Acid–base equilibria & buffers
The idea
Weak acids and bases only partially ionize, so their solutions are governed by equilibrium constants: Ka for an acid HA ⇌ H⁺ + A⁻, Kb for its conjugate base, with Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 25 °C. The pKa scale compresses these constants logarithmically and pairs naturally with pH. You already know pH = −log[H₃O⁺] and the strong/weak distinction; the university skill is computing pH for weak systems, conjugate pairs, and — most importantly — buffers.
A buffer is a deliberate mixture of a weak acid and its conjugate base in comparable amounts. Added strong acid is absorbed by the base member, added strong base by the acid member, so the pH moves only slightly. The Henderson–Hasselbalch equation, pH = pKa + log([base]/[acid]), makes the arithmetic nearly mental: when the two members are equal, pH = pKa, and each factor of ten in the ratio moves the pH one unit. Strong acid or base added to a buffer reacts essentially completely first — do that stoichiometry before any equilibrium math.
Correct the notion that a buffer freezes pH. It only moderates change, and only within its capacity: once added acid or base consumes one member entirely, the buffering collapses. Buffers work best within about one pH unit of pKa, where both members remain plentiful.
Worked example
A buffer is 0.250 M in acetic acid (Ka = 1.8 × 10⁻⁵) and 0.400 M in sodium acetate. Find the pH, and then the pH after 0.050 mol of HCl is added to 1.00 L of this buffer (assume no volume change).
- Find pKa = −log(1.8 × 10⁻⁵) = 4.74. Both buffer members are present in comparable concentrations, so Henderson–Hasselbalch applies directly.
- Initial pH = 4.74 + log(0.400/0.250) = 4.74 + log(1.60) = 4.74 + 0.20 = 4.94. With more base than acid, the pH sensibly sits above pKa.
- Treat the added HCl as reacting completely with the base member: acetate falls to 0.400 − 0.050 = 0.350 M while acetic acid rises to 0.250 + 0.050 = 0.300 M. This neutralization step always precedes the equilibrium step.
- Recompute: pH = 4.74 + log(0.350/0.300) = 4.74 + log(1.17) = 4.74 + 0.07 = 4.81.
- Interpret the protection: the same 0.050 mol of HCl in 1.00 L of pure water would give pH = −log(0.050) = 1.30, a plunge of 5.7 units from neutral; the buffer held the change to 0.13 units.
Answer. The buffer pH is 4.94, falling only to 4.81 after the acid — a drop of 0.13 units versus the crash to pH 1.30 the same acid causes in plain water.
Check your understanding
- Why does a buffer resist pH change in both directions, and what determines how much it can absorb before failing?
- How would you choose the best conjugate pair to buffer at pH 9.2, and what ratio of base to acid would you mix?
- Why must added strong acid be handled with stoichiometry before any equilibrium calculation, and what error results from skipping that order?
- What does the relation Ka × Kb = Kw imply about the basicity of the conjugate of a very weak acid?
Build the foundations first
Acid–base equilibria & buffers builds on these concepts. If any feel shaky, start there.