Question

The solubility product of a rare earth metal hydroxide m(oh)3 at room temperature is 4.32*10-14. Its solubility is

Answer 1

Answer:

So, the compound dissolves by the equation:

M(OH)3 <--> M3+ + 3 OH-

Ksp = [M3+][OH-]^3 = 4.32 X 10^-14

Let x = molar solubility of M(OH)3. Then, in a saturated solution, [M3+] = x and [OH-] = 3x. Plugging those into the expression for Ksp gives:

Ksp = 4.32 X 10^-14 = (x)(3x)^3 = 27x^4

x = 2X10^-4