Question

What is the molar solubility of Al(OH)₃ in 0.2 M NaOH Solution? Given that, solubility product Al(OH)₃ = 2.4 × 10^-24:
1. 3 × 10^-19
2. 12 × 10^-21
3. 3 × 10^-22
4. 12 × 10^-23

Answer 1

Correct Answer - Option 3 : 3 × 10^-22

Concept:

Let the solubility of

Al(OH)₃ ⇌ Al³⁺ + 3OH^-

s 3s

NaoH ⇌ Na⁺⁺ OH^-

0.2 M 0.2M

Calculation:

[(Al³⁺)] = s and [OH^-] = 3s + 0.2 ≈ 0.2

k_sp = 2.4 × 10^-24 = [Al³⁺] [OH^-]

2.4 × 10^-24 = s(0.2)³

\(s = \frac{{2.4 \times {{10}^{ - 24}}{\rm{\;}}}}{{8 \times {{10}^3}{\rm{\;}}}} = 3 \times {10^{ - 22}}\;{\rm{mol}}/L\)