3.1.12 Acids and Bases

A Brønsted–Lowry acid donates a proton, H⁺. In a reaction, identify it by following which species loses H⁺ and becomes its conjugate base.

Conjugate acid–base pairs differ by exactly one proton. Writing the pair side by side is a quick way to check that you have not changed anything else in the species.

A Brønsted–Lowry base accepts a proton, H⁺, usually by using a lone pair to form a new bond. After accepting H⁺ it becomes its conjugate acid.

When you identify a base in an equation, follow the proton transfer. The base is the species that gains H⁺; it need not contain OH⁻.

Acid–base equilibria are reversible proton transfers. One side contains an acid and base; the other contains their conjugate base and conjugate acid.

Pairs differ by one H⁺. If you cannot match each acid to a species on the other side that differs by exactly one proton, recheck the equation.

Compare pH values: remember that the scale is logarithmic. A change of one pH unit corresponds to a tenfold change in [H⁺], which is why pH can represent such a huge concentration range compactly.

So pH 2 has 100 times the [H⁺] of pH 4. Treating the pH numbers as an ordinary linear scale gives the wrong comparison.

Use pH = −log₁₀[H⁺]. The minus sign means a larger hydrogen-ion concentration gives a smaller pH value.

For [H⁺] = 1.0 × 10⁻³ mol dm⁻³, pH = 3.00. Keep the concentration in mol dm⁻³ and use the log key, not natural log.

For [H⁺] from pH, reverse the logarithm: [H⁺] = 10^(−pH). Put the whole negative pH in the exponent; calculator brackets matter here.

The concentration answer is in mol dm⁻³. A pH increase of 1 should make [H⁺] ten times smaller, which is a useful check on the result.

For a strong acid, find [H⁺] from the acid concentration and the stoichiometry of proton release after dealing with any dilution or reaction. Then use pH = −log₁₀[H⁺].

Do not apply weak-acid equilibrium approximations to a strong acid. The key chemical step is complete dissociation plus the correct mole ratio to H⁺.

For pure water, remember that it self-ionises only very slightly. In the simplified equilibrium H₂O ⇌ H⁺ + OH⁻, it produces equal concentrations of H⁺ and OH⁻.

Those concentrations are small because the equilibrium lies strongly towards molecular water. The equilibrium nevertheless gives the Kw relationship used in pH calculations.

Connect Kw directly to water self-ionisation: Kw = [H⁺][OH⁻]. The concentration of liquid water is effectively constant, so it is absorbed into the equilibrium constant.

At 25 °C, Kw is commonly 1.00 × 10⁻¹⁴ mol² dm⁻⁶. Use the value supplied if the temperature or question gives a different one.

If [H⁺] or [OH⁻] is known, find the other from Kw divided by the known concentration. You can then convert [H⁺] to pH if required.

Write Kw = [H⁺][OH⁻] before rearranging. It prevents the common mistake of using pH directly in the equilibrium expression.

When you use Kw, remember that its value changes with temperature. Neutral water still means [H⁺] = [OH⁻], but neutral pH is only exactly 7 when Kw makes each concentration 1.0 × 10⁻⁷ mol dm⁻³.

So “neutral” means equal hydrogen- and hydroxide-ion concentrations, not permanently pH 7 at every temperature.

For a strong base, find [OH⁻] first from the concentration and stoichiometry. Then use [H⁺] = Kw/[OH⁻] and finish with pH = −log₁₀[H⁺].

Do not put [OH⁻] directly into the pH expression. AQA’s route here is through Kw unless the question has supplied an equivalent shortcut explicitly.

Compare strong and weak acids or bases: keep strength separate from concentration. Strong species ionise essentially completely in water; weak ones establish equilibria with only partial ionisation.

Strength is about degree of ionisation, not concentration. A dilute strong acid can have lower [H⁺] than a concentrated weak acid while still being the stronger acid in the chemical sense.

For HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻]/[HA]. Larger Ka means the weak acid is more ionised and therefore stronger; pKa = −log₁₀Ka reverses that numerical ordering.

So stronger weak acid → larger Ka → smaller pKa. Keep that chain together when you compare acids.

