Question

Molar solubility of Ni(OH)2 in 0.5 M Ca(OH)2 is (Ksp of Ni(OH)2 = 2 x 10-15 M).​

Answer 1

Explanation:

Ksp=[Ni +2][OH − ] 2

2×10 −15 =[Ni +2][10 −1 ] 2

2×10 −3 =[Ni +2]

Molar Solubility of [Ni

+2]=2×10 −13M

Answer 2

Answer:

The molar solubility is $2$ × $10^{-15}mol/L$.

Explanation:

The $Ni(OH)_{2}$ is dissociated as follows-

                                        $Ni(OH)_{2}$ ⇄ $Ni^{-2} + 2OH^{-}$

Concentration:                s mol/L      s mol/L    2s mol/L

The $Ca(OH)_{2}$ is dissociated as follows-

                                      $Ca(OH)_{2}$ → $Ca^{2+} + 2OH^{-}$

Concentration:             0.5 mol/L   0.5 mol/L  2(0.5) mol/L

Total concentration of $OH^{-}$ = 2s + 2(0.5)

                                              = (2s + 1) mol/L

$K_{sp}$ = [$Ni^{+2}$][$OH^{-}$]²

2 × 10⁻¹⁵ = [s][2s+1]²

As $K_{sp}$small, 2s <<< 1, so we ignore the value of 2s.

2 × 10⁻¹⁵ = [s][1]²

2 × 10⁻¹⁵= s

Conclusion:

The molar solubility of $Ni(OH)_{2}$ is $2$ × $10^{-15} mol/L$.