Acid Dissociation of Hydrated Aluminum Ion

Steps:

• Recognize the reaction as acid hydrolysis: [Al(H2O)6]3+ acts as a Brønsted acid, donating H+ to H2O to form [Al(H2O)5OH]2+ (conjugate base) and H3O+ (conjugate acid).
• Apply Le Chatelier's principle: Adding base (OH-) neutralizes H3O+, decreasing its concentration and shifting equilibrium right to produce more H3O+.
• Confirm K expression: K = [[Al(H2O)5OH]2+][H3O+] / [[Al(H2O)6]3+], constant at fixed temperature, unaffected by pH.
• Eliminate non-conjugate pairs and roles: H2O is base, not conjugate to [Al(H2O)6]3+; H3O+ is product, not reactant acid.

Why C is correct:

• Adding base shifts equilibrium right per Le Chatelier's principle, as OH- consumes H3O+ (assuming "pH is added" means base addition to increase pH).

Why the others are wrong:

• A: Conjugate acid-base pair is [Al(H2O)6]3+ and [Al(H2O)5OH]2+; H2O is the base, not conjugate to the acid.
• B: H3O+ is the conjugate acid product, not acting as an acid in this forward reaction.
• D: Equilibrium constant K is fixed at constant temperature, independent of pH or concentrations.

Final answer: C