5.1.3 Acids, Bases and Buffers

A Brønsted–Lowry acid donates a proton, H⁺. I’d learn that wording exactly because it is short and it tells you what to look for in an equation.

When you are identifying the acid in a reaction, find the species that loses H⁺ as the reaction goes forward.

A Brønsted–Lowry base accepts a proton, H⁺. In an equation, look for the species that gains H⁺.

That definition is broader than simply calling a base an alkali. NH₃, for example, is a Brønsted–Lowry base because it accepts H⁺ to form NH₄⁺.

A conjugate acid–base pair differs by one H⁺. If you can turn one species into the other by adding or removing a single proton, you have found a conjugate pair.

For CH₃COOH ⇌ H⁺ + CH₃COO⁻, the pair is CH₃COOH/CH₃COO⁻. In NH₄⁺ ⇌ H⁺ + NH₃, it is NH₄⁺/NH₃.

The prefix tells you how many protons one acid molecule can donate: monobasic acids can donate one H⁺, dibasic acids two, and tribasic acids three.

Do not confuse this with acid strength. “Strong” tells you how far an acid dissociates; “dibasic” tells you how many protons each molecule can donate.

• monobasic → 1 H⁺ per molecule
• dibasic → 2 H⁺ per molecule
• tribasic → 3 H⁺ per molecule

When an acid reacts with a reactive metal, H⁺ is reduced to hydrogen gas. Keep the metal in the equation because it is also changing.

For magnesium, two H⁺ ions are needed to make one H₂ molecule and to balance the +2 charge on Mg²⁺.

For a carbonate, the useful ionic equation is one I would recognise on sight: carbonate ions take up H⁺ and end up as carbon dioxide and water.

Check the charge before you move on: 2H⁺ and CO₃²⁻ give zero overall charge on the left, matching the neutral products.

A metal oxide reacts with H⁺ to give the metal ion and water. For a solid oxide such as MgO, keep MgO intact in the ionic equation; there are no free O²⁻ ions in the solution.

This is one I’d learn exactly. In a strong acid–strong alkali neutralisation, cancel the spectator ions and you are left with H⁺ reacting with OH⁻ to make water.

Before you finish, check the state symbols and the charge: +1 and −1 cancel.

For a weak acid HA, write its dissociation first: HA ⇌ H⁺ + A⁻. The acid dissociation constant, Kₐ, measures the position of this equilibrium.

Put the equilibrium concentrations of H⁺ and A⁻ on top and HA underneath. A larger Kₐ means the acid dissociates further, so it is the stronger weak acid.

pKₐ puts Kₐ on a log scale. Use pKₐ = −log₁₀Kₐ and reverse it with Kₐ = 10⁻ᵖKₐ.

The scale runs the opposite way to Kₐ: a stronger acid has a larger Kₐ but a smaller pKₐ. That reversal is the bit worth checking before you compare two acids.

Once you know [H⁺], the pH calculation is direct. Put the hydrogen-ion concentration in mol dm⁻³ into pH = −log₁₀[H⁺].

As a quick check, a larger [H⁺] should give a lower pH.

To go back from pH to concentration, undo the logarithm: [H⁺] = 10⁻ᵖᴴ. Your answer is a concentration, so give it in mol dm⁻³.

Keep the unrounded calculator value if you need [H⁺] for another step; round at the end of the full calculation.

In any aqueous solution, [H⁺] and [OH⁻] are linked by the ionic product of water, Kᵥ. At a fixed temperature, their product has a fixed value.

At 298 K, Kᵥ = 1.00 × 10⁻¹⁴ mol² dm⁻⁶. This is the link you need when a question gives you [OH⁻] but asks for pH.

A strong monobasic acid fully dissociates and gives one H⁺ per acid molecule, so after any dilution or stoichiometry you can take [H⁺] as the acid concentration.

Then use the ordinary pH equation. If the question involves mixing or dilution, do that chemistry first — the logarithm is the last step.

For a strong base such as NaOH, start with [OH⁻], not [H⁺]. Use Kᵥ to convert [OH⁻] into [H⁺], then calculate pH.

I’d use that route because it keeps the sequence clear: [OH⁻] → [H⁺] → pH.

For a weak monobasic acid, only a small fraction dissociates. If the usual approximations are valid, [HA] at equilibrium is almost the starting acid concentration and [H⁺] ≈ [A⁻].

That turns the Kₐ expression into Kₐ ≈ [H⁺]²/[HA]. From there you can find [H⁺] with a square root and then calculate pH.

After a weak-acid calculation, compare the [H⁺] you found with the starting acid concentration. The approximation assumes dissociation is small, so [H⁺] must be much smaller than [HA].

