The solubility of Al(OH)₃ in water at 25 °C is (given K_sp = 8.5 x 10^-32): 

CONCEPT:

Solubility Product Constant (K_sp) and Solubility (S)

• The solubility product constant (K_sp) is the equilibrium constant for the dissolution of a sparingly soluble salt in water.
• For Al(OH)₃, the dissolution equation is:

  Al(OH)₃(s) → Al³⁺(aq) + 3OH^-(aq)

• The K_sp expression for Al(OH)₃ is:

  K_sp = [Al³⁺][OH^-]³ = (S)(3S)³ = 27S⁴

• To find the solubility (S), we solve for S using the given K_sp value:

  S = (K_sp / 27)^1/4

EXPLANATION:

• Given:

  K_sp = 8.5 × 10^-32

• The solubility equation is:

  S = ((8.5 × 10^-32) / 27)^1/4

• By calculating the value of S, we get:

  S = 7.49 × 10^-7 moles/liter

• Now, to convert the solubility into grams per liter, we multiply by the molar mass of Al(OH)₃ (78.00 g/mol):

  Solubility (g/liter) = S × 78.00 g/mol = 7.49 × 10^-7 × 78.00 = 5.84 × 10^-7 g/liter

Therefore, the solubility of Al(OH)₃ in water is approximately 5.8 × 10^-7 g/liter.