Acids and bases

In a nutshell

Acidity in water is caused by hydrogen ions, H⁺. Because their concentration spans a huge range, we measure it on a logarithmic scale, pH.

This note covers what acids and bases are (proton transfer), how to calculate pH for strong acids, strong bases and weak acids, the shapes of titration curves, how to choose an indicator, and how buffers resist changes in pH.

Almost every part of this topic is a calculation, and almost every calculation ends in a pH that must be quoted to two decimal places.

Assumed knowledge: Chemical equilibria, Le Chatelier's principle and Kc, Amount of substance.

Core content

Brønsted-Lowry acids and bases

AQA uses the Brønsted-Lowry definitions, which are about the transfer of a proton (an H⁺ ion):

• An acid is a proton (H⁺) donor.
• A base is a proton (H⁺) acceptor.

Acid-base reactions are therefore proton-transfer reactions. In water, hydrogen chloride donates a proton to a water molecule:

HCl(aq) + H₂O(l) ⟶ H₃O⁺(aq) + Cl^-(aq)

Here HCl is the acid (it donates H⁺) and H₂O is the base (it accepts H⁺). The H₃O⁺ ion is the oxonium ion, and we usually simplify it to H⁺(aq).

Every acid becomes its conjugate base when it loses a proton, and every base becomes its conjugate acid when it gains one. The two species that differ by one proton are a conjugate pair.

Defining and calculating pH

pH is defined mathematically, not in words:

$\text{pH} = -\log_{10}[H^+]$

where [H⁺] is the hydrogen ion concentration in mol dm^-3. Rearranging gives the reverse conversion:

$[H^+] = 10^{-\text{pH}}$

A strong acid is completely dissociated (ionised) into its ions in aqueous solution, so all of the acid has turned into ions:

HCl(aq) ⟶ H⁺(aq) + Cl^-(aq)

For a strong monoprotic acid (one that donates one proton, such as HCl or HNO₃), [H⁺] is simply equal to the concentration of the acid, so the pH follows directly.

For a diprotic strong acid such as H₂SO₄, each mole donates two moles of H⁺, so [H⁺] is twice the acid concentration.

The ionic product of water, Kw

Water is very slightly dissociated:

H₂O(l) ⇌ H⁺(aq) + OH^-(aq)

Because the concentration of undissociated water is effectively constant, it is folded into the constant, giving the ionic product of water:

$K_w = [H^+][OH^-]$

Note that water does not appear in the expression. At 25 °C, K_w = 1.00 × 10^-14 mol² dm^-6.

K_w lets you find [H⁺] for a strong base. A strong base (such as NaOH) is completely dissociated, so [OH^-] equals the base concentration; then rearrange K_w to find [H⁺]:

$[H^+] = \frac{K_w}{[OH^-]}$

K_w also explains why the pH of pure water changes with temperature. Dissociation of water is endothermic, so raising the temperature shifts the equilibrium to the right. This increases [H⁺], so the pH falls below 7, even though the water is still neutral (because [H⁺] still equals [OH^-]).

Weak acids and Ka

A weak acid is only slightly dissociated (ionised) in aqueous solution, so it sits as an equilibrium mixture of mostly undissociated acid with a little H⁺ and A^-:

HA(aq) ⇌ H⁺(aq) + A^-(aq)

The position of this equilibrium is measured by the acid dissociation constant, Ka:

$K_a = \frac{[H^+][A^-]}{[HA]}$

The larger the K_a, the further the equilibrium lies to the right, so the stronger the weak acid. K_a has units of mol dm^-3.

Because the numbers are tiny, K_a is often quoted as pKa:

$\text{p}K_a = -\log_{10}K_a \qquad K_a = 10^{-\text{p}K_a}$

To calculate the pH of a solution of a weak acid on its own, two approximations are made:

1. The acid dissociates in a 1:1 ratio, so [H⁺] = [A^-].
2. So little dissociates that the equilibrium concentration of the acid equals its starting concentration: [HA]_eqm ≈ [HA]_start.

These collapse the K_a expression to a form you can solve:

$K_a = \frac{[H^+]^2}{[HA]} \qquad \Rightarrow \qquad [H^+] = \sqrt{K_a \times [HA]}$

Still don't get it? · the weak-acid approximation

Imagine a huge crowd of acid molecules in a room. Only a tiny handful ever "let go" of their proton at any instant. So if you counted the molecules still holding their proton, it would be almost exactly the number you started with: the handful that let go is too small to notice.

That is assumption 2: the amount of acid that dissociates is so small that the acid concentration at equilibrium is still basically its starting concentration, so you can use the number on the bottle for [HA].

Assumption 1 is just bookkeeping: every molecule that lets go of a proton (H⁺) produces exactly one A⁻ at the same time, so in a solution of the acid alone the two must be present in equal amounts, [H⁺] = [A⁻].

