
Methyl orange, as its range is close to the pKa of ethanoic acid. — The indicator must change colour where the pH curve is steep, near equivalence. pH 4.76 lies in the buffer region, far before equivalence; methyl orange would change colour much too early.
Bromothymol blue, as the equivalence point of a titration is at pH 7. — Only strong acid–strong base titrations have equivalence at pH 7. Here the salt, CH₃COONa, is basic, so equivalence is at about pH 8.7. Bromothymol blue's range (6.0–7.6) lies mostly below the steep section, so its colour starts to change more than 1 cm³ before equivalence. Phenolphthalein (8.3–10.0) is the appropriate indicator.
Any of the three, as the pH passes through every range near equivalence. — With a weak acid the steep section is short, from about pH 7 to 11 here. The pH passes through methyl orange's range (3.1–4.4) in the buffer region, long before equivalence, so the choice matters.
18 Which statement correctly distinguishes the end point of a titration from its equivalence point? HL
Answer and reasoning
The end point and the equivalence point are the same point by definition, so the two terms can be used interchangeably in any titration. — They are defined differently: one by the amounts reacted, the other by the indicator's colour change. With an unsuitable indicator they can be several cm³ apart.
The equivalence point is the point where the pH equals 7; the end point is the pH at which the colour of the chosen indicator changes. — The equivalence point is where the amounts are in the stoichiometric ratio, and its pH depends on the salt formed: above 7 for a weak acid–strong base titration, below 7 for strong acid–weak base.
The end point is where the pH curve levels off after its steep section, which shows that the reaction between the acid and the base has finished. — The reaction is complete at the equivalence point, within the steep section. The level region beyond it is excess titrant; an indicator changing colour there would give too large a titre.
The end point is where the indicator changes colour; the equivalence point is where acid and base have reacted in the stoichiometric ratio. — The equivalence point is defined by stoichiometry; the end point by an observation. A well-chosen indicator makes the end point coincide closely with the equivalence point.
19 Which mixture, made up to 1.00 dm³ with water, is a buffer solution? HL
Answer and reasoning
CH₃COOH (0.30 mol) with NaOH (0.30 mol) — The amounts are equal, so the reaction CH₃COOH + NaOH → CH₃COONa + H₂O uses up all the acid. Only CH₃COO⁻ remains, so there is no weak acid to remove added OH⁻: not a buffer.
NH₃ (0.30 mol) with HCl (0.15 mol) — HCl reacts with half the ammonia: NH₃ + HCl → NH₄Cl, leaving 0.15 mol NH₃ and 0.15 mol NH₄⁺. A weak base with its conjugate acid in similar amounts is a basic buffer.
HCl (0.15 mol) with NaCl (0.30 mol) — HCl is a strong acid and Cl⁻ is too weak a base to remove added H⁺. A buffer needs a weak acid with its conjugate base (or a weak base with its conjugate acid).
NaCl (0.30 mol) with KNO₃ (0.15 mol) — These salts give a neutral solution but contain no weak acid or weak base, so nothing removes added H⁺ or OH⁻. A buffer is defined by resisting pH change, not by being at pH 7.
20 A buffer solution at 298 K contains 0.20 mol dm⁻³ CH₃COOH and 0.10 mol dm⁻³ CH₃COONa. Ka(CH₃COOH) = 1.74 × 10⁻⁵. What is the pH of the buffer? HL
Answer and reasoning
5.06 — 5.06 comes from Ka × [CH₃COO⁻]/[CH₃COOH], the ratio inverted. Rearranging Ka = [H⁺][A⁻]/[HA] gives [H⁺] = Ka[HA]/[A⁻] = 3.48 × 10⁻⁵ mol dm⁻³, pH 4.46.
4.76 — 4.76 is the pKa, which equals the pH only when [CH₃COOH] = [CH₃COO⁻]. Here the acid is twice as concentrated, so [H⁺] = 2Ka and pH = 4.76 − 0.30 = 4.46.
4.46 — [H⁺] = Ka × [CH₃COOH]/[CH₃COO⁻] = 1.74 × 10⁻⁵ × 0.20/0.10 = 3.48 × 10⁻⁵ mol dm⁻³, so pH = 4.46. With more acid than salt, the pH is below the pKa (4.76).
0.70 — 0.70 is −log₁₀(0.20), treating the ethanoic acid as fully ionized. It is weak, and in the buffer [H⁺] = Ka[HA]/[A⁻] = 3.48 × 10⁻⁵ mol dm⁻³, pH 4.46.
