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

The molar solubility of
`Ni(OH)_2` in water
`7.93*10^(-6) mol/L`.

Step-by-step explanation:

`Ni(OH)_2\rightleftharpoons Ni^(2+)+2OH^-`

S 2S

Solubility of Nickel hydroxide =
`K_(sp)=2.0* 10^(-15)`

`K_(sp)=S* 2S^(2)=4S^3`

`2.0* 10^(-15)=4S^3`

`S=7.93*10^(-6) mol/L`

The molar solubility of
`Ni(OH)_2` in water
`7.93*10^(-6) mol/L`.