Chemistry of Aqueous Solutions

A strong acid is almost fully ionised in dilute solution. A weak acid sets up an equilibrium in which only a small fraction is ionised. For a weak monoprotic acid, HA(aq) ⇌ H⁺(aq) + A^-(aq). In the same way, a strong base such as NaOH releases its hydroxide almost completely. A weak base such as NH₃ reacts only partly with water: NH₃ + H₂O ⇌ NH₄⁺ + OH^-.

At the same concentration, a weak monoprotic acid gives a lower hydrogen-ion concentration than a strong one, because less of it ionises. Diluting a weak acid can raise the fraction that ionises, even while [H⁺] itself falls. So the degree of ionisation and [H⁺] are different quantities. Do not treat them as the same.

At 298 K, K_w is about 1.00 × 10^-14 mol² dm^-6, so pH + pOH = 14.00. K_w changes with temperature. A neutral solution always has [H⁺] = [OH^-]. Its pH is exactly 7.00 only when pK_w = 14.00, that is, at 298 K.

For a conjugate pair HA/A^-, K_a(HA) × K_b(A^-) = K_w, so pK_a + pK_b = pK_w. This works only for a conjugate pair. For example, K_a of ethanoic acid times K_b of ammonia does not give K_w, because they are not a pair.

The weak-acid shortcut comes from K_a = x²/(c - x), where x = [H⁺] = [A^-]. If x is much smaller than c, you can replace c - x with c. Check by working out the fraction ionised: if it is small, the shortcut is fine. If it is not small, the shortcut is not justified. Syllabus questions will not ask you to solve a quadratic.

An acid-base indicator is usually a weak acid or base whose two forms have different colours. Its colour changes over a range of pH, often around pK_a ± 1. The endpoint is where you see the colour change. The equivalence point is where the amounts have reacted exactly. The two should be close, but they are not the same thing.

An indicator does not need to change colour at pH 7. What matters is how much titrant you add while the colour changes. On a steep part of the curve, this volume is tiny, so the endpoint error is small. If you are given unfamiliar indicators, compare their ranges with the steep part of the curve. Do not just pick a name you know.

An acidic buffer contains a good amount of a weak acid, HA, and of its conjugate base, A^-. You can make one by mixing the acid with its soluble salt, or by partly neutralising the acid. Added H⁺ reacts with A^-. Added OH^- reacts with HA. An alkaline buffer works the same way with a weak base and its conjugate acid, such as NH₃/NH₄⁺.

These reactions stop the added strong acid or base from staying free in the solution. If the addition is small compared with both components, their ratio changes only a little, so the pH changes only a little. A buffer does not keep the pH perfectly fixed. It stops working once one component is nearly used up.

Buffers keep a suitable pH for enzyme reactions, chemical measurements and calibrating pH meters. Diluting both components by the same factor leaves their ratio about the same, so the pH changes little. But there is less of each component in every cm³, so the buffer can absorb less acid or base. At extreme dilution the simple formula no longer works.

Rearranging K_a = [H⁺][A^-]/[HA] gives pH = pK_a + log₁₀([A^-]/[HA]). Both components are in the same volume, so their mole ratio equals their concentration ratio. Use the amounts left after any added strong acid or base has reacted. This works when the further ionisation of HA is small compared with those amounts.

For a weak base B and its conjugate acid BH⁺, you have two choices. Use K_a of BH⁺ in the same equation. Or use pOH = pK_b + log₁₀([BH⁺]/[B]), then pH = pK_w - pOH. Equal amounts of NH₃ and NH₄⁺, with K_b = 1.80 × 10^-5 mol dm^-3, give a pH of about 9.26 at 298 K.

The pair HCO₃^-/CO₃^2- is an acid and conjugate-base buffer. Carbonate accepts H⁺. Hydrogencarbonate can react with OH^- to form carbonate and water. Seawater has other buffering species too, but this pair links rising CO₂ in the air directly to how much carbonate is available.

Carbon dioxide in the air is rising fast, so surface seawater takes up more of it. The buffer resists the extra acid by using up carbonate. Its capacity is limited, so the equilibrium pH still falls. Ocean acidification means a fall in pH. Seawater does not have to drop below pH 7. With less carbonate available, it is also harder for sea creatures to form calcium carbonate shells and skeletons.

For AgCl(s) ⇌ Ag⁺(aq) + Cl^-(aq), K_sp = [Ag⁺][Cl^-]. The pure solid is omitted. For CaF₂(s) ⇌ Ca²⁺(aq) + 2F^-(aq), K_sp = [Ca²⁺][F^-]². The exponent follows the dissolution coefficient. Units follow from the expression: K_sp(AgCl) has units of mol² dm^-6, and K_sp(CaF₂) has units of mol³ dm^-9.

The ionic product, Q, has the same form as K_sp but uses the free-ion concentrations at that moment. If Q < K_sp, the solution is unsaturated. If Q = K_sp, it is saturated and at equilibrium. If Q > K_sp, a precipitate forms. When you mix solutions, first work out each concentration in the total volume.

Adding Cl^- to a saturated AgCl solution raises the concentration of one product ion. This favours precipitation, so less AgCl stays dissolved. Suppose the solution already has 0.0100 mol dm^-3 chloride, far more than AgCl adds. Then [Ag⁺] ≈ K_sp/[Cl^-] = (1.80 × 10^-10 mol² dm^-6)/(0.0100 mol dm^-3) = 1.80 × 10^-8 mol dm^-3. Check this shortcut when the added chloride is not much larger.

Ammonia binds free silver ions: Ag⁺(aq) + 2NH₃(aq) ⇌ [Ag(NH₃)₂]⁺(aq). This removes free Ag⁺, so [Ag⁺][Cl^-] falls below K_sp and more AgCl dissolves. The total dissolved silver can rise, while the free Ag⁺ stays small. K_sp uses only the free ion, not the silver held in the complex.

Nitric acid removes carbonate or hydroxide that would interfere, and it adds no halide. Hydrochloric acid would add chloride and give a false positive. The trend with ammonia comes from the balance between each very small solubility product and complex formation. It does not mean ammonia breaks every silver halide lattice equally easily.

Buffers resist small changes by reacting with what is added, but their capacity is limited. Ocean CO₂ uptake uses up carbonate and lowers pH. For solubility, compare the free-ion product with K_sp. A common ion reduces dissolving, while forming a complex with a dissolved ion can increase it. Give every K its units (K_a and K_b: mol dm^-3; K_sp: count the ions). A p value has no units.