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1. What the student is asking

The student sees that the aluminium ion, Al³⁺, is a “hard” Lewis/Brønsted‑acidic cation and therefore expects it to behave like the classic hydrolysis of a metal salt:

[ \ce{AlCl3 + 3 H2O -> Al(OH)3 + 3 HCl} ]

Is that equation correct for the reaction of anhydrous AlCl₃ with water?
If not, what really happens when AlCl₃ is put into water, and under what conditions (if any) does solid aluminium hydroxide, Al(OH)₃, appear?

2. Step‑by‑step reasoning

2.1 Dissolution and first hydration

Aluminium chloride is a strongly hygroscopic ionic solid. When it contacts water it first dissolves:

[ \boxed{\ce{AlCl3(s) -> Al^{3+}(aq) + 3Cl^{-}(aq)}} ]

Because Al³⁺ is a small, highly charged ion it is immediately solvated by six water molecules to give the octahedral aquo‑complex

[ \ce{[Al(H2O)6]^{3+}} \tag{A} ]

so the real first step is

[ \ce{AlCl3 + 6 H2O -> [Al(H2O)6]^{3+} + 3Cl^{-}} \qquad (1) ]

The three chloride ions stay in solution as spectators; no molecular HCl is formed.

2.2 Acidic hydrolysis of the hexaaqua‑ion

The hexaaqua‑ion is a Brønsted‑acid because one of the coordinated water molecules can donate a proton to bulk water:

[ \ce{[Al(H2O)6]^{3+} + H2O <=> [Al(H2O)5(OH)]^{2+} + H3O^{+}} \qquad (2) ]

The equilibrium constant for the first deprotonation is the first acidity constant,

[ K_{a1}=10^{-5.0}\;(pK_{a1}\approx5.0) ]

Thus in pure water only a few percent of the Al³⁺ is present as the monohydroxo‑complex; the solution is mildly acidic (pH≈3–4 for a 0.1 M AlCl₃ solution).

2.3 Successive deprotonations

Further loss of protons is possible:

[ \begin{aligned} \ce{[Al(H2O)5(OH)]^{2+} + H2O &<=> [Al(H2O)4(OH)2]^{+} + H3O^{+}} \qquad (3)\[2mm] \ce{[Al(H2O)4(OH)2]^{+} + H2O &<=> [Al(H2O)3(OH)3] + H3O^{+}} \qquad (4) \end{aligned} ]

The corresponding acidity constants are much smaller:

[ pK_{a2}\approx 10,\qquad pK_{a3}\approx 13 ]

Consequently, under neutral or acidic conditions the equilibria (3) and (4) lie far to the left; virtually no (\ce{[Al(H2O)3(OH)3]}) is formed.

2.4 When does solid Al(OH)₃ appear?

Aluminium hydroxide is sparingly soluble:

[ K_{sp}(\ce{Al(OH)3}) = 1.3\times10^{-33} ]

The solubility product can be expressed in terms of the free ion concentrations:

[ K_{sp}= [\ce{Al^{3+}}]\,[\ce{OH^{-}}]^3 ]

If the solution is neutral or basic (i.e. ([\ce{OH^{-}}]) is increased by adding a base), the product ([\ce{Al^{3+}}][\ce{OH^{-}}]^3) quickly exceeds (K_{sp}) and a precipitate of (\ce{Al(OH)3(s)}) forms.

In pure water the concentration of (\ce{OH^{-}}) is only (10^{-7}) M, and the corresponding ([\ce{Al^{3+}}]) from the dissolution of AlCl₃ is far below the level needed to reach (K_{sp}). Hence no solid Al(OH)₃ precipitates unless the pH is raised (typically pH > 5).

2.5 Why the “(\ce{AlCl3 + 3 H2O -> Al(OH)3 + 3 HCl})” equation is misleading

  • No molecular HCl is released; the chloride ions stay dissolved.
  • The reaction is not stoichiometrically complete in water; only a tiny fraction of the aluminium ions are deprotonated enough to give (\ce{Al(OH)3}).
  • The correct net ionic representation, valid only when the pH is high enough to precipitate the hydroxide, is

[ \boxed{\ce{AlCl3 + 3 H2O -> Al(OH)3(s) + 3 H^{+} + 3 Cl^{-}}} ]

or, more explicitly,

[ \ce{[Al(H2O)6]^{3+} + 3 H2O -> Al(OH)3(s) + 3 H3O^{+}} ]

but only under basic or at least neutral conditions.
In neutral water the equilibrium lies essentially on the left, so the written equation gives the wrong impression that the reaction “goes to completion”.

3. Final answer

  • In pure water aluminium chloride dissolves to give (\ce{[Al(H2O)6]^{3+}}) and (\ce{Cl^{-}}).
  • The hexaaqua‑ion undergoes partial hydrolysis (first (pK_a\approx5)), producing a mildly acidic solution; solid (\ce{Al(OH)3}) does not form.

  • The simple stoichiometric equation

    [ \ce{AlCl3 + 3 H2O -> Al(OH)3 + 3 HCl} ]

    is incorrect for the reaction in water because it implies complete conversion to solid hydroxide and molecular HCl.

  • The correct description is:

    [ \ce{AlCl3 + 6 H2O -> [Al(H2O)6]^{3+} + 3 Cl^{-}} \quad\text{(dissolution)} ]

    followed by the equilibrium hydrolysis steps (2)–(4).
    Only when the solution is made neutral or basic (e.g., by adding NaOH, Na₂S, etc.) does the equilibrium shift far enough right to precipitate (\ce{Al(OH)3(s)}).

4. Common mistakes

Mistake Why it’s wrong How to avoid it
Writing “(\ce{AlCl3 + 3 H2O -> Al(OH)3 + 3 HCl})” as a balanced equation Implies formation of molecular HCl and complete conversion to solid hydroxide, neither of which occurs in pure water. Remember that chloride remains as (\ce{Cl^{-}}) in solution and that only a fraction of Al³⁺ hydrolyzes.
Assuming Al³⁺ is a strong Brønsted acid (pKₐ ≈ 0) The first hydrolysis of (\ce{[Al(H2O)6]^{3+}}) has (pK_a) ≈ 5; it is weak‑acidic, so only limited H⁺ is released. Use the reported acidity constants (pKₐ₁ ≈ 5, pKₐ₂ ≈ 10, pKₐ₃ ≈ 13).
Neglecting the solubility product of (\ce{Al(OH)3}) Without considering (K_{sp}) one may think Al(OH)₃ precipitates automatically. Compare ([\ce{Al^{3+}}][\ce{OH^{-}}]^3) with (K_{sp}); precipitation only occurs when pH > ~5.
Treating the reaction as a “neutralization” that produces HCl gas HCl is a strong acid; in aqueous solution it exists as (\ce{H^{+}}) and (\ce{Cl^{-}}), not as a separate molecular species. Write net‑ionic equations and keep chloride as a spectator ion.
Skipping the stepwise hydrolysis equilibria Leads to the impression of a single, irreversible reaction. Show each deprotonation step and its equilibrium constant.

With these points in mind, you can correctly explain the behaviour of aluminium chloride in water and why the simple “AlCl₃ + 3 H₂O → Al(OH)₃ + 3 HCl” equation is not an accurate representation.

Original question: Reaction of aluminium trichloride with water on Chemistry Stack Exchange, licensed CC BY-SA.