Acids, bases and buffers: strong does not mean concentrated

Before calculating a pH, identify the chemical situation. A logarithm cannot repair an incorrect assumption about how many hydrogen ions the acid produces.

Strength, concentration and pH answer different questions

Quantity	What it describes	What it does not establish alone
Acid strength	Tendency to transfer a proton in a given medium	How much acid is in the solution
Concentration	Amount per volume	Fraction ionised
pH	Hydrogen-ion activity on a logarithmic scale	Total amount of acid present

In water, a strong acid such as HCl is essentially completely ionised under the usual introductory conditions. A weak acid establishes an equilibrium with a substantial undissociated fraction. For weak acids compared in the same solvent at the same temperature, larger Ka means greater acid strength.

A dilute strong acid can have a higher pH than a more concentrated weak acid. “Strong acid” therefore does not, by itself, tell you which of two unspecified solutions has the lower pH. OpenStax explains the equilibrium basis of acid strength.

Find conjugate pairs by tracking one proton

In the Brønsted–Lowry model, an acid donates H⁺ and a base accepts it. A conjugate acid–base pair differs by one proton, with a corresponding one-unit difference in charge.

For HA + H₂O ⇌ H₃O⁺ + A⁻, the pairs are HA/A⁻ and H₃O⁺/H₂O. HA and H₂O are the reacting acid and base, but they are not each other’s conjugates.

Another example is NH₄⁺/NH₃: ammonium donates a proton to become ammonia. The stronger an acid, the weaker its conjugate base relative to other pairs in the same medium. See Brønsted–Lowry acid–base reactions.

A buffer needs both partners

A weak acid and its conjugate base provide complementary reactions:

• Added H⁺ reacts with A⁻ to form HA.
• Added OH⁻ reacts with HA to form A⁻ and water.

The pH changes less than it would in an unbuffered solution, but it is not locked at one value. Nor must a buffer have pH 7. Its useful pH range depends on the conjugate pair. OpenStax’s buffer section explains both action and finite capacity.

Worked example: account for the reaction first

An original practice mixture contains 10 mmol HA and 10 mmol A⁻. Add 2 mmol of a strong acid, assuming it reacts essentially completely with A⁻ and neither buffer component is exhausted.

The new amounts are 12 mmol HA and 8 mmol A⁻. The base-to-acid ratio falls from 1 to 2/3, so the pH falls.

Under the usual buffer approximations, the Henderson–Hasselbalch relation gives:

pH ≈ pKa + log₁₀([A⁻]/[HA])

The new pH is approximately pKa − 0.18. Amounts can replace concentrations in this ratio because both species occupy the same final volume. Do the neutralisation step before applying the equation.

The limits that change the answer

Equal dilution of both partners approximately preserves their ratio, but reduces capacity per unit volume. At extreme dilution, the simple approximation can fail. Adding enough strong acid or base to exhaust one partner also invalidates the ordinary buffer calculation.

Finally, strong and weak are not synonyms for dangerous and safe. They describe chemical behaviour, not a complete assessment of a substance’s hazards.

If the chemistry is clear but the logarithm goes wrong, continue with the maths behind pH.