Write the weak-acid equilibrium first, then build Ka from equilibrium concentrations: for HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻]/[HA].

Liquid water does not appear in this expression. Use equilibrium concentrations, not the initial acid concentration unless an approximation has been justified.

For the usual weak monoprotic acid approximation, let [H⁺] = [A⁻] = x and take [HA] ≈ the initial acid concentration c when dissociation is small. Then Ka ≈ x²/c.

We find x = [H⁺] first and only then take −log₁₀ to obtain pH. If the approximation is questionable from the data supplied, use the equilibrium expression more exactly.

Use pKa = −log₁₀Ka and Ka = 10^(−pKa). A smaller pKa should correspond to a larger Ka and a stronger acid.

That ordering is an immediate calculator check: if both Ka and pKa increase together, the logarithm has been handled incorrectly.

At the equivalence point, I want you to think stoichiometry: the acid and base have reacted in the exact mole ratio from the balanced equation. That does not automatically mean the pH is 7.

The pH at equivalence depends on the strengths of the acid and base and on any conjugate species left in solution.

Use the measured titre to calculate moles of the known solution, apply the acid–base mole ratio, then divide by the relevant sample volume if the unknown concentration is required.

Keep the pipetted sample volume and the burette titre separate. They play different roles in n = cV and mixing them is a very easy way to obtain a plausible but wrong concentration.

Read titration curves by their starting pH, buffering region where present, steep section and equivalence pH. Strong acid–strong base has an equivalence point near pH 7 at 25 °C.

Weak acid–strong base has an equivalence point above 7 because the conjugate base hydrolyses; strong acid–weak base has an equivalence point below 7 because the conjugate acid is acidic.

When you sketch a titration curve, show the initial pH, the gradual region, any buffer region, the steep change and the equivalence point in chemically sensible positions.

The curve direction depends on what starts in the flask. The equivalence pH depends on acid/base strength, so do not force every vertical section to be centred on pH 7.

Choose an indicator whose transition range lies within the steep pH change around equivalence. Then a tiny volume change carries the indicator through its colour transition and the endpoint closely approximates equivalence.

Do not choose an indicator just because its midpoint equals the equivalence pH. What matters is that its whole useful transition range sits inside a sufficiently steep part of the curve.

A buffer limits large pH changes when small amounts of acid or base are added, and its pH also changes relatively little on dilution. It does not keep pH perfectly fixed.

Think of the pair as two chemical “mops”: the conjugate base removes a small addition of H⁺, while the weak acid removes a small addition of OH⁻. The equilibrium then shifts, so [H⁺] changes only modestly.

An acidic buffer contains a weak acid HA together with a source of its conjugate base A⁻. You can make one from the weak acid plus a soluble salt of A⁻, or by partially neutralising the weak acid.

Both HA and A⁻ must remain in significant amounts. Complete neutralisation would remove the weak acid and destroy the buffer pair.

For a basic buffer, picture a weak base B together with a source of its conjugate acid BH⁺. You often get the BH⁺ from a salt or by partially neutralising the weak base.

The pair works because B can remove added H⁺ while BH⁺ can replace H⁺ consumed by added OH⁻ through the relevant acid–base equilibria.

In a buffer question, name the conjugate pair and show which member deals with added H⁺ or OH⁻. That gives the chemistry behind the small pH change in a chemical or biological context.

If the context includes dilution, remember that both members are diluted by the same factor, so their ratio can remain similar even though their absolute concentrations fall.

For an acidic buffer HA/A⁻, added H⁺ is removed by A⁻ and added OH⁻ is removed by HA. For a basic buffer B/BH⁺, B removes added H⁺ and BH⁺ supplies H⁺ to neutralise added OH⁻.

Write the relevant proton-transfer reaction if the direction feels slippery. It is much clearer than describing the buffer as if it physically “absorbs” acid or base.

For an acidic buffer HA/A⁻, rearrange Ka to [H⁺] = Ka[HA]/[A⁻], use the relevant equilibrium quantities for the two buffer components and then convert [H⁺] to pH.

If the buffer was made by partial neutralisation, we do the stoichiometric reaction in moles first. The moles of HA left and A⁻ formed then determine the ratio used in the equilibrium calculation.