For a stronger weak acid, that may no longer be true. If a sizeable fraction has dissociated, [HA] at equilibrium is noticeably lower than its starting concentration and the shortcut becomes inaccurate. OCR can ask you to recognise that limitation, but it does not require you to solve a quadratic equation.

A buffer solution is a system that minimises changes in pH when small amounts of acid or base are added. A buffer only copes with a limited amount, which is why “small” matters.

For the acidic buffers in OCR, you need a weak acid and its conjugate base together: one deals with added OH⁻ and the other with added H⁺.

The simplest way to make an acid buffer is to mix a weak acid with a soluble salt containing its conjugate base. For ethanoic acid, CH₃COOH supplies the weak acid and CH₃COONa supplies CH₃COO⁻.

You now have both members of the conjugate pair in useful amounts: CH₃COOH can deal with added OH⁻ and CH₃COO⁻ can deal with added H⁺.

You can also make a buffer by adding enough strong alkali to neutralise only some of a weak acid. Do the neutralisation in moles first. Some HA is left and some A⁻ is formed, so the final mixture contains the conjugate pair.

This is where I’d pause before using Kₐ: write down the moles of HA left and A⁻ made, then convert to concentrations if you need them.

If you add a small amount of acid to an acid buffer, the extra H⁺ is removed mainly by the conjugate base, A⁻. It accepts the proton and forms more HA.

Because most of the added H⁺ is converted into weak acid, [H⁺] changes only a little and so does the pH.

Add a small amount of alkali and the weak acid, HA, reacts with the added OH⁻. That makes A⁻ and water.

The added OH⁻ is therefore largely removed before it can cause a large pH rise. You still have plenty of both HA and A⁻, so the buffer continues to work.

For a buffer, do not use the weak-acid square-root method. Rearrange the full Kₐ expression using the weak-acid concentration [HA] and conjugate-base concentration [A⁻].

I’d write the rearrangement before putting numbers in: [H⁺] = Kₐ[HA]/[A⁻]. That makes it much harder to swap the acid and salt concentrations.

The same buffer equation can be rearranged to find whatever the question has left unknown. Start by turning pH into [H⁺] if necessary, then write Kₐ = [H⁺][A⁻]/[HA] and rearrange for the missing quantity.

If the buffer was made by mixing or partial neutralisation, settle the moles first. Only then decide whether you need concentrations. If HA and A⁻ are in the same final volume, that common volume cancels in their ratio.

OCR uses the H₂CO₃/HCO₃⁻ conjugate pair as the blood-buffer example. Blood plasma is normally kept at about pH 7.35–7.45.

If extra H⁺ appears, HCO₃⁻ accepts it to form H₂CO₃. If OH⁻ is added, H₂CO₃ can donate H⁺ and the OH⁻ is converted into water. In each case the buffer removes most of the added acid or base before the pH can change very far.

When you sketch these, first decide what is in the flask and what is being added. If acid is in the flask and base is added, all four curves rise, but the starting pH, equivalence-point pH and steepness are different.

I’d learn these as one pattern. The acid strength affects the starting pH, and the acid/base strengths control the equivalence-point pH and how steep the change is.

• strong acid + strong base → equivalence about pH 7, very steep change
• weak acid + strong base → equivalence above pH 7, buffer region before equivalence
• strong acid + weak base → equivalence below pH 7, smaller steep region
• weak acid + weak base → no sharp vertical section

Read a pH curve in stages. The initial pH tells you about the solution in the flask, the shape before equivalence can reveal a weak acid or weak base, the steep section locates the equivalence region, and the final pH reflects excess titrant.

Do not assume equivalence means pH 7. Weak-acid/strong-base equivalence is above 7; strong-acid/weak-base equivalence is below 7.

Choose the indicator from the steep part of the titration curve around equivalence. Its transition range needs to lie inside that rapid pH change so a tiny change in added volume produces the visible colour change.

Do not choose an indicator from the starting pH of the acid or base. And for a weak-acid/weak-base titration there is no steep enough section, so OCR says no indicator is suitable.

Treat the indicator as a weak acid: HA ⇌ H⁺ + A⁻. HA and A⁻ have different colours, so changing pH changes their ratio and therefore the colour you see.

Adding acid increases [H⁺] and shifts the indicator equilibrium towards HA. Adding alkali removes H⁺ and shifts it towards A⁻. The colour change happens across a range of pH values because both forms are present during the transition.

Before you trust a pH reading, calibrate the meter with buffer solutions of known pH. Rinse the probe with distilled or deionised water between solutions so you do not carry one sample into the next, then place it in the solution and wait for a stable reading.

For a titration curve, record pH against the measured volume added. Use smaller additions through the steep region so you do not jump straight over the part of the curve you most need to locate.