Question

1. The solubility product, K_sp, for aluminum hydroxide, Al(OH)₃, is 1.9 ✕ 10⁻³³ at 25°C. What is the molar solubility of aluminum hydroxide in a solution containing 0.066 M KOHat 25°C?

Answer 1

Ans.       Al (OH)₃ ---------------> Al³⁺ + 3OH^-

[Al³⁺] = X               ; [OH^-] = 3X

KOH ----------------> K⁺ + OH^-          ; [OH^-] = 0.066 M

Total [OH^-] = 3X (from Al. hydroxide) + 0.066 M (from KOH) = (3X + 0.066) M

One mole Al(OH)₃ dissociates into one mole Al³⁺ and 3 moles OH^-. Being a strong base, each KOH also produces one mole OH^- in the solutions. Thus, the total [OH^-] is equal to the sum of OH^- ions released from both AL(OH)₃ and KOH.

Solubility product, Ksp = [Al³⁺] [OH^-]³

Or, 1.9 x 10^-33 = X x (3X + 0.066)³

Or, 1.9 x 10^-33 = X x (0.066)³                                         ; 3X << 0.066, thus adding 3X to 0.066 is insignificant.

Or, X = (1.9 x 10^-33) / (0.066)³ = 6.6 x 10^-30

Hence, molar solubility of Al(OH)₃ under given condition is 2.87 x 10^-30 M