Acid–Base Formulae

Introduction

Acid–base formulae are compact forms of equilibrium reasoning. pH is a logarithmic measure of hydrogen-ion activity; Ka and Kb measure dissociation tendencies; Kw connects conjugate hydronium and hydroxide amounts in water. The Henderson–Hasselbalch equation is useful for a buffer but can fail near exhaustion of one component or in strongly nonideal solutions.

Core explanation

Thermodynamically pH = −log₁₀ a H⁺, where a H⁺ is a dimensionless activity based on a chosen convention. In dilute classroom calculations, a H⁺ is often approximated by [H₃O⁺]/c°, giving pH ≈ −log₁₀([H₃O⁺] in mol L⁻¹). If [H₃O⁺] ≈ 1.0 × 10⁻³ M, pH ≈ 3.00. A negative logarithm turns tenfold concentration changes into one pH unit.

For HA ⇌ H⁺ + A⁻, Ka = a H⁺a A⁻/a HA. A small Ka means the acid remains relatively undissociated under the standard comparison; pKa = −log₁₀Ka. For a base B + H₂O ⇌ BH⁺ + OH⁻, Kb describes proton acceptance in water. A conjugate acid–base pair obeys KaKb = Kw at the same temperature when thermodynamic constants and conventions are consistent.

Water autoionization gives Kw = a H⁺a OH⁻, approximately 1.0 × 10⁻¹⁴ at 25 °C in common dilute approximations. Thus pH + pOH ≈ 14.00 at 25 °C, but 14 is not universal across temperature because Kw varies. “Neutral pH” means equal hydronium and hydroxide activities under the model, not necessarily a number of exactly seven at every temperature.

For a weak monoprotic acid initially at concentration C with negligible initial A⁻, the usual dilute approximation gives [H⁺] ≈ √(Ka C) when dissociation x is much smaller than C and water autoionization is negligible. Test x/C after calculating. If that ratio is not small, solve Ka = x²/(C − x) more exactly. The approximation also fails for extremely dilute acid where water's contribution matters.

A buffer contains appreciable HA and A⁻. Rearranging Ka gives the Henderson–Hasselbalch form pH ≈ pKa + log₁₀([A⁻]/[HA]) under activity-to-concentration approximation. It is most reliable when both components remain present, total concentrations are sufficient and ionic-strength effects are limited. Before using it after added strong acid or base, perform stoichiometric neutralization first: added H⁺ consumes A⁻, and added OH⁻ consumes HA.

Buffer capacity depends on absolute amounts as well as ratio. Two buffers with the same [A⁻]/[HA] have similar initial pH but the more concentrated one can absorb more acid or base before its ratio changes greatly. A 1:1 pair has pH near pKa and typically strong local resistance to pH change, although detailed capacity also depends on conditions.

Polyprotic acids have multiple Ka values and species, while very strong acids or concentrated electrolytes need activity treatment. One pKa or one square-root expression may not describe the whole system. Write the relevant dissociation and mass balance before choosing a shortcut.

Step-by-step reasoning

Identify acid and conjugate base species. Write the dissociation equilibrium and any strong-acid/base stoichiometric reaction first. Select pH, weak-acid or buffer expression based on which species remain. Calculate with dimensionless logarithm arguments, then test small-dissociation and concentration assumptions.

Visual explanation

Draw two containers connected by a reversible arrow: HA on one side and H⁺ plus A⁻ on the other. The Ka expression is written over the arrow. Beneath, show a buffer pair with acid addition converting A⁻ into HA and shifting their ratio.

Real-world analogy

A buffer resembles a pair of counters with two exchangeable forms of the same token. Added acid converts base-form tokens to acid-form tokens, and added base does the reverse. The analogy helps with stoichiometry but not with the logarithmic activity relationship.

Real-world example

Acetate buffers are prepared by combining acetic acid and acetate salt. Their pH is estimated from pKa and the acetate-to-acid ratio, but a measured pH can differ if ionic strength, temperature or activity coefficients matter. A pH electrode provides the operational check.

Why?

These formulae support titration, buffer preparation, environmental chemistry and biochemical experiments. The key skill is knowing when a convenient approximate expression follows from the full equilibrium model.

Common misconception

“pH + pOH is always exactly 14” ignores temperature and activity conventions. Another error applies Henderson–Hasselbalch after one buffer component has been completely consumed by added strong acid or base.

Worked example

A buffer has pKa = 4.76, [A⁻] = 0.20 M and [HA] = 0.10 M. Its approximate pH is 4.76 + log₁₀(2) ≈ 5.06. If strong acid consumes 0.05 mol A⁻ per litre and makes 0.05 mol HA per litre, new amounts are 0.15 and 0.15 M, so pH ≈ 4.76. This assumes final-volume change is negligible and both species remain substantial.

Quick check

1. What is the approximate pH when a buffer has equal conjugate-base and acid activities? Answer: pH is approximately pKa because the logarithm of their ratio is zero.

Exam focus

Use activities conceptually, specify dilute approximations, and perform strong-reagent stoichiometry before buffer equilibrium. Check whether a weak-acid x/C approximation is small and whether 25 °C is assumed for Kw.

Advanced insight

A glass-electrode pH reading is operationally calibrated with standard buffers; single-ion activity cannot be measured independently without a convention. This is why rigorous pH and equilibrium work distinguishes activity-based definitions from simple concentration calculations.

Summary

pH, Ka, Kb and Kw describe linked aqueous acid–base equilibria. Henderson–Hasselbalch follows from Ka for a mixture containing both acid and conjugate base. Temperature, activities and component depletion limit shortcuts.

Practice questions

1. What pH corresponds approximately to [H₃O⁺] = 10⁻⁵ M in a dilute solution? Answer: About 5.00. 2. What is the relation between pKa and Ka? Answer: pKa = −log₁₀Ka. 3. Why neutralize added strong acid before using a buffer formula? Answer: It changes the amounts of A⁻ and HA, which determine the new equilibrium ratio. 4. Does a 1:1 buffer ratio guarantee large capacity? Answer: No. Capacity also depends on total buffer concentration or amount.

Sources

- OpenStax Chemistry 2e: Buffers. - OpenStax Chemistry 2e: Acids and Bases.