Working Ka = [H⁺][CH₃COO⁻]/[CH₃COOH], so [H⁺] = Ka × [CH₃COOH]/[CH₃COO⁻] = 1.74 × 10⁻⁵ × (0.20/0.10) = 3.48 × 10⁻⁵ mol dm⁻³. pH = −log₁₀(3.48 × 10⁻⁵) = 4.46.
21 A buffer containing 0.20 mol dm⁻³ CH₃COOH and 0.10 mol dm⁻³ CH₃COO⁻ has pH 4.46. It is diluted with an equal volume of water. To a first approximation, what happens to its pH? HL
Answer and reasoning
It rises by about 0.30, as [H⁺] halves when the volume doubles. — That would be true for a strong acid. In a buffer [H⁺] = Ka[HA]/[A⁻], and dilution changes [HA] and [A⁻] by the same factor, so [H⁺] and the pH stay essentially the same.
It stays at about 4.46, as the ratio [CH₃COOH]/[CH₃COO⁻] is unchanged. — Both concentrations halve (to 0.10 and 0.050 mol dm⁻³), so the ratio stays 2 : 1 and [H⁺] = Ka × 2 is unchanged: pH 4.46. Dilution does reduce the buffer's capacity.
It rises to pH 7, as the added water neutralizes the acid present. — Water does not neutralize the acid; both buffer components are still present in the same ratio. The pH stays at about 4.46.
It falls, as dilution makes a larger fraction of CH₃COOH ionize. — A larger fraction ionized does not mean a higher [H⁺]. In the buffer, [H⁺] = Ka[HA]/[A⁻], and the ratio is unchanged by dilution, so the pH stays about 4.46.
22 Solid magnesium oxide, MgO, is added to dilute nitric acid, HNO₃(aq). Which statement about what happens is correct?
Answer and reasoning
MgO and HNO₃ react in a 1 : 1 mole ratio, as NaOH and HCl do too. — NaOH and HCl react 1 : 1 because OH⁻ accepts one proton. O²⁻ accepts two (O²⁻ + 2H⁺ → H₂O), and Mg²⁺ needs two NO₃⁻: MgO + 2HNO₃ → Mg(NO₃)₂ + H₂O, so the ratio is 1 : 2.
No reaction occurs, as MgO is not a base: it has no hydroxide ions. — A Brønsted–Lowry base need not contain OH⁻. The oxide ion accepts protons, forming water, so MgO is a base and neutralizes nitric acid: MgO + 2HNO₃ → Mg(NO₃)₂ + H₂O.
Hydrogen gas is released, as acids give off H₂ in their reactions. — Hydrogen forms when an acid reacts with a reactive metal. With a metal oxide the protons are accepted by O²⁻, forming water: MgO + 2HNO₃ → Mg(NO₃)₂ + H₂O. No gas is produced.
MgO and HNO₃ react in a 1 : 2 mole ratio, as O²⁻ accepts two H⁺. — MgO(s) + 2HNO₃(aq) → Mg(NO₃)₂(aq) + H₂O(l). The oxide ion accepts two H⁺ (O²⁻ + 2H⁺ → H₂O), and the salt Mg(NO₃)₂ needs two NO₃⁻ for each Mg²⁺, so 1 mol of MgO reacts with 2 mol of HNO₃.
Working MgO(s) + 2HNO₃(aq) → Mg(NO₃)₂(aq) + H₂O(l). Atoms: Mg 1 = 1; N 2 = 2; O 1 + 6 = 6 + 1; H 2 = 2. O²⁻ + 2H⁺ → H₂O, so n(HNO₃) : n(MgO) = 2 : 1.
23 Solid sodium hydrogencarbonate, NaHCO₃, is added to dilute hydrochloric acid, HCl(aq). Which statement about the reaction is correct?
Answer and reasoning
NaHCO₃ and HCl react in a 1 : 1 mole ratio, as HCO₃⁻ accepts one H⁺. — NaHCO₃(s) + HCl(aq) → NaCl(aq) + H₂O(l) + CO₂(g). HCO₃⁻ accepts one H⁺ and the product breaks down to water and carbon dioxide, which is released as a gas: HCO₃⁻ + H⁺ → H₂O + CO₂.
NaHCO₃ and HCl react in a 1 : 2 mole ratio, as carbonates do. — The 1 : 2 ratio belongs to the carbonate ion, which accepts two protons (CO₃²⁻ + 2H⁺ → H₂O + CO₂). HCO₃⁻ already carries one H, so it accepts only one: NaHCO₃ + HCl → NaCl + H₂O + CO₂.
Only NaCl and water form, as neutralization gives salt and water. — 'Acid + base → salt + water' is incomplete for hydrogencarbonates. HCO₃⁻ + H⁺ → H₂O + CO₂, so carbon dioxide is also produced and the mixture fizzes.
