R3.1 Proton transfer reactions | IB DP Chemistry

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🎯What you need to be able to do

Deduce the Brønsted–Lowry acid and base in a reaction, and the conjugate acid or base of a given species.
Interpret and formulate equations for amphiprotic species.
Perform calculations relating pH and \( [\mathrm{H^{+}}] \), and use \( K_{\mathrm{w}} \) to recognise solutions as acidic, neutral or basic.
Distinguish strong from weak acids and bases, and strong from concentrated.
Formulate equations for the reactions of acids with metals, metal oxides, metal hydroxides, hydrogencarbonates and carbonates, and identify the parent acid and base of a salt.
Sketch and interpret the pH curve for a strong acid–strong base titration.
AHL Interconvert \( [\mathrm{H^{+}}] \), \( [\mathrm{OH^{-}}] \), pH and pOH.
AHL Interpret \( K_{\mathrm{a}} \), \( K_{\mathrm{b}} \), \( \mathrm{p}K_{\mathrm{a}} \) and \( \mathrm{p}K_{\mathrm{b}} \), and use \( K_{\mathrm{a}} \times K_{\mathrm{b}} = K_{\mathrm{w}} \).
AHL Predict the pH of a salt solution and construct equations for the hydrolysis of its ions.
AHL Interpret pH curves for all four strong/weak combinations, explain how indicators work, choose an appropriate indicator, and explain and calculate the pH of buffer solutions.

📚The chemistry

Brønsted–Lowry acids and bases

A Brønsted–Lowry acid is a proton donor; a Brønsted–Lowry base is a proton acceptor. Note that this is a definition about a reaction: a species is an acid because of what it does, not because of what it is.

A proton in aqueous solution may be written as \( \mathrm{H^{+}}(aq) \) or as \( \mathrm{H_3O^{+}}(aq) \), the hydronium (oxonium) ion; both are accepted. And distinguish two words that are often used loosely: a base is any proton acceptor, while an alkali is a base that is soluble in water. All alkalis are bases; copper(II) oxide is a base that is not an alkali.

Conjugate pairs

Two species differing by a single proton form a conjugate acid–base pair. Every Brønsted–Lowry reaction contains exactly two such pairs:

\[ \underbrace{\mathrm{HCl}}_{\text{acid 1}} + \underbrace{\mathrm{H_2O}}_{\text{base 2}} \rightarrow \underbrace{\mathrm{Cl^{-}}}_{\text{base 1}} + \underbrace{\mathrm{H_3O^{+}}}_{\text{acid 2}} \]

To find a conjugate base, remove one \( \mathrm{H^{+}} \) (the charge becomes one unit more negative); to find a conjugate acid, add one. So the conjugate base of \( \mathrm{H_2SO_4} \) is \( \mathrm{HSO_4^{-}} \), and the conjugate acid of \( \mathrm{NH_3} \) is \( \mathrm{NH_4^{+}} \).

A useful principle: acid–base equilibria lie in the direction of the weaker conjugate. The stronger an acid, the weaker its conjugate base.

Amphiprotic species

An amphiprotic species can act as either a Brønsted–Lowry acid or a base — it can both donate and accept a proton. The species that must be recognised: \( \mathrm{H_2O} \), \( \mathrm{HCO_3^{-}} \), \( \mathrm{HSO_4^{-}} \), \( \mathrm{H_2PO_4^{-}} \) and amino acids.

\[ \mathrm{HCO_3^{-}} + \mathrm{H^{+}} \rightarrow \mathrm{H_2CO_3} \qquad \mathrm{HCO_3^{-}} + \mathrm{OH^{-}} \rightarrow \mathrm{CO_3^{2-}} + \mathrm{H_2O} \]

Water doing this to itself is the origin of \( K_{\mathrm{w}} \).

pH and the ionic product of water

\[ \mathrm{pH} = -\log_{10}[\mathrm{H^{+}}] \qquad\qquad [\mathrm{H^{+}}] = 10^{-\mathrm{pH}} \]

Because the scale is logarithmic, a change of one pH unit is a tenfold change in \( [\mathrm{H^{+}}] \). A solution of pH 2 is a hundred times more acidic than one of pH 4, not twice.

