text: calculate the ph required in an aqueous solution in order to reduce the solubility? calculate the ph required in an aqueous solution in order to reduce the solubility of al(oh)3 to 2.0x10^-14 m. ksp = 1.9x10^-33 for al(oh)3 given: ksp = 1.9 x 10-33  s_al(oh)3 = 2 x 10-14 [al3+] = 3[oh-]= s_as(oh)3  al(oh)3--> al^3+ + 3 oh^- formula: ksp = [al] x [oh]^3 = 3[oh-]^4 = 1.9x10-33 you can solve then the [oh-] of the solution in h2o only; eqtn 1 if you want to reduce solubility, then count on the common ion effect- you are to add oh- in your solution;  therefore       oh initial         from the ksp value change          +x equation       [oh-] value  +x              1.9 x 10-33 = 3([oh-] value + x)^4  solve for x then plug it in the equation below  ph = -log [h+] =14 - log ([oh-]+x)