Hydrogen gas is released, as acids give H₂ when they react. — Hydrogen forms when an acid reacts with a reactive metal. Here the proton is accepted by HCO₃⁻, and the gas released is carbon dioxide: NaHCO₃ + HCl → NaCl + H₂O + CO₂.
Working NaHCO₃(s) + HCl(aq) → NaCl(aq) + H₂O(l) + CO₂(g). Atoms: Na 1 = 1; H 1 + 1 = 2; C 1 = 1; O 3 = 1 + 2; Cl 1 = 1. HCO₃⁻ + H⁺ → H₂O + CO₂, so n(HCl) : n(NaHCO₃) = 1 : 1.
24 Which equation represents the hydrolysis reaction that determines the pH of aqueous sodium ethanoate, CH₃COONa(aq)? HL
Answer and reasoning
CH₃COONa(aq) + H₂O(l) ⇌ CH₃COOH(aq) + NaOH(aq) — Hydrolysis is not neutralization in reverse. NaOH is a strong base and does not exist as a molecular product in water, and Na⁺ takes no part; only the CH₃COO⁻ ion reacts: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻.
CH₃COONa(s) + aq → CH₃COO⁻(aq) + Na⁺(aq) — This is the salt dissolving. It produces no H₃O⁺ or OH⁻, so it cannot explain the pH. Hydrolysis is the next step: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻.
CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq) — CH₃COO⁻ is the conjugate base of the weak acid CH₃COOH, so it accepts H⁺ from water, releasing OH⁻. This makes CH₃COONa(aq) basic. Na⁺, the cation of the strong base NaOH, does not hydrolyse measurably.
CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq) — This is the ionization of ethanoic acid, which is not what is dissolved. Sodium ethanoate supplies CH₃COO⁻, which accepts H⁺ from water: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, making the solution basic.
25 Which mixture, made up to 1.00 dm³ with water, is an acidic buffer solution? HL
Answer and reasoning
HNO₃ (0.10 mol) with NaNO₃ (0.20 mol) — HNO₃ is a strong acid, and NO₃⁻ is too weak a base to remove added H⁺. A buffer needs a weak acid with its conjugate base (or a weak base with its conjugate acid).
CH₃COOH (0.10 mol) with NaOH (0.20 mol) — NaOH is in excess, so all the ethanoic acid is converted to CH₃COO⁻ and 0.10 mol of OH⁻ remains. With no weak acid left to remove added OH⁻, this is not a buffer.
NH₃ (0.20 mol) with NH₄Cl (0.10 mol) — This is a buffer, but a basic one: it is built on the weak base NH₃ with its conjugate acid. Its pH is 9.25 + log₁₀(0.20/0.10) = 9.55. NH₄Cl being an acidic salt does not make the buffer acidic.
CH₃COOH (0.20 mol) with KOH (0.10 mol) — KOH converts half the ethanoic acid into its conjugate base: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O, leaving 0.10 mol CH₃COOH and 0.10 mol CH₃COO⁻. A weak acid with its conjugate base in similar amounts is an acidic buffer (pH = pKa = 4.76).
26 25.0 cm³ of 0.100 mol dm⁻³ CH₃COOH(aq) (pKa = 4.76) is titrated with 0.100 mol dm⁻³ NH₃(aq) (pKb = 4.75) at 298 K. Which statement describes the pH curve? HL
Answer and reasoning
It rises almost vertically near 25.0 cm³, from about pH 3 to about pH 11. — A long vertical section from about pH 3 to 11 is typical of strong acid–strong base titrations. With a weak acid and a weak base the pH changes gradually, by less than 2 units between 24.0 and 26.0 cm³.
It has no near-vertical section near 25.0 cm³, as acid and base are both weak. — With both partners weak, the pH changes only gradually through the equivalence point (from about 6.1 at 24.0 cm³ to about 7.9 at 26.0 cm³), passing about pH 7 at 25.0 cm³ because CH₃COO⁻ and NH₄⁺ hydrolyse to similar extents.
It starts near pH 11, as that is the pH of the aqueous ammonia added. — The intercept is the pH of the solution in the flask before any titrant is added: 0.100 mol dm⁻³ CH₃COOH, pH 2.88. About pH 11 is the pH of the ammonia solution in the burette.
It reaches equivalence well above pH 7, as the acid being titrated is weak. — Equivalence is above 7 for a weak acid with a strong base. Here the base is also weak: NH₄⁺ in the salt makes the solution acidic just as CH₃COO⁻ makes it basic, and with pKa 4.76 and pKb 4.75 the equivalence pH is about 7.
