[orthoformate]

If for Ni(OH)₂ the K_sp = 8.0·10^-18 then the expression used to calculate molar solubility (S) of Ni(OH)₂ is what?

Since K_sp=(Ni)·(OH)·(OH) and (Ni)=(OH)=(OH)then K_sp=(I)³

So (S)=(K_sp)^1/3

which means (S)=(8.0·10^-18)^1/3

On question 61 of a GRE Chemistry practice exam (https://www.ets.org/s/gre/pdf/practice_book_chemistry.pdf) they list the answer as (S)=(2.0·10^-18)^1/3

They take the cube root of 8, but they do not remove it from the cube root.

Where am I going wrong!

[Borek]

(Ni)=(OH)=(OH)

Think it over.

[confusedstud]

If for Ni(OH)₂ the K_sp = 8.0·10^-18 then the expression used to calculate molar solubility (S) of Ni(OH)₂ is what?

Since K_sp=(Ni)·(OH)·(OH) and (Ni)=(OH)=(OH)then K_sp=(I)³

So (S)=(K_sp)^1/3

which means (S)=(8.0·10^-18)^1/3

On question 61 of a GRE Chemistry practice exam (https://www.ets.org/s/gre/pdf/practice_book_chemistry.pdf) they list the answer as (S)=(2.0·10^-18)^1/3

They take the cube root of 8, but they do not remove it from the cube root.

Where am I going wrong!

I think you should say that x is the concentration of the  Ni(OH)2 salt so the concentration of Ni2+ is also x and OH- is 2x so Ksp=x(2x)^2 which is 4x^3 instead.

the equilibrium reaction Ni(OH)2 Ni2+ +2OH- so that it would be clear that the concentration of the OH- is twice of the entire salt in solution.

[orthoformate]

(OH)=2(Ni)

Thanks guys!