Water self-ionises slightly:

\[ \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H^{+}}(aq) + \mathrm{OH^{-}}(aq) \qquad K_{\mathrm{w}} = [\mathrm{H^{+}}][\mathrm{OH^{-}}] = 1.00 \times 10^{-14} \text{ at 298 K} \]

\( K_{\mathrm{w}} \) is an equilibrium constant, so it varies with temperature. Self-ionisation is endothermic, so raising the temperature shifts the equilibrium right, increases \( K_{\mathrm{w}} \) and lowers the pH of pure water — but the water remains neutral, because \( [\mathrm{H^{+}}] \) still equals \( [\mathrm{OH^{-}}] \). Neutral means the two are equal, not that the pH is 7.

Acidic
\( [\mathrm{H^{+}}] > [\mathrm{OH^{-}}] \)

Neutral
\( [\mathrm{H^{+}}] = [\mathrm{OH^{-}}] \)

Basic
\( [\mathrm{H^{+}}] < [\mathrm{OH^{-}}] \)

pH can be estimated with universal indicator and measured precisely with a pH meter or probe. A digital probe gives a continuous numerical reading, so it is preferable whenever a curve is to be plotted or small changes detected; indicator paper is quicker but only approximate and subjective.

Strong and weak

Strong and weak refer to the extent of ionisation; concentrated and dilute refer to the amount of solute per unit volume. The two are independent, and confusing them is the classic error of this sub-topic: a concentrated solution of a weak acid can have a lower pH than a dilute solution of a strong one.

Strong acids ionise essentially completely. You must know the list: HCl, HBr, HI, HNO₃ and H₂SO₄.
Strong bases: the group 1 hydroxides.
Weak acids and bases ionise only partially, so an equilibrium is set up, and the equation is written with \( \rightleftharpoons \). Ethanoic acid, carbonic acid, ammonia.

Experimental ways to distinguish a strong from a weak acid of the same concentration: the strong acid has a lower pH, a higher electrical conductivity (more ions), and reacts faster with a carbonate or a metal (a greater \( [\mathrm{H^{+}}] \)). Note what does not distinguish them: the volume of alkali needed to neutralise them is identical, since that depends on the total amount of acid present, not on how much of it has ionised.

Reactions of acids

acid + metal
→ salt + hydrogen
\( \mathrm{Mg} + 2\mathrm{HCl} \rightarrow \mathrm{MgCl_2} + \mathrm{H_2} \)
(a redox reaction — R3.2)

acid + metal oxide
→ salt + water
\( \mathrm{CuO} + \mathrm{H_2SO_4} \rightarrow \mathrm{CuSO_4} + \mathrm{H_2O} \)

acid + metal hydroxide
→ salt + water
\( \mathrm{NaOH} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{H_2O} \)

acid + carbonate
→ salt + water + carbon dioxide
\( \mathrm{CaCO_3} + 2\mathrm{HCl} \rightarrow \mathrm{CaCl_2} + \mathrm{H_2O} + \mathrm{CO_2} \)

acid + hydrogencarbonate
→ salt + water + carbon dioxide
\( \mathrm{NaHCO_3} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{H_2O} + \mathrm{CO_2} \)

acid + ammonia / amine
→ salt only
\( \mathrm{NH_3} + \mathrm{HCl} \rightarrow \mathrm{NH_4Cl} \)

To identify the parent acid and base of a salt, split the formula: the cation came from the base, the anion from the acid. Ammonium sulfate came from ammonia and sulfuric acid; sodium ethanoate from sodium hydroxide and ethanoic acid.

pH curves

For a strong acid titrated with a strong base, monoprotic (which is all that is assessed at SL), the curve starts low, rises slowly, then rises almost vertically through the equivalence point at pH 7, then flattens again at a high pH.

Features to identify and label: the initial pH (the intercept on the pH axis, set by the concentration and strength of the acid), the equivalence point (where the acid and base are present in exactly the stoichiometric ratio), and the vertical section. The equivalence point is the midpoint of the vertical section, not simply the steepest point you can see by eye.

