Determine the molar solubility of al(oh)3 in a solution containing 0.0500 m alcl3. ksp (al(oh)3) = 1.3 × 10-33.

When the initial [Al+3] = 0.05 M

and according to the reaction equation:

by using the ICE table:

             Al(OH)3(s) → Al3+  +  3OH-

initial                            0.05         0

change                         +X           +3X

Equ                            0.05+X         3X

when Ksp = [Al3+] [OH-]^3

and Ksp = 1.3 x 10^-33

and [Al3+] = (0.05+X)

[OH-]^3 = 3X

when we assume that X = the molar solubility of Al(OH)3
 
by substitution:

1.3 x 10^-33 = (0.05 + X) (3X)^3 by solving for X

∴ X = 8.6 X 10^-12 M

∴ the molar solubility of Al(OH)3 = 8.6 x 10^-12 M  

Abbreviated as mg-25.