posted by Anonymous on .
Calculate the solubility (in grams per 1.00 102 mL of solution) of magnesium hydroxide in a solution buffered at pH = 9
the ksp of Mg(oh)2 = 2.06e-13
........Mg(OH)2 ==> Mg^2+ + 2OH^-
Ksp = 2.06E-13 = (Mg^2+)(OH^-)
For (Mg^+) substitute x
For (OH^-)^2 substitute (2x + 1E-5)^2 [note: 2x for the Mg(OH)2 and 1E-5 for the OH^- concn of a soln with pH = 9
Solve for x = solubility in mols/L, then correct that to mols/100 mL, then to g/100 mL.