A solution has [Al^3+]=8.0×10−2 M and [HC2H3O2]= 1.10 M. What is the maximum quantity of NaHC2H3O2 that can be added to 200.0 mL of this solution before precipitation of Al(OH)3 begins?
Ksp = [Al^3+][OH^-]^3
Enter Al&3+ and solve for OH.
Convert OH^- to H^+.
I'll call acetic acid HAc and Ac^- acetate.
Ka = (H^+)(Ac^-)/(HAc)
You know Ka, H^+, and HAc, solve for Ac^- which gives you the molarity of the Ac^-, multiply that by molar mass NaAc to find moles sodium acetate (I assume that is sodium acetate with a typo of an extra H) in a liter and divide by 5 to obtain grams for 200 mL of the soln.