IMAT Chemistry · Acids, Bases and pH

Neutralisation, Titration, Indicators and Buffers

Acids and bases neutralise each other mole for mole of H⁺ and OH⁻; an indicator shows when that point is reached, and a buffer is a mixture that resists changes in pH.

All Acids, Bases and pH notes

Why this matters

Only one past question, from 2019, is on this page: the colours of universal indicator as an alkali is neutralised. Titration calculations and buffers are on the syllabus but have not been asked, so learn them as likely new questions.

Concept 1 of 3: Neutralisation and titration calculations

However the acid and base are dressed up, the reaction underneath is one

\mathrm{H^+}

meeting one

\mathrm{OH^-}

to make water. So neutralisation is complete when the moles of

\mathrm{H^+}

supplied equal the moles of

\mathrm{OH^-}

supplied. A titration measures the volume needed to reach that point.

Definition

Neutralisation: acid + base $\rightarrow$ salt + water; the ionic equation is $\mathrm{H^+ + OH^- \rightarrow H_2O}$ .
In a titration, a solution of known concentration is added from a burette until the reaction is just complete.
The equivalence point is where moles of $\mathrm{H^+}$ equal moles of $\mathrm{OH^-}$ . The end point is where the indicator changes colour; a good indicator makes them coincide.
Moles = concentration × volume in litres. Count 2 for each $\mathrm{H_2SO_4}$ or $\mathrm{Ba(OH)_2}$ .
If one reagent is in excess, the leftover moles divided by the TOTAL volume give the final concentration and pH.

At the equivalence point

n_a\, c_a V_a = n_b\, c_b V_b

$c_a, V_a$ concentration and volume of the acid
$c_b, V_b$ concentration and volume of the base
$n_a, n_b$ $\mathrm{H^+}$ per acid formula, $\mathrm{OH^-}$ per base formula

Worked example

20.0 mL of sulfuric acid is exactly neutralised by 25.0 mL of 0.10 mol/L sodium hydroxide. Find the concentration of the sulfuric acid.

Practice this conceptself-check · 3 quick reps

Use the total volume after mixing

When an acid and a base are mixed and one is in excess, the leftover

\mathrm{H^+}

\mathrm{OH^-}

is spread through BOTH volumes added together. Dividing by only the acid's volume gives a concentration that is too high and a pH that is off.

Concept 2 of 3: Titration curves, equivalence point pH and indicators

As base is added to an acid, the pH creeps up slowly, then leaps by several units around the equivalence point, then levels off. Where that leap is centred depends on the salt formed, since the salt itself may be acidic or basic. An indicator is a weak acid whose two forms have different colours, and it is useful only if its colour change falls inside the leap.

Definition

The pH at the equivalence point is the pH of the salt solution formed (see the salt rules on the previous page).
An acid-base indicator changes colour over about two pH units. Choose one whose range lies on the steep part of the curve.
Litmus: red in acid, blue in alkali.
Phenolphthalein: colourless below about pH 8.2, pink above about pH 10.
Methyl orange: red below about pH 3.1, yellow above about pH 4.4.
Universal indicator is a mixture: red at low pH, then orange, yellow, green at pH 7, blue, and violet (purple) at high pH.

Titration	pH at equivalence	Salt formed	Suitable indicator
Strong acid + strong base	7	Neutral, e.g. $\mathrm{NaCl}$	Phenolphthalein or methyl orange (the jump is very large)
Weak acid + strong base	Above 7, about 8 to 9	Basic, e.g. $\mathrm{CH_3COONa}$	Phenolphthalein
Strong acid + weak base	Below 7, about 5	Acidic, e.g. $\mathrm{NH_4Cl}$	Methyl orange
Weak acid + weak base	Near 7	Depends on Ka and Kb	None works well: there is no sharp jump

Practice this conceptself-check · 3 quick reps

The end point is not always at pH 7

The equivalence point sits at pH 7 only for a strong acid with a strong base. With a weak acid it is basic, with a weak base it is acidic, because the salt formed hydrolyses. Pick the indicator by where the jump is, not by where 7 is.