27 Solution X is 0.0010 mol dm⁻³ KOH(aq). Solution Y is 2.0 mol dm⁻³ NH₃(aq). Which statement is correct?
Answer and reasoning
NH₃ is the stronger base, as solution Y is more concentrated. — Concentration and strength are different ideas. Y is more concentrated, but NH₃ is a weak base: fewer than 1 in 100 of its molecules have reacted with water. KOH is fully ionized, so it is the stronger base.
NH₃ is not a base, as its formula contains no OH group. — A Brønsted–Lowry base need not contain OH. The lone pair on the nitrogen of NH₃ accepts H⁺ from water, forming NH₄⁺ and OH⁻, so NH₃ is a (weak) base.
KOH is the stronger base, as it is fully ionized. — Strength is the extent of ionization, not concentration. KOH, a group 1 hydroxide, is fully ionized even in this dilute solution; NH₃ reacts only partly with water (NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ lies well to the left), even at 2.0 mol dm⁻³.
Both are strong bases, as both release OH⁻ ions in water. — Releasing OH⁻ makes both of them alkalis, not both strong. NH₃ produces OH⁻ only by reacting partly with water, and NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ lies well to the left. Only KOH is fully ionized.
28 10.0 cm³ of hydrochloric acid of pH 1.70 is diluted with distilled water to a total volume of 1.00 dm³ at 298 K. What is [H⁺] in the diluted solution?
Answer and reasoning
0.00020 mol dm⁻³ — [H⁺] = 10⁻¹·⁷⁰ = 0.0200 mol dm⁻³ in the original acid. Amount of H⁺ = 0.0200 × 0.0100 dm³ = 2.00 × 10⁻⁴ mol, in 1.00 dm³, so [H⁺] = 0.00020 mol dm⁻³ (a 100-fold dilution).
0.020 mol dm⁻³ — 0.020 mol dm⁻³ is [H⁺] in the original acid, 10⁻¹·⁷⁰. Diluting 10.0 cm³ to 1.00 dm³ makes the solution 100 times less concentrated: [H⁺] = 0.00020 mol dm⁻³.
0.50 mol dm⁻³ — This uses 10¹·⁷⁰ = 50 mol dm⁻³ for the original acid, dropping the minus sign: [H⁺] = 10⁻ᵖᴴ. A solution of pH above 0 has [H⁺] below 1 mol dm⁻³; here 0.020 mol dm⁻³, diluted to 0.00020 mol dm⁻³.
0.0018 mol dm⁻³ — This uses e⁻¹·⁷⁰ = 0.18, the natural antilogarithm. pH is defined with log₁₀, so [H⁺] = 10⁻¹·⁷⁰ = 0.020 mol dm⁻³, diluted 100-fold to 0.00020 mol dm⁻³.
Working [H⁺] in the original acid = 10⁻ᵖᴴ = 10⁻¹·⁷⁰ = 0.0200 mol dm⁻³. n(H⁺) = 0.0200 mol dm⁻³ × (10.0/1000) dm³ = 2.00 × 10⁻⁴ mol. [H⁺] after dilution = 2.00 × 10⁻⁴ mol ÷ 1.00 dm³ = 0.00020 mol dm⁻³ (2.0 × 10⁻⁴ mol dm⁻³).
29 Which is a correct equation for HCO₃⁻ acting as a Brønsted–Lowry base in the forward reaction?
Answer and reasoning
HCO₃⁻(aq) + H₃O⁺(aq) ⇌ H₂CO₃⁻(aq) + H₂O(l) — Gaining H⁺ raises the charge by one unit, so HCO₃⁻ becomes H₂CO₃, not H₂CO₃⁻. As written, the charge does not balance: 0 on the left, −1 on the right.
HCO₃⁻(aq) + OH⁻(aq) ⇌ CO₃²⁻(aq) + H₂O(l) — Here HCO₃⁻ loses a proton to OH⁻, becoming CO₃²⁻: that is acid behaviour. A Brønsted–Lowry base accepts a proton, as HCO₃⁻ does with H₃O⁺.
H₂CO₃(aq) + CH₃COO⁻(aq) ⇌ HCO₃⁻(aq) + CH₃COOH(aq) — In the forward reaction HCO₃⁻ is formed, when H₂CO₃ donates a proton to CH₃COO⁻; it is the conjugate base of H₂CO₃ but is not acting as a base. It accepts a proton only in the reverse reaction.
