Everything from the 4 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
Water ionises slightly: H2O ⇌ H+ + OH−, and the ionic product of water Kw = [H+][OH−] = 1.00 × 10−14 mol2 dm−6 at 298 K. pH = −log[H+]; for a strong acid [H+] equals the acid concentration, and for a strong base [H+] = Kw ÷ [OH−], so 0.0100 mol dm−3 NaOH has pH 12.0. A weak acid HA is only partly ionised, described by the acid dissociation constant Ka = [H+][A−] ÷ [HA], and pKa = −log Ka (smaller pKa, stronger acid). Assuming [H+] = [A−] and that very little HA ionises, [H+] = √(Ka × c): 0.100 mol dm−3 ethanoic acid (Ka 1.74 × 10−5 mol dm−3) has pH 2.88.
Do not set [H+] equal to the concentration of a weak acid; that is only true for strong acids. For a strong base, find [OH−] first, then use Kw; quoting −log[OH−] as the pH is the classic slip (that is pOH). Remember that Ba(OH)2 releases two OH− per formula unit, and that pure water is neutral at every temperature even when its pH is not 7.
A buffer solution is a solution that resists changes in pH when small amounts of acid or alkali are added. An acidic buffer contains a weak acid and its conjugate base, for example CH3COOH with CH3COONa: added H+ is removed by CH3COO− + H+ → CH3COOH, and added OH− is removed by CH3COOH + OH− → CH3COO− + H2O. An alkaline buffer such as NH3 with NH4Cl works the same way using NH3 and NH4+. Because both reservoirs are large, their ratio hardly changes, so the pH, given by [H+] = Ka × [HA] ÷ [A−], hardly changes. Blood is buffered by H2CO3 / HCO3−.
A buffer is not a solution whose pH never changes: it resists change from small additions, and fails once one reservoir is used up. In buffer calculations use the salt concentration for [A−], not [H+]; the √(Ka c) shortcut is for a weak acid on its own. When acid and alkali are mixed to make a buffer, work in moles after reaction before taking the ratio.
In a saturated solution of a sparingly soluble salt, the solid is in dynamic equilibrium with its ions: AgCl(s) ⇌ Ag+(aq) + Cl−(aq). The solubility product is the product of the ion concentrations, each raised to the power of its coefficient: Ksp = [Ag+][Cl−] (mol2 dm−6), Ksp = [Pb2+][I−]2 (mol3 dm−9). The solid does not appear. From a solubility s: AgCl has s = 1.34 × 10−5 mol dm−3, so Ksp = s2 = 1.80 × 10−10 mol2 dm−6; for PbI2, Ksp = 4s3. Adding a common ion shifts the equilibrium to the left and lowers the solubility. If the ionic product of mixed solutions exceeds Ksp, a precipitate forms.
Square or cube the right terms: for PbI2, [I−] = 2s and Ksp = 4s3, not s3 or s2. When two solutions are mixed, each concentration is diluted (halved for equal volumes) before you work out the ionic product. The common ion lowers the solubility, not Ksp. Units of Ksp depend on the formula: mol2 dm−6 for AB, mol3 dm−9 for AB2.
When a solute X is shaken with two immiscible solvents, such as hexane and water, it distributes itself between them and reaches a dynamic equilibrium, X(aq) ⇌ X(organic). The partition coefficient is the ratio of the equilibrium concentrations, Kpc = [X]organic ÷ [X]aqueous, which is constant at a fixed temperature and has no units. A non-polar solute such as iodine is more soluble in a non-polar solvent, so Kpc (organic over water) is greater than 1; a polar solute such as ammonia, which hydrogen bonds to water, has Kpc less than 1. Because the amount extracted depends on the volumes as well as Kpc, several small portions of solvent extract more than one large portion.
Kpc is a ratio of concentrations, not masses: divide each amount by its own layer's volume. Keep the order as stated (organic over aqueous here); inverting it gives 1 ÷ Kpc. The amount left in water after extraction is the starting amount minus x, and each new portion starts from what is left. A large Kpc means the solute prefers the organic layer, which needs a non-polar solute.