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-

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

Start Free Mock LoginAlready registered? Login here

Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductEasy2 minPYQ_2020

ChemistryEasysingle choice

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

Options:

Answer:

Solution:

Ni(OH)₂ $⇌$ Ni²⁺+ 2OH^-
s 2s

Total [OH^-] = 2s + 0.1

Ionic product = [s] [2s + 0.1]²

$\approx S (0.01)$ $= 2 \times 10^{-15}$

0.01 s = 2 × 10^-15

s = 2 × 10^-13

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...