Reactions with Bases

Charge Density Influence: Metal cations with a higher charge and smaller ionic radius, such as $M^{3+}$ ions, possess a significantly higher charge density compared to $M^{2+}$ ions. This high charge density allows the metal ion to exert a strong polarizing force on the water ligands surrounding it.

O-H Bond Weakening: The central metal ion pulls electron density away from the oxygen atoms in the water ligands, which in turn weakens the $O-H$ bonds within the water molecules. This polarization makes the hydrogen atoms more $\delta+$ and easier to lose as protons ($H^+$).

Hydrolysis Equilibrium: Because of this polarization, metal-aqua ions act as Brønsted-Lowry acids by donating protons to the solvent. The general equilibrium can be represented as $[M(H_2O)_6]^{n+} + H_2O \rightleftharpoons [M(H_2O)_5(OH)]^{(n-1)+} + H_3O^+$, where the solution becomes acidic.

Stepwise Deprotonation: When a base like $OH^-$ or $NH_3$ is added, it removes protons from the water ligands in a series of steps. For a $+2$ ion, this continues until a neutral complex like $[M(H_2O)_4(OH)_2]$ is formed; for a $+3$ ion, it reaches $[M(H_2O)_3(OH)_3]$.

Precipitation: These neutral complexes are insoluble in water because they lack an overall charge to interact with polar water molecules. Consequently, they drop out of solution as solid metal hydroxides, often appearing as characteristic colored precipitates.

Ammonia as a Ligand: While $NH_3$ initially acts as a base to form precipitates, in some cases (like copper), excess ammonia can act as a ligand. It undergoes ligand substitution to replace water molecules, forming a soluble charged complex and causing the precipitate to redissolve.

Formation of Carbonates ($M^{2+}$): Metal ions with a $+2$ charge have lower acidity and do not polarize water ligands enough to release $H^+$ ions that would react with carbonate. Instead, they react directly with carbonate ions to form insoluble metal carbonates: $M^{2+}(aq) + CO_3^{2-}(aq) \rightarrow MCO_3(s)$.

Hydroxide and Gas Formation ($M^{3+}$): Highly acidic $+3$ ions react with carbonate ions by transferring protons. The carbonate ion acts as a base, accepting protons to form $CO_2$ gas and water, while the metal ion precipitates as a hydroxide: $2[M(H_2O)_6]^{3+} + 3CO_3^{2-} \rightarrow 2[M(H_2O)_3(OH)_3] + 3CO_2 + 3H_2O$.

Visual Observations: Reactions with $+3$ ions are distinguished by the observation of effervescence (bubbles of $CO_2$), whereas reactions with $+2$ ions only produce a solid precipitate without gas evolution.

Definition of Amphoterism: Some metal hydroxides, most notably aluminium hydroxide, are amphoteric, meaning they can react as both an acid and a base. This allows them to dissolve in both acidic and strongly alkaline conditions.

Reaction with Excess Base: In the presence of excess strong base (like $NaOH$), the neutral hydroxide precipitate $[Al(H_2O)_3(OH)_3]$ undergoes further deprotonation. It loses another proton to form the soluble aluminate ion, $[Al(OH)_4]^-$, causing the solid to redissolve into a colorless solution.

Contrast with Other Ions: Most other transition metal hydroxides, such as those of iron, are not amphoteric and will remain as solid precipitates even if excess sodium hydroxide is added.

Identify the Charge First: Always check the oxidation state of the metal ion before predicting a reaction with carbonates. If it is $+3$, you must include $CO_2$ in your products and observations.

Observation Precision: When describing reactions, mention both the color of the initial solution and the color/state of the final product (e.g., "pale green solution forms a dark green precipitate").

Equation Balancing: Ensure that the number of $OH^-$ ions added matches the charge of the metal ion to produce a neutral, insoluble precipitate. For example, $Fe^{3+}$ requires $3OH^-$ to precipitate.

Check for Redissolving: Remember that only $Al(OH)_3$ redissolves in excess $NaOH$ (amphoteric), and only $Cu(OH)_2$ redissolves in excess $NH_3$ (ligand substitution) among the common ions studied.