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NTA Abhyas NEET2020ChemistryIonic EquilibriumPractice

The ionic product of Ni(OH) 2 is 2.0 × 10 -15 . The molar solubility of Ni(OH) 2 in 0.10 M NaOH is-

Options

  1. A3.2 × 10 -12
  2. B2.0 × 10 -13
  3. C​4.34 × 10 -12
  4. D​0.58 × 10 -4

Correct answer

B. 2.0 × 10 -13

Step-by-step solution

Ni(OH) 2 ⇌ Ni 2 + + 2OH - s 2s Total [OH - ] = 2s + 0.1 Ionic product = [s] [2s + 0.1] 2 ≈ S ( 0.01 ) = 2 × 10 -15 0.01 s = 2 × 10 -15 s = 2 × 10 -13

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