Question

Find out the solubility of Ni(OH)₂ in 0.1M NaOH. Given that the ionic product of Ni(OH)₂ is 2 × 10^–15 . 

(1) 1 × 10⁸ M 

(2) 2 × 10^–13 M 

(3) 2 × 10^–8 M 

(4) 1 × 10^–13 M A

Answer 1

(2) 2 × 10^–13 M 

\(\alpha\) = 1 for NaoH

Ionic product = (S') (0.1 + 2S')² 

2 × 10^–15 = S'(0.1)² 

S' = 2 × 10^–13 M