HCO₃⁻(aq) + H₃O⁺(aq) ⇌ H₂CO₃(aq) + H₂O(l) — HCO₃⁻ gains a proton from H₃O⁺, becoming H₂CO₃: gaining H⁺ is what a Brønsted–Lowry base does. Its charge rises by one, from −1 to 0, and the equation balances in atoms and charge.
30 What is the formula of the conjugate base of ethanoic acid, CH₃COOH?
Answer and reasoning
CH₃COO — A proton is H⁺, not a hydrogen atom. Removing H⁺ from a neutral molecule leaves a species with a charge of −1: CH₃COO⁻, not the neutral CH₃COO.
CH₃COO⁻ — The conjugate base is formed when the acid donates one proton, H⁺. Ethanoic acid loses the hydrogen of its –COOH group, so the formula loses one H and the charge falls by one: CH₃COOH → CH₃COO⁻ (the ethanoate ion).
CH₃COOH₂⁺ — CH₃COOH₂⁺ is formed when ethanoic acid GAINS a proton, so it is the conjugate acid of CH₃COOH. The conjugate base is the species left after the acid has donated its proton, CH₃COO⁻.
CH₂COOH⁻ — Only the hydrogen of the –COOH group is released as a proton; the C–H hydrogens of the methyl group are not acidic. Losing the O–H proton gives CH₃COO⁻.
31 Barium hydroxide, Ba(OH)₂, is a strong base that dissociates completely in water. What is the pH of 0.050 mol dm⁻³ Ba(OH)₂(aq) at 298 K? HL
Answer and reasoning
12.70 — This takes [OH⁻] = 0.050 mol dm⁻³, the concentration of the base. Ba(OH)₂ releases two OH⁻ ions per formula unit, so [OH⁻] = 0.10 mol dm⁻³, pOH = 1.00 and pH = 13.00.
14.00 — This divides 14.00 by pOH. The p-values ADD: pH + pOH = 14.00 at 298 K, because taking −log₁₀ of Kw = [H⁺][OH⁻] turns the product into a sum. pH = 14.00 − 1.00 = 13.00.
13.00 — Each formula unit gives two OH⁻ ions, so [OH⁻] = 0.10 mol dm⁻³ and pOH = 1.00. pH = 14.00 − 1.00 = 13.00.
11.70 — This uses the natural logarithm: −ln(0.10) = 2.30. pOH is defined with log₁₀, so pOH = −log₁₀(0.10) = 1.00 and pH = 13.00.
Working Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻(aq), so [OH⁻] = 2 × 0.050 = 0.10 mol dm⁻³. pOH = −log₁₀(0.10) = 1.00. At 298 K, pH + pOH = 14.00, so pH = 14.00 − 1.00 = 13.00.
32 The table shows data at 298 K for four weak bases.
W: Kb = 4.4 × 10⁻⁴
X: pKb = 4.75
Y: Kb = 1.6 × 10⁻⁶
Z: pKb = 9.4
Which statement is correct? HL
Answer and reasoning
Z is a stronger base than Y, as pKb(Z) is larger than pKb(Y). — pKb = −log₁₀Kb, so a LARGER pKb means a SMALLER Kb and a weaker base. pKb(Y) = 5.80 and pKb(Z) = 9.4, so Z is the weaker of the two.
Y is a stronger base than W if Y is the more concentrated. — Strength is the extent of ionization and is fixed by Kb, not by how much base is dissolved. W (Kb = 4.4 × 10⁻⁴) is a stronger base than Y (Kb = 1.6 × 10⁻⁶) at any concentration; a concentrated solution of Y is concentrated, not strong.
X and Y are equally strong, as both are weak bases. — Weak bases differ in strength, and Kb or pKb measures it. Kb(X) = 1.8 × 10⁻⁵ is about ten times Kb(Y) = 1.6 × 10⁻⁶, so X is the stronger base of the two.
W is a stronger base than X, as Kb(W) is larger than Kb(X). — Kb(X) = 10⁻⁴·⁷⁵ = 1.8 × 10⁻⁵, which is smaller than Kb(W) = 4.4 × 10⁻⁴. A larger Kb (smaller pKb: 3.36 for W against 4.75 for X) means the base ionizes to a greater extent, so W is the stronger base.
33 25.0 cm³ of 0.100 mol dm⁻³ NaOH(aq) is placed in a flask and 0.100 mol dm⁻³ HCl(aq) is added from a burette. The figure shows four sketched pH curves, A to D. Which curve shows how the pH in the flask changes?
Answer and reasoning
Curve C — The flask starts with 0.100 mol dm⁻³ NaOH(aq), pH 13. Added acid is neutralized while the pH stays high, then at 25.0 cm³ (n(HCl) = n(NaOH)) the pH falls almost vertically through 7, and excess HCl(aq) takes the pH down towards 1.5. Only panel C has this shape.