AHL pOH, K_a and K_b

\[ \mathrm{pOH} = -\log_{10}[\mathrm{OH^{-}}] \qquad [\mathrm{OH^{-}}] = 10^{-\mathrm{pOH}} \qquad \mathrm{pH} + \mathrm{pOH} = 14.00 \ \text{ at 298 K} \]

For a weak acid \( \mathrm{HA} \rightleftharpoons \mathrm{H^{+}} + \mathrm{A^{-}} \):

\[ K_{\mathrm{a}} = \frac{[\mathrm{H^{+}}][\mathrm{A^{-}}]}{[\mathrm{HA}]} \qquad \mathrm{p}K_{\mathrm{a}} = -\log_{10}K_{\mathrm{a}} \]

Large \( K_{\mathrm{a}} \) means a strong acid; small \( \mathrm{p}K_{\mathrm{a}} \) means a strong acid. The logarithm reverses the direction, and that reversal is the most common source of confusion here. For a conjugate pair:

\[ K_{\mathrm{a}} \times K_{\mathrm{b}} = K_{\mathrm{w}} \qquad\qquad \mathrm{p}K_{\mathrm{a}} + \mathrm{p}K_{\mathrm{b}} = 14.00 \]

To find the pH of a weak acid: since \( [\mathrm{H^{+}}] = [\mathrm{A^{-}}] \) and very little \( \mathrm{HA} \) ionises, \( K_{\mathrm{a}} \approx \dfrac{[\mathrm{H^{+}}]^2}{[\mathrm{HA}]_{\text{initial}}} \), so \( [\mathrm{H^{+}}] = \sqrt{K_{\mathrm{a}}\,[\mathrm{HA}]} \). Quadratics are not expected, so this approximation is always available.

AHL Salt hydrolysis

The pH of a salt solution depends on the relative strengths of its parent acid and base, because an ion derived from a weak parent reacts with water:

strong acid + strong base
e.g. NaCl
neutral — neither ion hydrolyses

strong acid + weak base
e.g. NH₄Cl
acidic
\( \mathrm{NH_4^{+}} + \mathrm{H_2O} \rightleftharpoons \mathrm{NH_3} + \mathrm{H_3O^{+}} \)

weak acid + strong base
e.g. CH₃COONa
basic
\( \mathrm{CH_3COO^{-}} + \mathrm{H_2O} \rightleftharpoons \mathrm{CH_3COOH} + \mathrm{OH^{-}} \)

weak acid + weak base
depends on the relative values of \( K_{\mathrm{a}} \) and \( K_{\mathrm{b}} \)

The examples the guide names are \( \mathrm{NH_4^{+}} \), \( \mathrm{RCOO^{-}} \), \( \mathrm{CO_3^{2-}} \) and \( \mathrm{HCO_3^{-}} \). The acidity of hydrated transition element ions and of \( \mathrm{Al^{3+}}(aq) \) is not required.

AHL The four pH curves

Strong acid + strong base
equivalence at pH 7
long vertical section, roughly pH 3–11

Weak acid + strong base
equivalence above pH 7
higher initial pH; shorter vertical section; a buffer region where the curve is flat, with pH = \( \mathrm{p}K_{\mathrm{a}} \) at the half-equivalence point

Strong acid + weak base
equivalence below pH 7
shorter vertical section; buffer region after the equivalence point

Weak acid + weak base
equivalence near pH 7 but with no vertical section at all — which is why no indicator works and such titrations are not done

That pH = \( \mathrm{p}K_{\mathrm{a}} \) at half-equivalence result is worth learning: at that point exactly half the weak acid has been converted to its conjugate base, so \( [\mathrm{HA}] = [\mathrm{A^{-}}] \) and the \( K_{\mathrm{a}} \) expression reduces to \( K_{\mathrm{a}} = [\mathrm{H^{+}}] \). It is the standard experimental method for finding \( \mathrm{p}K_{\mathrm{a}} \) from a titration curve.

AHL Indicators

An indicator is a weak acid whose conjugate acid–base forms are different colours:

\[ \underbrace{\mathrm{HInd}}_{\text{colour A}} \rightleftharpoons \mathrm{H^{+}} + \underbrace{\mathrm{Ind^{-}}}_{\text{colour B}} \]

Add acid and Le Châtelier shifts the equilibrium left, giving colour A; add alkali, which removes \( \mathrm{H^{+}} \), and it shifts right, giving colour B. The colour changes over a range of about two pH units centred on the indicator’s \( \mathrm{p}K_{\mathrm{a}} \).

