Question

Calculate the solubility of magnesium hydroxide (Mg(OH)2), ksp=1.1x10^-11?

i have done this problem so many times and keep getting it wrong, i know the answer is 1.4x10^-4 but how do i get there? i did an ice chart and tried doing the cube root but its wrong. help!

Answer 1

Mg(OH)_2 ---------> Mg^2+ + 2OH^-

I. X 0 0

C. -X X 2X


K_sp <=========> [Mg^2+] [OH^-]^2


1.1*10^-11 = X (2X)^2

X = 3_/ 2.75*10^-12

X= 1.4*10^-4

hope i helped you :)

Cheers!

Answer 2

I have calculated by different cheap method and found answer to be correctly as 1.44x10^-4