Concept 3 of 3: Buffer solutions and the Henderson-Hasselbalch equation

A buffer holds a weak acid and its conjugate base side by side. Added

\mathrm{H^+}

is taken up by the base; added

\mathrm{OH^-}

is neutralised by the acid. Either way the change becomes a small shift in the acid to base ratio, and the pH depends only on the logarithm of that ratio, so it barely moves.

Definition

A buffer is a solution that resists changes in pH when small amounts of acid or base are added, or when it is diluted.
It is made from a weak acid and its conjugate base ( $\mathrm{CH_3COOH}$ with $\mathrm{CH_3COONa}$ ) or a weak base and its conjugate acid ( $\mathrm{NH_3}$ with $\mathrm{NH_4Cl}$ ). A strong acid with its salt is NOT a buffer.
Added acid: $\mathrm{A^- + H^+ \rightarrow HA}$ . Added base: $\mathrm{HA + OH^- \rightarrow A^- + H_2O}$ .
Henderson-Hasselbalch: pH equals pKa plus the log of (base over acid). With equal amounts, pH = pKa. A buffer works best within about one pH unit of its pKa.
A weak acid half-neutralised by a strong base is a buffer with pH = pKa.
Blood is kept at pH 7.35 to 7.45, mainly by the carbonic acid and hydrogencarbonate buffer: $\mathrm{CO_2 + H_2O \rightleftharpoons H_2CO_3 \rightleftharpoons H^+ + HCO_3^-}$ . The lungs remove $\mathrm{CO_2}$ and the kidneys adjust $\mathrm{HCO_3^-}$ . Phosphate and proteins (including haemoglobin) buffer too. Blood pH below the range is acidosis, above it alkalosis.

Henderson-Hasselbalch equation

\text{pH} = \text{p}K_a + \log_{10}\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}

$[\mathrm{A^-}]$ concentration (or moles) of the conjugate base
$[\mathrm{HA}]$ concentration (or moles) of the weak acid

Worked example

A buffer contains 0.20 mol of a weak acid HA (pKa 4.8) and 0.020 mol of its sodium salt NaA in 1.0 L. Find its pH, and say what happens to the pH if the buffer is diluted to 2.0 L.

Practice this conceptself-check · 4 quick reps

A strong acid with its own salt is not a buffer

HCl with NaCl cannot resist added acid, because

\mathrm{Cl^-}

does not take up protons. A buffer needs a WEAK acid with its conjugate base, or a weak base with its conjugate acid.

A buffer slows pH change; it does not stop it

A buffer absorbs small additions. Add more acid than it has base to react with and its capacity is used up, after which the pH falls sharply. Options claiming the pH can never change are wrong.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Neutralisation and titration calculations
At the equivalence point
$n_a\, c_a V_a = n_b\, c_b V_b$
Buffer solutions and the Henderson-Hasselbalch equation
Henderson-Hasselbalch equation
$\text{pH} = \text{p}K_a + \log_{10}\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}$

Reference tables (1)

Titration curves, equivalence point pH and indicators4 rows

Titration	pH at equivalence	Salt formed	Suitable indicator
Strong acid + strong base	7	Neutral, e.g. $\mathrm{NaCl}$	Phenolphthalein or methyl orange (the jump is very large)
Weak acid + strong base	Above 7, about 8 to 9	Basic, e.g. $\mathrm{CH_3COONa}$	Phenolphthalein
Strong acid + weak base	Below 7, about 5	Acidic, e.g. $\mathrm{NH_4Cl}$	Methyl orange
Weak acid + weak base	Near 7	Depends on Ka and Kb	None works well: there is no sharp jump

Watch out for (4)

Use the total volume after mixing
→ Neutralisation and titration calculations
The end point is not always at pH 7
→ Titration curves, equivalence point pH and indicators
A strong acid with its own salt is not a buffer
→ Buffer solutions and the Henderson-Hasselbalch equation
A buffer slows pH change; it does not stop it
→ Buffer solutions and the Henderson-Hasselbalch equation

Drill the past-year questions on this topic

Open the bank with this topic pre-filtered.

Open in the bank