25. Equilibria

In Brønsted–Lowry chemistry, an acid donates a proton and a base accepts one. The conjugate acid of a base is the species after it gains H+, while the conjugate base of an acid is what remains after it loses H+.

Conjugate pairs differ by exactly one proton. Water can act as either acid or base, so identify the direction from the equation.

NH₃/NH₄⁺ is a base/conjugate-acid pair; H₂O/OH⁻ is an acid/conjugate-base pair.

Conjugate does not mean “opposite charge” alone; track the proton transfer.

In HA + B ⇌ A⁻ + BH⁺, HA/A⁻ and B/BH⁺ are conjugate pairs. Each pair differs by one H⁺ and the charges change accordingly.

Mark the proton donor and acceptor first, then pair each reactant with its product. This prevents confusing the acid with its conjugate acid.

HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻ contains HCO₃⁻/H₂CO₃ and H₂O/OH⁻ pairs; bicarbonate is amphiprotic.

Do not pair species that differ by two protons or by an unrelated atom.

pH = −log[H⁺], Ka = [H⁺][A⁻]/[HA], pKa = −log Ka, and Kw = [H⁺][OH⁻] at a specified temperature. Concentrations are treated consistently with the syllabus approximation.

A larger Ka or smaller pKa means a stronger weak acid. Convert logs and powers carefully and state the temperature when using Kw.

If [H⁺]=1.0×10⁻³ mol dm⁻³, pH=3.00. If Ka=1.0×10⁻⁵, pKa=5.00.

pH is not [H⁺] itself, and pKa is not a concentration. Do not introduce Kb or Kw=KaKb when outside the assessed scope.

For a strong monoprotic acid, [H⁺]≈c; for a strong alkali, [OH⁻]≈c and use Kw to obtain [H⁺]. For a weak acid, use Ka and the equilibrium concentration rather than assuming complete dissociation.

Include stoichiometric H⁺/OH⁻ numbers for polyprotic or multi-hydroxide species when appropriate. Check that the weak-acid approximation is small relative to the initial concentration.

0.010 mol dm⁻³ HCl has pH 2.00; 0.010 mol dm⁻³ NaOH has pOH 2.00 and pH≈12.00 at 25 °C.

Do not use strong-acid shortcuts for weak acids or forget that pH + pOH depends on temperature.

A buffer contains a weak acid and its conjugate base, or a weak base and its conjugate acid. Added H⁺ is consumed by the base component and added OH⁻ by the weak acid component.

For the carbonic buffer, H₂CO₃ ⇌ H⁺ + HCO₃⁻; HCO₃⁻ removes added H⁺ and H₂CO₃ neutralises added OH⁻. Buffer action is finite and works best when both components are present in comparable amounts.

Blood bicarbonate helps moderate pH changes, but ventilation and kidney regulation continuously alter CO₂/HCO₃⁻ balance.

A buffer does not keep pH absolutely constant and is not just any neutral salt solution.

For a buffer, use the weak-acid equilibrium to relate [H⁺], Ka, [HA] and [A⁻]. The pH depends mainly on the ratio of conjugate base to weak acid, not their common scale alone.

Account for dilution or neutralisation before substituting. Keep concentrations in the same units and use pH = pKa + log([A⁻]/[HA]) when that form is permitted.

If [A⁻]=[HA], pH≈pKa. Adding a small amount of strong acid consumes A⁻ and forms HA, so the ratio changes only modestly.

A buffer is not strongest when one component is absent, and pH is not determined by total concentration alone.

The solubility product Ksp is the equilibrium constant for a solid dissolving into its aqueous ions. Pure solids are omitted; aqueous-ion concentrations are raised to their stoichiometric powers.

A small Ksp generally indicates low solubility, but the numerical comparison is meaningful only for salts with comparable dissolution stoichiometry.

For AgCl(s) ⇌ Ag⁺ + Cl⁻, Ksp=[Ag⁺][Cl⁻]. For CaF₂(s) ⇌ Ca²⁺ + 2F⁻, the expression has [F⁻]².

Ksp is not the concentration of the solid and cannot be compared blindly across different ion powers.

Write the saturated dissolution equilibrium first, then multiply aqueous-ion concentrations according to the coefficients. The undissolved solid has activity one and is not included.

Use parentheses for polyatomic ions and distinguish charges from stoichiometric powers. The expression is tied to the chosen dissolution direction.

Al(OH)₃(s) ⇌ Al³⁺ + 3OH⁻ gives Ksp=[Al³⁺][OH⁻]³; PbI₂(s) gives Ksp=[Pb²⁺][I⁻]².

Do not include [Al(OH)₃] or write powers from ionic charges instead of equation coefficients.

Let the molar solubility be s, express each ion concentration as its stoichiometric multiple of s, then substitute into Ksp. Conversely, solve the expression for s from measured ion concentrations.

Use the correct power and units; for salts producing several ions, the solubility is not equal to every ion concentration.

For AgCl, if s=1.0×10⁻⁵ mol dm⁻³ then Ksp=s²=1.0×10⁻¹⁰. For CaF₂, [F⁻]=2s, so Ksp=4s³.

Do not use s² for every salt or forget that common ions change the initial concentration before equilibrium.

Adding an ion already present in a dissolution equilibrium shifts the equilibrium toward the solid, reducing the solubility. Ksp remains constant at fixed temperature; the ion concentrations change.

Set up the common-ion concentration before adding the small solubility contribution, then substitute into Ksp. The approximation is valid only when the added ion dominates.

AgCl is less soluble in NaCl solution because added Cl⁻ shifts AgCl(s) ⇌ Ag⁺ + Cl⁻ left. Calculate [Ag⁺] from Ksp/[Cl⁻] when [Cl⁻] is known.

The common ion does not change Ksp itself and does not always make precipitation instantaneous.

For a solute distributed between two immiscible solvents, Kpc = concentration in one named solvent divided by concentration in the other at equilibrium and fixed temperature.

State which solvent is numerator and keep phases/units consistent. Kpc describes equilibrium distribution, not the total amount extracted in one operation.

If Kpc(organic/aqueous)=4, the equilibrium concentration in the organic phase is four times that in the aqueous phase under the stated conditions.

Do not invert Kpc without changing the definition, and do not assume a large Kpc means one extraction removes all solute.

For a solute in the same physical state in two immiscible solvents, Kpc = concentration in solvent 1 divided by concentration in solvent 2 at equilibrium.

State the numerator solvent, use matched units and measure after the two phases have equilibrated. If the solute associates or reacts in one phase, the simple expression may not apply.

If [solute]organic=0.80 mol dm⁻³ and [solute]aqueous=0.20 mol dm⁻³, Kpc(organic/aqueous)=4.0.

Do not invert the ratio without changing the label, and do not calculate from initial concentrations before equilibrium.

A solute partitions according to its relative affinity for the two phases. Similar polarity and intermolecular forces favour dissolution in a solvent, while a polarity mismatch favours the other phase.

Hydrogen bonding, ionisation and temperature can alter Kpc. Predict direction qualitatively, but do not treat polarity as the only possible factor.

A non-polar hydrocarbon generally partitions more into an organic solvent than water; an ionised acid may remain preferentially in the aqueous phase.

A larger Kpc is not an intrinsic label independent of solvent order; reversing numerator and denominator gives the reciprocal.