Put both into Ka = [H⁺][A⁻]/[HA]: the top becomes [H⁺] × [H⁺] = [H⁺]², and the bottom is just the starting concentration. Rearranging gives [H⁺] = √(Ka × [HA]). The exam mark is for showing you used both assumptions, then taking the square root at the end.

pH curves and titrations

A pH (titration) curve plots the pH of the mixture against the volume of base added. The equivalence point is the volume at which the acid and base have reacted in exactly the ratio of the equation; it lies at the midpoint of the steep, near-vertical section.

The shape depends on whether each reactant is strong or weak. You must be able to sketch and explain all four combinations of monoprotic acid and base.

Strong acid and strong base (for example HCl and NaOH). The pH starts very low, stays low, then rises almost vertically through a long steep section (roughly pH 3 to 11), with the equivalence point at pH 7.

Weak acid and strong base (for example ethanoic acid and NaOH). The pH starts higher (around 3), rises quickly then flattens into a buffer region, then climbs through a shorter steep section, with the equivalence point above pH 7. At the half-equivalence point (half the volume needed to reach equivalence), [HA] = [A^-], so pH = pKa. This is how K_a can be found from a curve.

Strong acid and weak base (for example HCl and NH₃). The pH starts very low and rises through a shorter steep section, with the equivalence point below pH 7; it then levels off at a modest pH because the weak base cannot push the pH high.

Weak acid and weak base (for example ethanoic acid and NH₃). There is no steep section: the pH rises gradually and smoothly through the equivalence point, so there is no sharp change and no suitable indicator.

Required practical 9: investigating pH changes in a titration

Required practical 9 is to investigate how pH changes when a weak acid reacts with a strong base and when a strong acid reacts with a weak base. The method builds the curve point by point:

1. Pipette a known volume (for example 25.0 cm³) of the acid into a conical flask and measure its pH with a pH meter.
2. Add the base from a burette in small portions, stirring after each addition, and record the pH and the total volume added.
3. Near the equivalence point (where the pH changes fastest), add the base in smaller portions so the steep part is captured in detail.
4. Continue until the base is in excess and the pH levels off, then plot pH against volume added.

Two practical points earn credit:

• Calibrate the pH meter against a buffer of known pH before use, because stored electrodes drift. Keeping the temperature constant also improves reliability.
• Reading the pH with a meter (not an indicator) is what lets you follow the whole curve rather than just spot the end point.

Choosing an indicator

An acid-base indicator is itself a weak acid, HIn, whose acid form and conjugate base form are different colours:

HIn(aq) ⇌ H⁺(aq) + In^-(aq)

By Le Chatelier's principle, added H⁺ pushes the equilibrium left (colour of HIn) and added base pulls it right (colour of In^-). The colour changes over a narrow pH range.

The rule: choose an indicator whose pH range lies entirely on the steep (near-vertical) part of the titration curve, so the colour changes sharply at the equivalence point.

• A strong acid and strong base curve has a long steep part spanning both ranges, so either indicator works.
• A weak acid and weak base has no steep part, so neither works; a pH meter must be used instead.

Buffer action

A buffer solution maintains an approximately constant pH when small amounts of acid or base are added, or on dilution.

• An acidic buffer is made from a weak acid and the salt of that weak acid (for example ethanoic acid and sodium ethanoate).
• A basic buffer is made from a weak base and the salt of that weak base (for example ammonia and ammonium chloride).

In an ethanoic acid buffer there is a large reservoir of both the weak acid and its conjugate base (from the salt):

CH₃COOH(aq) ⇌ CH₃COO^-(aq) + H⁺(aq)

• Add acid: the extra H⁺ is removed by the large reservoir of ethanoate ions, CH₃COO^- + H⁺ ⟶ CH₃COOH, so the equilibrium shifts left.
• Add base: the OH^- reacts with H⁺ to form water, and the equilibrium shifts right as more CH₃COOH dissociates to replace the H⁺ removed. (Equivalently, OH^- + CH₃COOH ⟶ CH₃COO^- + H₂O.)

Because both reservoirs are large, the ratio [CH₃COOH]/[CH₃COO^-] barely changes, so [H⁺] and therefore the pH stay almost constant. On dilution the two concentrations fall by the same factor, so the ratio, and the pH, are unchanged.

Still don't get it? · how a buffer holds pH steady

Think of a bus with lots of empty seats and lots of standing passengers. If a few more people get on (adding acid), standing passengers sit down into the spare seats; if a few get off (adding base), seated passengers stand up to fill the gap. Either way the bus looks about the same, because there is a big reserve of both seated and standing people to absorb the change.

