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.
- 2 10-8 M
- 1 10-13 M
- 1 108 M
- 2 10-13 M
Answer
(4) 2 10-13 M
Solution
Ni(OH)2(s) Ni2+(aq) + 2OH-(aq)
Let the solubility be s. NaOH gives [OH-] = 0.1 M (common ion), and 2s 0.1.
Ksp = [ Ni2+][OH-]2 = s(0.1)2
2 10-15 = s 10-2
s = 2 10-13 M
Answer: (4)