Choosing an indicator: its end point range must lie within the vertical section of the pH curve, which is where a single drop of titrant produces a large pH change. Phenolphthalein (range about 8.3–10) suits a weak acid with a strong base; methyl orange (about 3.1–4.4) suits a strong acid with a weak base; either works for strong with strong. Universal indicator is a mixture of many indicators, giving a gradual change over a wide range, which makes it useful for estimating pH and useless for a titration.

Distinguish two terms: the equivalence point is where the reactants are present in exactly the stoichiometric ratio; the end point is where the indicator changes colour. A good choice of indicator makes them coincide closely.

AHL Buffers

A buffer solution resists change in pH on the addition of small amounts of acid or alkali. There are two kinds:

Acidic buffer
a weak acid and its conjugate base (its salt)
e.g. \( \mathrm{CH_3COOH} \) with \( \mathrm{CH_3COONa} \)

Basic buffer
a weak base and its conjugate acid
e.g. \( \mathrm{NH_3} \) with \( \mathrm{NH_4Cl} \)

How it works — a large reservoir of each half of the conjugate pair: added \( \mathrm{H^{+}} \) is removed by the conjugate base (\( \mathrm{CH_3COO^{-}} + \mathrm{H^{+}} \rightarrow \mathrm{CH_3COOH} \)); added \( \mathrm{OH^{-}} \) is removed by the weak acid (\( \mathrm{CH_3COOH} + \mathrm{OH^{-}} \rightarrow \mathrm{CH_3COO^{-}} + \mathrm{H_2O} \)). This is also why a buffer must be made from a weak conjugate system: a strong acid is fully ionised, so there is no reservoir of undissociated molecules to mop up added alkali.

Rearranging the \( K_{\mathrm{a}} \) expression gives the pH:

\[ [\mathrm{H^{+}}] = K_{\mathrm{a}} \times \frac{[\mathrm{acid}]}{[\mathrm{salt}]} \qquad\Longrightarrow\qquad \mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log_{10}\frac{[\mathrm{salt}]}{[\mathrm{acid}]} \]

Two consequences: the pH depends on the ratio of the two, so diluting a buffer does not change its pH (both concentrations fall by the same factor) — though it does reduce its capacity. And a buffer works best when \( [\mathrm{acid}] = [\mathrm{salt}] \), where \( \mathrm{pH} = \mathrm{p}K_{\mathrm{a}} \), so choose an acid whose \( \mathrm{p}K_{\mathrm{a}} \) is close to the pH you want.

✏️Worked example

(a) Identify the two conjugate acid–base pairs in \( \mathrm{HCO_3^{-}} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_2CO_3} + \mathrm{OH^{-}} \), and state what this shows about \( \mathrm{HCO_3^{-}} \).
(b) Calculate the pH of 0.0500 mol dm⁻³ HCl, and of 0.0500 mol dm⁻³ NaOH at 298 K.
(c) AHL Calculate the pH of 0.100 mol dm⁻³ ethanoic acid, \( K_{\mathrm{a}} = 1.74 \times 10^{-5} \).
(d) AHL A buffer is made by dissolving 0.200 mol of sodium ethanoate in 1.00 dm³ of 0.100 mol dm⁻³ ethanoic acid. Calculate its pH, and explain what happens when a small amount of HCl is added.

(a) \( \mathrm{HCO_3^{-}} \) accepts a proton from water to become \( \mathrm{H_2CO_3} \), so it acts as the base and \( \mathrm{H_2CO_3} \) is its conjugate acid — that is pair 1. Water donates the proton, so it is the acid, and \( \mathrm{OH^{-}} \) is its conjugate base — pair 2.

Since \( \mathrm{HCO_3^{-}} \) also donates a proton in \( \mathrm{HCO_3^{-}} + \mathrm{OH^{-}} \rightarrow \mathrm{CO_3^{2-}} + \mathrm{H_2O} \), it can act as both an acid and a base and is therefore amphiprotic.

