NTA Abhyas NEET2020ChemistryIonic EquilibriumPractice
If the ionic product of N i O H 2 is 1.9 × 10 - 15 , the molar solubility of N i O H 2 in 1.0 M NaOH is
Options
- A1.9 × - 18 M
- B1.9 × 10 - 13 M
- C1.9 × 10 - 15 M
- D1.9 × 10 - 14 M
Correct answer
C. 1.9 × 10 - 15 M
Step-by-step solution
NaOH ⇌ Na + + OH - C M CM Ni OH 2 ⇌ Ni 2 + + 2 OH - x x 2x ∴ K s p = N i 2 + O H - 2 = x x + C 2 K s p = x C 2 (neglecting higher power of x ) x = K s p C 2 = 1.9 × 10 - 15 1 2