NEET2020Chemistry-Ionic Equilibrium
NEET 2020 Chemistry Solubility Product MCQ Question
Type: MCQ-numerical-Medium-Class 11
Find out the solubility of Ni(OH)₂ in 0.1 M NaOH. Given that the ionic product of Ni(OH)₂ is 2 × 10⁻¹⁵.
A
1 × 10⁻¹³ M
B
1 × 10⁸ M
C
2 × 10⁻¹³ M
D
2 × 10⁻⁸ M
Correct Answer
Option C
Detailed Explanation
The solubility product expression for Ni(OH)₂ is Ksp = [Ni²⁺][OH⁻]². In 0.1 M NaOH, [OH⁻] = 0.1 M. Solving 2 × 10⁻¹⁵ = s × (0.1)² gives s = 2 × 10⁻¹³ M.
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