Answer to Question #225163 in Chemistry for Junior

Question

The solubility of Al(OH)3 is 5.15 x 10-7 g/ℓ. Calculate the Ksp of Al(OH)3.

Answer 1

Solution:

Convert g/L to mol/L:

The molar mass of Al(OH)₃ is 78 g/mol.

Therefore,

(5.15×10^-7 g Al(OH)₃ / 1 L) × (1 mol Al(OH)₃ / 78 g Al(OH)₃) = 6.6×10^-9 mol/L

S(Al(OH)₃) = 6.6×10^-9 mol/L

Al(OH)₃(s) ⇌ Al³⁺(aq) + 3OH^-(aq)

According to the equation above:

[Al³⁺] = S(Al(OH)₃) = 6.6×10^-9 mol/L

[OH^-] = 3 × S(Al(OH)₃) = 3 × (6.6×10^-9 mol/L) = 1.98×10^-8 mol/L

The K_sp expression for Al(OH)₃(s) is:

K_sp = [Al³⁺] × [OH^-]³

K_sp = (6.6×10^-9) × (1.98×10^-8)³ = 5.12×10^-32

K_sp = 5.12×10^-32

Answer: K_sp of Al(OH)₃ is 5.12×10^-32