Types of salts

A salt is the product of neutralisation. When an acid reacts completely with a base, the replaceable hydrogen ions of the acid are exchanged for the metal (or ammonium) ion of the base:

where: the acid supplies replaceable H atoms, the base supplies the metal or $\text{NH}_4^+$ ion that takes their place, and water is always the second product of this exchange.

Classification by degree of hydrogen replacement. Three categories follow directly from how many of the acid's replaceable hydrogens have actually been swapped out:

• A normal salt has every replaceable hydrogen of the acid replaced by the metal or ammonium ion, with no leftover hydrogen and no extra hydroxide. Examples: $\text{NaCl}$, $\text{Na}_2\text{SO}_4$, $\text{CaCl}_2$.
• An acidic salt forms when only some of the replaceable hydrogens of a polybasic acid are exchanged, so at least one acidic hydrogen remains attached and can react further with more base. Examples: $\text{NaHSO}_4$, $\text{NaHCO}_3$.
• A basic salt forms when a polyacidic base is only partly neutralised, so the salt retains some hydroxide (or oxide) group alongside the acid's anion. Example: basic lead carbonate, roughly $\text{Pb(OH)}_2 \cdot 2\text{PbCO}_3$.

Classification by water content. A completely separate axis of classification concerns whether water is locked into the crystal:

• A hydrated (crystalline) salt contains a fixed number of water molecules chemically bound within its crystal lattice — its water of crystallisation. Example: $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$, blue copper sulphate crystals with 5 water molecules per formula unit.
• An anhydrous salt has no water of crystallisation at all, e.g. the white $\text{CuSO}_4$ left behind once that water is driven off by heat.

Three related moisture behaviours all follow from this same idea of bound versus absorbed water:

• Efflorescence — a hydrated crystal spontaneously loses some or all of its own water of crystallisation to the air and crumbles to a powder, e.g. washing soda, $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$.
• Deliquescence — a substance absorbs so much moisture from the air that it dissolves in that absorbed water, forming a solution, e.g. anhydrous calcium chloride.
• Hygroscopy — a substance absorbs moisture from air without dissolving, so it can serve as a drying agent (physically removing water vapour from a gas), e.g. anhydrous $\text{CaCl}_2$ or silica gel. This differs from a dehydrating agent (e.g. concentrated $\text{H}_2\text{SO}_4$), which chemically strips water out of a compound, sometimes breaking it down in the process, as when it chars sugar by ripping out hydrogen and oxygen as water.

There is no numeric formula behind salt classification — it is a counting-and-observation scheme, so the reasoning that ties it together matters more than any equation.

Why hydrogen count decides normal versus acidic. A polybasic acid like $\text{H}_2\text{SO}_4$ has more than one replaceable hydrogen. Neutralisation is not all-or-nothing: exactly how much base is supplied decides how many of those hydrogens actually get exchanged. Supply enough base for every hydrogen and none is left — a normal salt, with no acidic character remaining. Supply less, and at least one hydrogen stays attached to the acid's anion; that leftover hydrogen is still capable of reacting with more base later, which is precisely why an acidic salt still behaves somewhat like an acid even though it was made from a base-and-acid reaction.

Why basic salts retain hydroxide. The mirror-image case happens when the base is polyacidic (has more than one replaceable $\text{OH}$) but is only partly neutralised by the acid supplied. The unreacted hydroxide or oxide portion stays bound into the same compound as the acid's anion, giving a salt with genuinely two different anion-type groups attached to the metal.

Why water of crystallisation is chemically bound, not just damp. In a hydrated salt, water molecules occupy fixed positions in the crystal lattice, held there by definite chemical/electrostatic forces to the metal ion or the anion — that is why the ratio (5 water molecules per $\text{CuSO}_4$, 10 per $\text{Na}_2\text{CO}_3$) is always the same fixed number, never a variable amount. This is fundamentally different from a substance simply being wet on the outside; heating removes water of crystallisation only at a specific temperature where the lattice bond to that water breaks, converting the compound cleanly into its anhydrous form. Efflorescence, deliquescence, and hygroscopy are just three different directions this water-lattice interaction can run: losing bound water to dry air, gaining atmospheric water and dissolving in it, or gaining atmospheric water while staying solid.

Question: $\text{H}_2\text{SO}_4$ reacts with NaOH. Show, by equations, how the same pair of reactants can give (a) a normal salt and (b) an acidic salt, classify each, and then state what is observed when the normal salt product is crystallised as $\text{Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O}$ and left exposed to dry air.

Step 1. Sulphuric acid is dibasic, so it carries two replaceable hydrogens. With enough NaOH to replace both (2 mol NaOH per mol acid): $\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$. Both hydrogens are replaced by sodium, so $\text{Na}_2\text{SO}_4$ is a normal salt.

Step 2. With only enough NaOH to replace one hydrogen (1 mol NaOH per mol acid): $\text{H}_2\text{SO}_4 + \text{NaOH} \rightarrow \text{NaHSO}_4 + \text{H}_2\text{O}$. Only one hydrogen is replaced; the other stays on the sulphate group, so $\text{NaHSO}_4$ is an acidic salt, and it can react further: $\text{NaHSO}_4 + \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}$.

Step 3. Crystallised from solution, the normal salt takes up a fixed water of crystallisation to become $\text{Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O}$, a hydrated salt (Glauber's salt). Left exposed to dry air, it spontaneously loses part of that bound water and crumbles into a powder — this is efflorescence, the same behaviour shown by washing soda, $\text{Na}_2\text{CO}_3 \cdot 10\text{H}_2\text{O}$.

Answer: the same acid-base pair gives a normal salt (Na2SO4, full replacement) or an acidic salt (NaHSO4, partial replacement) depending on the proportion of NaOH used, and the crystallised hydrated form of the normal salt effloresces on exposure to dry air.

• Calling an acidic salt "acidic" just because of its name — its acidity comes from a genuinely leftover replaceable hydrogen, not merely from being derived from an acid.
• Confusing deliquescence (absorbs moisture and dissolves) with efflorescence (loses its own water and crumbles) — they are opposite processes.
• Mixing up drying agents (physically remove water vapour, e.g. anhydrous $\text{CaCl}_2$) with dehydrating agents (chemically strip water from a compound, e.g. concentrated $\text{H}_2\text{SO}_4$ on sugar).
• Treating "water of crystallisation" as loose water trapped between crystals — it is chemically bonded in a fixed, definite proportion, part of the compound's formula, and requires heating to a specific temperature to remove.

Ask the tutor: Want to practise writing equations for normal versus acidic salt formation from other polybasic acids, or need help sorting more examples into efflorescent, deliquescent, and hygroscopic categories.