Curve B — The intercept on the pH axis is the pH of the solution in the FLASK before any titrant is added, not that of the HCl(aq) in the burette. The flask holds NaOH(aq), so the curve must start near pH 13 and fall, as in panel C.
Curve A — pH is logarithmic, so it does not change in proportion to the volume added. While NaOH is in excess the pH changes little; near 25.0 cm³ one drop changes it by several units, giving the steep section of panel C, not a straight line.
Curve D — After equivalence the added HCl(aq) is in excess and the solution becomes acidic, so the pH keeps falling below 7 and levels off near 1.5, as in panel C. It does not stay at 7.
34 The graph shows the pH curve obtained when 0.100 mol dm⁻³ NaOH(aq) is added to 25.0 cm³ of HCl(aq). What is the concentration of the HCl(aq)?
Answer and reasoning
1.25 × 10⁻¹ mol dm⁻³ — This inverts the volume ratio (25.0/20.0). The acid needed the LARGER volume of the two solutions, so it is the less concentrated: 25.0 cm³ of acid contains the same amount as 20.0 cm³ of 0.100 mol dm⁻³ NaOH, so c(HCl) = 0.100 × 20.0/25.0 = 0.0800 mol dm⁻³.
8.00 × 10⁻² mol dm⁻³ — Equivalence is at the middle of the steep section, 20.0 cm³. n(NaOH) = 0.100 × 0.0200 = 2.00 × 10⁻³ mol = n(HCl), so c(HCl) = 2.00 × 10⁻³ / 0.0250 = 0.0800 mol dm⁻³. The intercept at pH 1.1 confirms it: 10⁻¹·¹ ≈ 0.08 mol dm⁻³.
1.00 × 10⁻¹ mol dm⁻³ — Equivalence means equal AMOUNTS (mol), not equal concentrations or volumes. 20.0 cm³ of NaOH(aq), not 25.0 cm³, neutralizes the acid, so the acid is less concentrated than the base: 0.0800 mol dm⁻³.
4.44 × 10⁻² mol dm⁻³ — This divides 2.00 × 10⁻³ mol by the total volume at equivalence, 45.0 cm³. The acid was all in the 25.0 cm³ placed in the flask, so c(HCl) = 2.00 × 10⁻³ / 0.0250 = 0.0800 mol dm⁻³.
Working The equivalence point is at the middle of the near-vertical section, at 20.0 cm³ of NaOH(aq). n(NaOH) = 0.100 mol dm⁻³ × 0.0200 dm³ = 2.00 × 10⁻³ mol. HCl + NaOH → NaCl + H₂O, so n(HCl) = 2.00 × 10⁻³ mol. c(HCl) = 2.00 × 10⁻³ mol / 0.0250 dm³ = 0.0800 mol dm⁻³. Check: the intercept, pH 1.1, gives [H⁺] = 10⁻¹·¹ = 0.08 mol dm⁻³, consistent with a fully ionized 0.0800 mol dm⁻³ HCl(aq).
35 The graph shows the pH curve for the titration of 20.0 cm³ of 0.100 mol dm⁻³ of a weak monoprotic acid, HA, with 0.100 mol dm⁻³ NaOH(aq) at 298 K. Three points on the curve are labelled with their pH. What is the Ka of HA? HL
Answer and reasoning
1.6 × 10⁻⁹ — This uses the pH at the equivalence point (8.8, at 20.0 cm³). pH = pKa at HALF-equivalence, 10.0 cm³, where [HA] = [A⁻]; the equivalence pH is set by hydrolysis of A⁻. pKa = 4.8, so Ka = 1.6 × 10⁻⁵.
1.3 × 10⁻³ — This uses the initial pH, 2.9. Before any base is added [HA] ≫ [A⁻], so the pH of the acid alone is not the pKa; pH = pKa only where [HA] = [A⁻], at 10.0 cm³ (pH 4.8). Ka = 1.6 × 10⁻⁵.
8.2 × 10⁻³ — This is e⁻⁴·⁸. pKa is defined with log₁₀, so Ka = 10⁻ᵖᴷᵃ = 10⁻⁴·⁸ = 1.6 × 10⁻⁵.
1.6 × 10⁻⁵ — At half-equivalence (10.0 cm³, half of the 20.0 cm³ equivalence volume) [HA] = [A⁻], so pH = pKa = 4.8 and Ka = 10⁻⁴·⁸ = 1.6 × 10⁻⁵.