The weak acid is the "standing" reserve and its salt ion A⁻ is the "seated" reserve. Add H⁺ and the A⁻ mops it up (A⁻ + H⁺ → HA). Add OH⁻ and the HA tops the H⁺ back up (the equilibrium shifts right). Because you started with a lot of both HA and A⁻, a small addition barely changes the ratio [HA]/[A⁻].

The exam version: [H⁺] = Ka × [HA]/[A⁻]. The pH depends on the ratio, not the actual amounts, and a small addition hardly changes that ratio, so the pH stays almost constant. For the marks you must name both components and show the equation shifting to remove the added acid or base.

Applications. Buffers hold biological and industrial systems at a set pH: blood is held near pH 7.4 by a carbonic acid / hydrogencarbonate buffer, and buffers are used in shampoos, foods and fermentation.

To calculate the pH of an acidic buffer, rearrange the K_a expression, taking [A^-] as coming from the salt and [HA] as the (approximately unchanged) acid:

$[H^+] = K_a \times \frac{[HA]}{[A^-]}$

Worked examples

1. Interconverting [H⁺] and pH.

(a) Find the pH of a solution with [H⁺] = 4.0 × 10^-3 mol dm^-3.

$\text{pH} = -\log_{10}(4.0\times10^{-3}) = 2.40$

(b) Find [H⁺] in a solution of pH 10.46.

$[H^+] = 10^{-10.46} = 3.47\times10^{-11}\ \text{mol dm}^{-3}$

The pH answer carries two decimal places; the two digits after the point are what give the precision.

2. pH of a strong acid. Calculate the pH of 0.15 mol dm^-3 nitric acid, HNO₃.

HNO₃ is a strong monoprotic acid, so it is fully dissociated and [H⁺] = 0.15 mol dm^-3.

$\text{pH} = -\log_{10}(0.15) = 0.82$

If the acid were diprotic (H₂SO₄), [H⁺] would be 2 × 0.15 = 0.30 mol dm^-3.

3. pH of a strong base. Calculate the pH of 0.0500 mol dm^-3 sodium hydroxide at 25 °C (K_w = 1.00 × 10^-14 mol² dm^-6).

NaOH is fully dissociated, so [OH^-] = 0.0500 mol dm^-3.

$[H^+] = \frac{K_w}{[OH^-]} = \frac{1.00\times10^{-14}}{0.0500} = 2.00\times10^{-13}\ \text{mol dm}^{-3}$

$\text{pH} = -\log_{10}(2.00\times10^{-13}) = 12.70$

Watch for dibasic bases: 0.0500 mol dm^-3 Ba(OH)₂ would give [OH^-] = 0.100 mol dm^-3, because each formula unit releases two OH^-.

4. Converting Ka and pKa.

(a) Ethanoic acid has K_a = 1.74 × 10^-5 mol dm^-3. Find pK_a.

$\text{p}K_a = -\log_{10}(1.74\times10^{-5}) = 4.76$

(b) An acid has pK_a = 3.75. Find K_a.

$K_a = 10^{-3.75} = 1.78\times10^{-4}\ \text{mol dm}^{-3}$

5. pH of a weak acid. Calculate the pH of 0.100 mol dm^-3 ethanoic acid (K_a = 1.74 × 10^-5 mol dm^-3).

Using the two approximations:

$[H^+] = \sqrt{K_a \times [HA]} = \sqrt{1.74\times10^{-5} \times 0.100}$

$[H^+] = \sqrt{1.74\times10^{-6}} = 1.32\times10^{-3}\ \text{mol dm}^{-3}$

$\text{pH} = -\log_{10}(1.32\times10^{-3}) = 2.88$

The commonest slip is forgetting the square root at the end, or using the number of moles instead of the concentration.

6. A titration calculation from experimental results. 25.0 cm³ of 0.200 mol dm^-3 HCl is mixed with 30.0 cm³ of 0.250 mol dm^-3 NaOH. Calculate the pH of the resulting mixture.

Find the amount of each reactant:

$n(H^+) = 0.200 \times 0.0250 = 5.00\times10^{-3}\ \text{mol}$

$n(OH^-) = 0.250 \times 0.0300 = 7.50\times10^{-3}\ \text{mol}$

The base is in excess (H⁺ + OH^- ⟶ H₂O):

$n(OH^-)_{\text{excess}} = 7.50\times10^{-3} - 5.00\times10^{-3} = 2.50\times10^{-3}\ \text{mol}$

Divide by the total volume, 25.0 + 30.0 = 55.0 cm³ = 0.0550 dm³:

$[OH^-] = \frac{2.50\times10^{-3}}{0.0550} = 0.04545\ \text{mol dm}^{-3}$

$[H^+] = \frac{K_w}{[OH^-]} = \frac{1.00\times10^{-14}}{0.04545} = 2.20\times10^{-13}\ \text{mol dm}^{-3}$

$\text{pH} = -\log_{10}(2.20\times10^{-13}) = 12.66$

Two marks are lost here again and again: using the volume of just one solution instead of the combined volume, and (with excess base) stopping at [OH^-] instead of converting to [H⁺] through K_w.

