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Question:

What is the molar solubility of Al(OH)3 in 0.2 M NaOH solution?

3×10⁻²

12×10⁻²

12×10⁻¹

3×10⁻⁹

Solution:

Al(OH)₃ ⇌ Al³⁺ + 3OH⁻
Ksp = [Al³⁺] × [OH⁻]³
Now, Let S' be the solubility.
[OH⁻] = 3(S') + 0.2 M ≈ 0.2 M
Ksp = S' × (0.2)³ = 2.4 × 10⁻²⁴
S' = 3 × 10⁻²² M