Working Equivalence is at 20.0 cm³ (steep section). At half-equivalence, 10.0 cm³, half the HA has been converted to A⁻, so [HA] = [A⁻] and Ka = [H⁺][A⁻]/[HA] = [H⁺]; hence pH = pKa. The curve gives pH 4.8 at 10.0 cm³, so pKa = 4.8 and Ka = 10⁻⁴·⁸ = 1.6 × 10⁻⁵.
36 The figure shows four sketched pH curves, A to D, each for 25.0 cm³ of a 0.100 mol dm⁻³ solution titrated with a 0.100 mol dm⁻³ solution. Which curve shows a weak acid titrated with a strong base? HL
Answer and reasoning
Curve A — Panel A starts at pH 1.0, which is −log₁₀(0.100): it treats the acid as fully ionized. A weak acid releases only a small fraction of its H⁺, so 0.100 mol dm⁻³ of it starts near pH 3, and its equivalence point lies above 7, as in panel B. Panel A, starting at pH 1 with equivalence at 7, is a strong acid with a strong base.
Curve C — The steep section is produced by the STRONG titrant: near equivalence a single drop of NaOH switches the flask from slight excess acid to slight excess strong base. A weak acid with a strong base therefore still has a near-vertical section (panel B). A curve with no steep section, panel C, is a weak acid with a weak base.
Curve B — A weak acid starts at a moderate pH (about 3, as it is only partly ionized), rises through a buffer region as its conjugate base forms, then jumps steeply at 25.0 cm³ with the equivalence point above 7 (the salt of a weak acid and strong base is basic) and levels off in excess strong base. Panel B has all these features.
Curve D — The intercept on the pH axis is the pH of the solution in the flask, not of the NaOH in the burette. A curve starting at pH 13 and falling (panel D) is a strong base being titrated with acid. A weak acid in the flask starts near pH 3 and rises, as in panel B.
37 25.0 cm³ of 0.100 mol dm⁻³ of a weak acid, HA (pKa = 4.8), is titrated with 0.100 mol dm⁻³ NaOH(aq) at 298 K. The graph shows the pH curve, with the colour-change range of methyl red shaded. Which statement about using methyl red as the indicator for this titration is correct? HL
Answer and reasoning
It is unsuitable, as it changes colour gradually in the buffer region, before the equivalence point. — The curve crosses the shaded band (pH 4.4–6.2) between about 7 cm³ and 24 cm³, in the gently sloping buffer region, so the colour changes gradually and is complete before the steep section at 25.0 cm³. An indicator must change within the steep section, where the equivalence pH (about 8.7) lies, for example phenolphthalein (8.3–10.0).
It is suitable, as its range includes the pKa of HA, where the indicator should change. — The end point should match the pH at EQUIVALENCE, not the pKa of the acid being titrated. pH 4.8 is reached at half-equivalence, 12.5 cm³, far before the steep section; an indicator changing there would give a titre about half the true value.
It is suitable, as the pH of the flask passes through its range during the titration. — The pH passes through 4.4–6.2 in the buffer region, between about 7 and 24 cm³, not in the steep section at 25.0 cm³. The colour change must coincide with the equivalence point, so only an indicator whose range lies within the steep section (about pH 7–11) is suitable.
It is unsuitable, as its range does not include pH 7, the pH at the equivalence point. — The equivalence point of a weak acid–strong base titration is above 7 (about 8.7 here, read from the middle of the steep section), because the ethanoate-like anion A⁻ is a base. Methyl red is unsuitable because its range lies in the buffer region, not because it misses pH 7.
38 The graph shows the pH curve for 25.0 cm³ of 0.100 mol dm⁻³ NH₃(aq) titrated with 0.100 mol dm⁻³ HCl(aq) at 298 K. pKa(NH₄⁺) = 9.25. The four bars beside the curve show the colour-change ranges of four indicators, 1 to 4, on the same pH scale. Which indicator is suitable for this titration? HL
Answer and reasoning
Indicator 4 — Bar 4 (8.3–10.0) includes 9.25, the pKa of NH₄⁺ given in the stem, which is the pH at half-equivalence (12.5 cm³). But the indicator must change at the EQUIVALENCE pH, about 5.3. Indicator 4 would change colour in the buffer region, about halfway through the titration.
Indicator 2 — The equivalence point is at the middle of the steep section at 25.0 cm³, about pH 5.3 (the salt NH₄Cl is acidic, as NH₄⁺ is a weak acid). Indicator 2 (pH 4.4–6.2) is the only range that lies within the steep section and includes the equivalence pH, so its end point coincides with the equivalence point.
