NEET Chemistry Solubility Product MCQ Question
Calculate the solubility (S) of Ni(OH)2 in mol/L given its Ksp value is 2.0 × 10^-15.
0.58 × 10^-4
5.8 × 10^-5
7.8 × 10^-9
0 × 10^-5
Correct Answer
Detailed Explanation
For Ni(OH)2, Ksp = [Ni^2+][OH^-]^2 = 2.0 × 10^-15. Let [Ni^2+] = S and [OH^-] = 2S. Solving (S)(2S)^2 = 2.0 × 10^-15 gives S = 0.58 × 10^-4 mol/L.
Found an issue with this question?
Related Questions
More from Solubility Product
Assertion (A): The solubility of a sparingly soluble salt is affected by the presence of a common ion. | Reason (R): The solubility product constant (...
Match Column-I with Column-II. Column-I Column-II (a) Solubility Product Constant (K sp) (i) Represents maximum concentration of ions in s...
The solubility product of BaCl₂ is 3.2 × 10⁻⁹. What will be its solubility in mol L⁻¹?
More from
Assertion (A): For a dibasic acid H2X, the equilibrium constant for the first ionization step (Ka1) is generally greater than that for the second ioni...
Match Column-I with Column-II. Column-I Column-II (a) K c for H 2 + I 2 2HI (i) 57.0 (b) K c for N 2 + O 2 2NO (ii) 4.8 × 10 –31 (c) K...