Question

calculate the molar solubility of Ni(OH)2 in 0.1 M NaOH . The ionic product of Ni(OH)2 is 2 x 10-15

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

Answer 2

Answer : The molar solubility of $Ni(OH)_2$ is, $2\times 10^{-13}mole/L$

Explanation :

The equation for the reaction will be as follows :

$Ni(OH)_2\leftrightharpoons Ni^{2+}+2OH^-$

1 mole of $Ni(OH)_2$ gives 2 moles of $OH^{-}$ and 1 mole of $Ni^{2+}$.

Thus if solubility of $Ni(OH)_2$ is 's' moles/liter, solubility of $Ni^{2+}$ is 's' moles/liter and solubility of $OH^{-}$ is (2s+0.1) moles/liter

Therefore, the expression for solubility of equilibrium constant will be,

$K_{sp}=[Ni^{2+}][2OH^-]^2$

Now put all the given values in this expression,we get

$2\times 10^{-15}=(s)\times (2s+0.1)^2$

By solving the term 's' we get,

$s=2\times 10^{-13}mole/L$

Therefore, the molar solubility of $Ni(OH)_2$ is, $2\times 10^{-13}mole/L$