Question

The values of KSP of a sparingly soluble salt Ni(OH)2 is 2.0×10-15. What is its solubility? Explain.

Answer 1

for the dissociation:
Ni(OH)2<------> Ni2+ +2OH-
conc. s. 2s
Ksp= s*(2s)^2= 4s^3
2*10^-15= 4s^3.
now solve for s . that will be ur answer.

Answer 2

Answer: $7.34\times 10^{-4}grams/liter$

Explanation: The equation for the reaction will be as follows:

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

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

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 moles/liter.

Therefore,  

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

$2.0\times 10^{-15}=[s][2s]^2$

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

$s=0.79\times 10^{-5}moles/liter$

Solubility in grams/liter=$\text{solubility in moles/liter}\times Molar mass$

Solubility in grams/liter=$0.79\times 10^{-5}moles/liter\times 93g/mol=7.34\times 10^{-4}grams/liter$