Indicator 3 — Equivalence is at pH 7 only for a strong acid with a strong base. Here the salt is NH₄Cl, whose cation NH₄⁺ is a weak acid, so the equivalence pH is about 5.3. Bar 3 (6.8–8.4) lies above the steep section and would change colour before equivalence.
Indicator 1 — Bar 1 (1.2–2.8) matches the flat region after the steep section, but that region is excess HCl(aq), well past equivalence. The end point must lie in the steep section around 25.0 cm³, where the pH is about 5.3, so indicator 2 is required.
Working Equivalence is at 25.0 cm³ (n(HCl) = n(NH₃) = 2.50 × 10⁻³ mol). The flask then contains NH₄Cl(aq), [NH₄⁺] = 2.50 × 10⁻³ mol / 0.0500 dm³ = 0.0500 mol dm⁻³; Ka(NH₄⁺) = 10⁻⁹·²⁵ = 5.6 × 10⁻¹⁰, so [H⁺] = √(5.6 × 10⁻¹⁰ × 0.0500) = 5.3 × 10⁻⁶ mol dm⁻³ and pH ≈ 5.3, read from the middle of the steep section. The steep section spans roughly pH 3.5–7. Only bar 2 (4.4–6.2) lies within it and includes 5.3.
39 The chart shows the concentrations of CH₃COOH and CH₃COO⁻ in four buffer solutions, P to S, all at 298 K. Which buffer has the lowest pH? HL
Answer and reasoning
Buffer Q — Q contains the most ethanoic acid, but it contains equally much ethanoate, so [CH₃COOH]/[CH₃COO⁻] = 1 and pH = pKa (4.76). The pH depends on the ratio, not on the amount of acid alone; P's ratio of 2 gives a lower pH.
Buffer R — R has the largest [CH₃COO⁻]/[CH₃COOH] ratio (3), but [H⁺] = Ka × [CH₃COOH]/[CH₃COO⁻], with the ACID on top. A large proportion of conjugate base makes [H⁺] small, so R has the highest pH (about 5.2), not the lowest.
Buffer P — [H⁺] = Ka × [CH₃COOH]/[CH₃COO⁻], so the lowest pH belongs to the largest acid : base ratio. P has 0.20/0.10 = 2; Q and S have a ratio of 1 (pH = pKa) and R has 0.10/0.30 = 0.33. P has the largest ratio and so the lowest pH (about 4.5).
Buffer S — S is the most dilute buffer, but [H⁺] = Ka × [CH₃COOH]/[CH₃COO⁻] depends only on the ratio, which is 1 for S. Its pH equals pKa (4.76), the same as Q; dilution does not lower a buffer's pH. P, with a ratio of 2, has the lowest pH.
40 At 298 K, 0.10 mol dm⁻³ aqueous solutions have these pH values: HCl 1.0; CH₃COOH 2.9; NH₃ 11.1; NaOH 13.0. Which conclusion is supported by these data?
Answer and reasoning
HCl is only partly ionized in solution, as its pH is 1.0 rather than 0.0. — pH 1.0 corresponds to [H⁺] = 0.10 mol dm⁻³, exactly the concentration of the acid, so every HCl molecule has ionized. pH 0.0 would need [H⁺] = 1.0 mol dm⁻³, ten times more than is present.
NH₃ is a strong base, as its solution has a pH well above 7, at 11.1. — An alkaline pH shows only that OH⁻ is present. pH 11.1 gives [H⁺] = 7.9 × 10⁻¹² mol dm⁻³, so [OH⁻] = Kw/[H⁺] ≈ 1.3 × 10⁻³ mol dm⁻³, about 1% of the 0.10 mol dm⁻³ ammonia; the strong base NaOH at the same concentration reaches pH 13.0. NH₃ is a weak base.
NH₃(aq) needs less HCl(aq) than an equal volume of NaOH(aq) does. — Equal volumes of 0.10 mol dm⁻³ NH₃(aq) and NaOH(aq) contain equal amounts of base and need equal amounts of HCl. As the acid reacts, the NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ equilibrium shifts right until all the ammonia has reacted; these pH data say nothing about the titre.
CH₃COOH is a weak acid, as its [H⁺] is far below 0.10 mol dm⁻³. — pH 2.9 means [H⁺] = 10⁻²·⁹ ≈ 1.3 × 10⁻³ mol dm⁻³, only about 1% of the 0.10 mol dm⁻³ acid. Most of the CH₃COOH is un-ionized, which is what 'weak' means. HCl at the same concentration gives pH 1.0, [H⁺] = 0.10 mol dm⁻³: complete ionization.
That was your twenty minutes. Real practice on R3.1 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Chemistry guide (first assessment 2025) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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