Verified answer

The solubility product of Cr(OH)3 is l.2 x l0^-l5

and the equilibrium expression would be

Cr(OH)3(s) at equilib with Cr+3 (aq) plus 3 OH-

and the Ksp expression would be Ksp= l.2 x l0^-l5 = (Cr+++)( OH)^3

the coefficient of 3 in the balanced equation becomes the superscript 3 in the Ksp expression.

so if you were solving for the (OH) concentration in a saturated solution of chromium III hydroxide, you would let the (Cr) concentration = X and the (OH) concentration would be 3X

so you would have l.2 x l0^-l5 = (X) (3X)^3

Then solve for X, then triple the value to get the concentration of OH- in the solution.