(b) HCl is a strong monoprotic acid, so it ionises completely and \( [\mathrm{H^{+}}] = 0.0500\ \mathrm{mol\,dm^{-3}} \):

\[ \mathrm{pH} = -\log_{10}(0.0500) = 1.30 \]

NaOH is a strong base, so \( [\mathrm{OH^{-}}] = 0.0500 \). Route via \( K_{\mathrm{w}} \):

\[ [\mathrm{H^{+}}] = \frac{K_{\mathrm{w}}}{[\mathrm{OH^{-}}]} = \frac{1.00 \times 10^{-14}}{0.0500} = 2.00 \times 10^{-13} \]

\[ \mathrm{pH} = -\log_{10}(2.00 \times 10^{-13}) = 12.70 \]

(Or, faster: \( \mathrm{pOH} = 1.30 \), so \( \mathrm{pH} = 14.00 - 1.30 = 12.70 \).)

(c) Ethanoic acid is weak, so use the approximation:

\[ [\mathrm{H^{+}}] = \sqrt{K_{\mathrm{a}}[\mathrm{HA}]} = \sqrt{(1.74 \times 10^{-5})(0.100)} = \sqrt{1.74 \times 10^{-6}} = 1.32 \times 10^{-3} \]

\[ \mathrm{pH} = -\log_{10}(1.32 \times 10^{-3}) = 2.88 \]

Compare with a strong acid of the same concentration, which would have pH 1.00. The difference of almost two units is the whole meaning of “weak”.

(d) \( [\mathrm{salt}] = 0.200 \) and \( [\mathrm{acid}] = 0.100 \). \( \mathrm{p}K_{\mathrm{a}} = -\log_{10}(1.74 \times 10^{-5}) = 4.76 \).

\[ \mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log_{10}\frac{[\mathrm{salt}]}{[\mathrm{acid}]} = 4.76 + \log_{10}\!\left(\frac{0.200}{0.100}\right) = 4.76 + 0.30 = 5.06 \]

On adding a small amount of HCl, the added \( \mathrm{H^{+}} \) is removed by the large reservoir of ethanoate ions:

\[ \mathrm{CH_3COO^{-}}(aq) + \mathrm{H^{+}}(aq) \rightarrow \mathrm{CH_3COOH}(aq) \]

The ratio \( [\mathrm{salt}]/[\mathrm{acid}] \) falls slightly, so the pH falls slightly — but almost all of the added \( \mathrm{H^{+}} \) is converted into undissociated ethanoic acid rather than remaining free, so the change is far smaller than the same addition to unbuffered water would cause.

Check it. Four quick checks. pH direction: an acid must give pH below 7 and a base above — 1.30 and 12.70 ✓. Weak against strong: the pH of a weak acid must be higher than that of a strong acid of the same concentration, because less of it ionises; 2.88 against 1.00 ✓. Buffer: the pH must lie close to \( \mathrm{p}K_{\mathrm{a}} \) — within about one unit — and 5.06 against 4.76 ✓. Since there is more salt than acid, the pH must be above \( \mathrm{p}K_{\mathrm{a}} \), which it is. Log check: in (b), pH 1.30 for 0.05 mol dm⁻³ is sensible because 0.05 lies between \( 10^{-1} \) and \( 10^{-2} \), so the pH must lie between 1 and 2.

Treating “strong” as though it meant “concentrated”. They are independent. In (c), 0.100 mol dm⁻³ ethanoic acid has pH 2.88, while 0.0100 mol dm⁻³ HCl — ten times more dilute — has pH 2.00 and is the more acidic solution. Strong is about the extent of ionisation; concentrated is about the amount of solute per unit volume. And note the property that is not affected: 25 cm³ of 0.100 mol dm⁻³ ethanoic acid and of 0.100 mol dm⁻³ HCl require exactly the same volume of alkali to neutralise, because neutralisation depends on the total amount of acid present, not on how much has ionised at any instant.

📝Practise

Work through these on paper, then reveal the answer.

1. For \( \mathrm{NH_3} + \mathrm{H_2O} \rightleftharpoons \mathrm{NH_4^{+}} + \mathrm{OH^{-}} \), identify the acid, the base and the two conjugate pairs. Then give the conjugate base of \( \mathrm{H_2SO_4} \) and the conjugate acid of \( \mathrm{HCO_3^{-}} \).

