I will do an experiment that includes 4 grams of aluminum hydroxide and iron (III) oxide mixture, where 75% of it is Al(OH)3, and 15 ml of water and 25 ml of 2M KOH will be added and then boiled so that the Al(OH)3 dissolves. It will be filtered and then sulfuric acid will be added. How can I calculate the volume of sulfuric acid that I will need? What does the chemical equation include?

Question

I will do an experiment that includes 4 grams of aluminum hydroxide and iron (III) oxide mixture, where 75% of it is   Al(OH)3, and 15 ml of water and 25 ml of 2M KOH will be added and then boiled so that the Al(OH)3 dissolves. It will be filtered and then sulfuric acid will be added. How can I calculate the volume of sulfuric acid that I will need? What does the chemical equation include?

DrBob222 · Answer

75% of the 4 g mixture will be Al(OH)3 or 3 g. Al(OH)3 + OH^- ==> Al(OH)4^- Al(OH)4^- + 2H2SO4 ==> Al^3+ + 4H2O + 2SO4^2-

Tiffany · Answer

What about the iron (II) oxide? and the KOH?

DrBob222 · Answer

If the iron oxide is in solution then it will form Fe(OH)3 when KOH is added but adding excess KOH will NOT dissolve Fe(H)2 as it does Al(OH)3. That's how iron and aluminum are separated industrially; i.e., ppt both as the hydroxide, add excess to dissolve the Al, separate, then add acid to reppt the Al. But the way your problem is worded, the iron oxide is never placed in soln and I don't think appreciable iron oxide will dissolve in KOH. I think all you are doing is dissolving the Al(OH)3 in KOH and filtering the solution to remove the iron oxide.