Ground water intially contains 1.800mg/L of iron as Fe3+. What must the pH be raised to in order to precipitat all but 0.30mg/L of iron. The tempertaure is 25 degrees celcius

Answer 1

Fe(OH)3 ==> Fe^+3 + 3OH^-

Ksp = (Fe^+3)(OH^-)^3

You want (Fe^+3) to be 0.3 mg/L. Change that to moles/L, plug into the Ksp expression, and solve for OH^-, then convert to pH. Post your work if you get stuck.