Question

Calculate the molar solubility of Ni(OH)2 in water. Use 2.0 * 10^-15 as the solubility product constant of Ni(OH)2.

Answer 1

Ni(OH)₂(s) ⇄ Ni²⁺(aq) + 2OH⁻(aq)

Ksp=2.0*10⁻¹⁵

Ksp=[Ni²⁺][OH⁻]²

c=[Ni²⁺]=[OH⁻]/2

Ksp=c×(2c)²=4c³

c=∛(Ksp/4)

c=∛(2.0×10⁻¹⁵/4)=0.01995 mol/L ≈ 0.02 mol/l

Answer 2

Answer:

The molar solubility of $Ni(OH)_2$ in water $7.93\times10^{-6} mol/L$.

Explanation:

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

                        S        2S

Solubility of Nickel hydroxide =$K_{sp}=2.0\times 10^{-15}$

$K_{sp}=S\times 2S^{2}=4S^3$

$2.0\times 10^{-15}=4S^3$

$S=7.93\times10^{-6} mol/L$

The molar solubility of $Ni(OH)_2$ in water $7.93\times10^{-6} mol/L$.