Water donates a proton, so \( \mathrm{H_2O} \) is the acid and \( \mathrm{OH^{-}} \) is its conjugate base — that is one pair. Ammonia accepts the proton, so \( \mathrm{NH_3} \) is the base and \( \mathrm{NH_4^{+}} \) is its conjugate acid — the second pair. Note that water is behaving as an acid here and as a base when it reacts with HCl, which is what makes it amphiprotic. Conjugate base of \( \mathrm{H_2SO_4} \): remove one \( \mathrm{H^{+}} \) — \( \mathrm{HSO_4^{-}} \). Conjugate acid of \( \mathrm{HCO_3^{-}} \): add one \( \mathrm{H^{+}} \) — \( \mathrm{H_2CO_3} \). Watch the charge: adding a proton makes the charge one unit more positive, removing one makes it one unit more negative.

2. Calculate: (a) the pH of \( 2.5 \times 10^{-3} \) mol dm⁻³ HNO₃; (b) \( [\mathrm{H^{+}}] \) in a solution of pH 4.60; (c) the pH of 0.0200 mol dm⁻³ Ba(OH)₂.

(a) Nitric acid is strong and monoprotic, so \( [\mathrm{H^{+}}] = 2.5 \times 10^{-3} \) and \( \mathrm{pH} = -\log_{10}(2.5 \times 10^{-3}) = \mathbf{2.60} \). (b) \( [\mathrm{H^{+}}] = 10^{-\mathrm{pH}} = 10^{-4.60} = \mathbf{2.5 \times 10^{-5}\ mol\,dm^{-3}} \). (c) The trap here is that barium hydroxide is a strong base providing two \( \mathrm{OH^{-}} \) per formula unit: \( [\mathrm{OH^{-}}] = 2 \times 0.0200 = 0.0400\ \mathrm{mol\,dm^{-3}} \). Then \( \mathrm{pOH} = -\log_{10}(0.0400) = 1.40 \), so \( \mathrm{pH} = 14.00 - 1.40 = \mathbf{12.60} \). Forgetting the factor of two gives 12.30, which looks entirely plausible — always check the formula of the base before taking a logarithm.

3. Describe three experiments that would distinguish 0.1 mol dm⁻³ hydrochloric acid from 0.1 mol dm⁻³ ethanoic acid, and one measurement that would not.

Three that work. (i) Measure the pH with a pH meter: the hydrochloric acid is fully ionised so \( [\mathrm{H^{+}}] = 0.1 \) and pH = 1.0, while the ethanoic acid is only partially ionised, giving a much lower \( [\mathrm{H^{+}}] \) and a pH of about 2.9. (ii) Measure the electrical conductivity: the strong acid contains far more ions per unit volume, so it conducts considerably better. (iii) Compare the rate of reaction with magnesium ribbon or with a carbonate: the strong acid has a much higher \( [\mathrm{H^{+}}] \), so effervescence is visibly faster. One that does not work: titration with a standard alkali. Both solutions contain the same total amount of acid per unit volume, and as \( \mathrm{H^{+}} \) is neutralised the weak acid equilibrium shifts to replace it, so exactly the same volume of alkali is required to reach the equivalence point. What differs is the shape of the titration curve — the weak acid starts higher, has a buffer region, and reaches equivalence above pH 7 — but not the titre.

4. AHL Predict, with equations, whether aqueous solutions of NH₄Cl, CH₃COONa, NaNO₃ and Na₂CO₃ are acidic, basic or neutral.

NH₄Cl — acidic. Parent acid HCl (strong), parent base \( \mathrm{NH_3} \) (weak). The ammonium ion hydrolyses: \( \mathrm{NH_4^{+}} + \mathrm{H_2O} \rightleftharpoons \mathrm{NH_3} + \mathrm{H_3O^{+}} \), releasing \( \mathrm{H_3O^{+}} \). The chloride ion, being the conjugate base of a strong acid, is far too weak a base to hydrolyse. CH₃COONa — basic. Parent acid ethanoic (weak), parent base NaOH (strong). The ethanoate ion hydrolyses: \( \mathrm{CH_3COO^{-}} + \mathrm{H_2O} \rightleftharpoons \mathrm{CH_3COOH} + \mathrm{OH^{-}} \). NaNO₃ — neutral. Both parents are strong (HNO₃ and NaOH), so neither ion hydrolyses to any appreciable extent. Na₂CO₃ — basic, and strongly so. Carbonic acid is weak, so the carbonate ion hydrolyses: \( \mathrm{CO_3^{2-}} + \mathrm{H_2O} \rightleftharpoons \mathrm{HCO_3^{-}} + \mathrm{OH^{-}} \) — which is why washing soda solution feels soapy and is used as a cleaning alkali. The rule: the ion derived from the weak parent is the one that reacts with water, and it produces the ion that determines the pH.