7. pH of an acidic buffer. A buffer is made by dissolving ethanoic acid and sodium ethanoate so that it contains 0.100 mol of ethanoic acid and 0.0750 mol of sodium ethanoate in the same solution (K_a = 1.74 × 10^-5 mol dm^-3). Calculate its pH.

Both are in the same volume, so the volume cancels and the ratio of amounts can be used directly:

$[H^+] = K_a \times \frac{[HA]}{[A^-]} = 1.74\times10^{-5} \times \frac{0.100}{0.0750}$

$[H^+] = 2.32\times10^{-5}\ \text{mol dm}^{-3}$

$\text{pH} = -\log_{10}(2.32\times10^{-5}) = 4.63$

[A^-] comes from the salt, [HA] from the acid; do not treat a buffer as if only the weak acid is present.

Common exam mistakes

• Quoting pH to one decimal place (or as a whole number). AQA requires two decimal places: the digits after the point carry the precision, while the digit(s) before it only give the power of ten in [H⁺].
• Defining a weak acid as "not fully dissociated". This is rejected, because even strong acids may not be fully dissociated; the mark needs "only slightly dissociated/ionised".
• Including water in the K_w expression. It is K_w = [H⁺][OH^-] with no water term.
• Forgetting that diprotic acids and dibasic bases give two moles of ions: H₂SO₄ gives 2 H⁺ per mole; Ca(OH)₂ and Ba(OH)₂ give 2 OH^- per mole. Missing the factor of 2 is one of the most reported errors.
• In titration-mixture calculations, dividing by the volume of only one solution. The excess ions are spread through the combined volume of acid plus base.
• For an excess of strong base, stopping at [OH^-] and taking its negative log. You must convert to [H⁺] using K_w first.
• Writing the specific weak-acid expression [H⁺]²/[HA] when asked for the general K_a expression, which must be [H⁺][A^-]/[HA].
• Using moles instead of concentration in a K_a calculation, or forgetting to take the square root to get [H⁺].
• Treating a buffer made by partial neutralisation as if it were a solution of the weak acid alone. Once some salt has formed, use [H⁺] = K_a × [HA]/[A^-].
• Adding acid or base to a buffer the wrong way round: adding acid increases the moles of HA and decreases the moles of A^-; adding base does the reverse.
• Giving a vague buffer explanation ("it resists change by Le Chatelier"). The marks need you to name both components and show the equation that removes the added H⁺ or OH^-.
• Using a single indicator for a weak acid and weak base titration. There is no steep section, so no indicator changes colour sharply.
• Missing the equilibrium arrow ⇌ when writing a weak-acid dissociation, or losing the wrong proton from a carboxylic acid (it is the O-H proton that is lost).

Key definitions

• Acid (Brønsted-Lowry): a proton (H⁺) donor.
• Base (Brønsted-Lowry): a proton (H⁺) acceptor.
• Strong acid: an acid that is completely dissociated (ionised) into its ions in aqueous solution.
• Weak acid: an acid that is only slightly dissociated (ionised) in aqueous solution.
• pH: pH = −log₁₀[H⁺].
• Ionic product of water (K_w): K_w = [H⁺][OH^-] (= 1.00 × 10^-14 mol² dm^-6 at 25 °C).
• Acid dissociation constant (K_a): K_a = [H⁺][A^-]/[HA] for the equilibrium HA ⇌ H⁺ + A^-; pK_a = −log₁₀K_a.
• Buffer solution: a solution that resists change in pH (maintains an approximately constant pH) when small amounts of acid or base are added, or on dilution.

Specification

• I can state that an acid is a proton donor and a base is a proton acceptor, and that acid-base equilibria involve proton transfer.
• I can convert between [H⁺] and pH in both directions.
• I can calculate the pH of a strong acid from its concentration.
• I can write the expression for K_w and use it to calculate the pH of a strong base from its concentration.
• I can construct an expression for K_a, and convert between K_a and pK_a.
• I can calculate the pH of a weak acid from its concentration and K_a.
• I can perform pH calculations for acid-base titration mixtures from experimental results.
• I can sketch and explain the shapes of pH curves for all combinations of strong and weak monoprotic acids and bases.
• I can use a pH curve to select an appropriate indicator.
• I can describe required practical 9: following pH changes for a weak acid with a strong base, and a strong acid with a weak base.
• I can explain qualitatively how acidic and basic buffers work, and calculate the pH of an acidic buffer.

Related notes

• Chemical equilibria, Le Chatelier's principle and Kc
• Amount of substance