Question

Calculate the molar solubility of ni(oh)2 in 0.10 m naoh.The ionic ptoduct of ni(oh)2 is 2.0×10^-6.

Answer 1

Ni(OH)₂ (s) ⇔ Ni ²⁺ (aq) + 2OH⁻ (aq)

Ionic product: [ Ni²⁺ ][ OH⁻ ]²

Let molar solubility of Ni²⁺ be x M

Then concentration of OH⁻ = (2x + 0.1) M

2 × 10⁻¹⁵ = x (2x + 0.1)²

As x is very small compared to 0.1M, we ignore 2x [ x <<< 0.1]

2 × 10⁻¹⁵ = 0.01x

x = 2 × 10⁻¹³ M

Molar solubility of Ni(OH)₂ is 2 × 10⁻¹³ M

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