5. AHL Sketch the pH curve for titrating 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid (\( \mathrm{p}K_{\mathrm{a}} = 4.76 \)) with 0.100 mol dm⁻³ NaOH. Label the initial pH, the buffer region, the half-equivalence point, the equivalence point, and state which indicator you would use.

Initial pH about 2.9 — not 1.0, because the acid is weak and only partly ionised. The curve then rises gently and almost flatly through the buffer region, where both \( \mathrm{CH_3COOH} \) and \( \mathrm{CH_3COO^{-}} \) are present in quantity and resist pH change. At the half-equivalence point, after 12.5 cm³ of alkali, exactly half the acid has been converted to its conjugate base, so \( [\mathrm{HA}] = [\mathrm{A^{-}}] \) and pH = \( \mathrm{p}K_{\mathrm{a}} \) = 4.76. The curve then rises steeply through the equivalence point at 25.0 cm³ and a pH of about 8.7 — above 7, because the solution at that point is sodium ethanoate, whose anion hydrolyses to give \( \mathrm{OH^{-}} \). Beyond it the curve flattens towards pH 13. The vertical section is shorter than for a strong acid, spanning roughly pH 7 to 11. Indicator: phenolphthalein, range about 8.3–10.0, which lies entirely within that vertical section. Methyl orange would be useless, since its range of 3.1–4.4 lies in the buffer region, where it would change colour gradually over many cubic centimetres of added alkali.

6. AHL A buffer contains 0.150 mol dm⁻³ propanoic acid (\( K_{\mathrm{a}} = 1.35 \times 10^{-5} \)) and 0.100 mol dm⁻³ sodium propanoate. (a) Calculate its pH. (b) Explain why diluting the buffer with water does not change its pH. (c) Explain why a mixture of HCl and NaCl cannot act as a buffer.

(a) \( \mathrm{p}K_{\mathrm{a}} = -\log_{10}(1.35 \times 10^{-5}) = 4.87 \). \( \mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log_{10}\dfrac{[\mathrm{salt}]}{[\mathrm{acid}]} = 4.87 + \log_{10}\dfrac{0.100}{0.150} = 4.87 + (-0.176) = \mathbf{4.69} \). The pH is below \( \mathrm{p}K_{\mathrm{a}} \), as it must be when there is more acid than salt. (b) The pH depends only on the ratio \( [\mathrm{salt}]/[\mathrm{acid}] \). Diluting the solution reduces both concentrations by the same factor, so the ratio — and hence the logarithm, and hence the pH — is unchanged. What dilution does reduce is the buffer capacity: there are fewer moles of each species per unit volume, so a smaller addition of acid or alkali is enough to exhaust it. (c) A buffer needs a reservoir of a weak acid and its conjugate base. HCl is a strong acid, so it is completely ionised — there are no undissociated HCl molecules in solution to react with added \( \mathrm{OH^{-}} \). And \( \mathrm{Cl^{-}} \), being the conjugate base of a strong acid, is far too weak a base to accept an added proton. So added alkali is not mopped up, added acid is not mopped up, and the pH changes freely. Buffering requires a conjugate pair in which both members are present in appreciable quantity, which only a weak system provides.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

PhET — Acid–Base Solutions and pH Scale, which show the difference between strong and weak (extent of ionisation) and between concentrated and dilute side by side, on the same screen.
RSC Learn Chemistry — titration protocols and the use of a pH probe with a data logger to plot a full curve, which is the best route to a strong internal assessment on this topic.
Your data booklet — the pH, pOH and \( K_{\mathrm{w}} \) relationships, and the table of indicators with their pH ranges. The indicator table in particular should never be memorised.