posted by jessie on .
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!
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
hope i helped you :)