When anhydrous aluminium chloride reacts with water, solution Y is formed that contains a complex aluminium ion, Z, and chloride ions - AQA - A-Level Chemistry - Question 7 - 2018 - Paper 1

The complex ion Z, which is [Al(H₂O)₆]³⁺, can act as a Brønsted–Lowry acid by donating a proton (H⁺) to water, forming:

$[\text{Al(H}_2\text{O)}_6]^{3+} + \text{H}_2\text{O} \rightarrow [\text{Al(H}_2\text{O)}_5(\text{OH})]^{2+} + \text{H}_3\text{O}^+$

1. Observation 1: Effervescence or bubbles of gas are observed, indicating the formation of carbon dioxide gas.

2. Observation 2: A white precipitate (aluminum hydroxide) forms in the solution.

Equation for the Reaction

When sodium carbonate is added to the solution Y, the following reaction occurs:

$[\text{Al(H}_2\text{O)}_6]^{3+} + 3\text{Na}_2\text{CO}_3 \rightarrow [\text{Al(OH)}_3]_{(s)} + 3\text{Na}^+ + 3\text{CO}_2 + 3\text{H}_2\text{O}$

1. Observation 1: A white precipitate forms initially, which may dissolve upon adding excess potassium hydroxide, forming a clear solution. This indicates that aluminum hydroxide is forming and dissolving in excess.

2. Observation 2: The solution becomes colorless upon the complete dissolution of the precipitate.

Equation 1

The initial reaction for the formation of aluminum hydroxide when potassium hydroxide is added is:

$[\text{Al(H}_2\text{O)}_6]^{3+} + 3\text{OH}^- \rightarrow [\text{Al(OH)}_3]_{(s)} + 6\text{H}_2\text{O}$

Equation 2

Upon adding excess potassium hydroxide, the aluminum hydroxide dissolves to form a soluble complex:

$[\text{Al(OH)}_3]_{(s)} + 2\text{OH}^- \rightarrow [\text{Al(OH)}_4]^{-}$