write the balanced equation for its solubility equilibrium. Use symbols and indicate charges.

Al(OH)₃(s) <----> Al³⁺(aq) + 3 OH^-(aq)

write the Ksp expression

K_sp = [Al³⁺][OH^-]³

calculate the molar solubility of aluminum hydroxide, then find the concentrations of each ion in solution

We know that whatever the equilibrium concentration of Al³⁺ is, the equilibrium concentration of OH^- will be three times as much because of the 3 : 1 ratio as shown in the equation above.

K_sp = [Al³⁺][OH^-]³

1.9 x 10^-33 = [Al³⁺][OH^-]³

1.9 x 10^-33 = (x)(3x)³

1.9 x 10^-33 = 27x⁴

x = 2.9... x 10^-9 mol/L

The molar solubility of aluminum hydroxide is 2.9 x 10^-9 mol/L. The concentration of Al³⁺ is 2.9 x 10^-9 M and the concentration of OH^- is 8.7 x 10^-9 M

what happens to the solubility of AI(OH)3 when AICI3 is added? briefly explain.

The solubility of Al(OH)₃ decreases. This is because of the common-ion effect, the general observation that if a weak electrolyte (like Al(OH)₃) is in the presence of a strong electrolyte containing a common ion (like AlCl₃ which has the common ion Al³⁺), the weak electrolyte will dissociate less than what it normally would. Since there is a constraint on the concentrations of Al³⁺ and OH^- at equilibrium and Al³⁺ is already being contributed by the strong electrolyte AlCl₃, the solubility of Al(OH)₃ decreases.

what happens to the solubility of AI(OH)3 when acid is added? briefly explain

The solubility of Al(OH)₃ increases. This is because the hydroxide ions react with the acid and are consumed in the process. Le Chatelier's principle says that if a system at equilibrium is disturbed, such as by a change in concentration, the system will shift in a way to counteract that disturbance. Since the hydroxide ions react with any acid present, the concentration of OH^- decreases so the dissolution reaction will shift to the right to produce more OH^- ions (and more Al³⁺ ions). As a result, the solubility of Al(OH)₃ increases when acid is added.