# How can I mask the Chromium (III) and Chromium (VI) ions in my solution?

I want to conduct a quantitative analysis on my acid etching solution for my Aluminum pre-treatment process. The easiest way for me to conduct my quantitative analysis of Al is through complexometric titration with EDTA and back-titration with standardized Zinc solution.

My problem is that there's also Chromium ions in my solution and it can bind with the EDTA which reduces the volume of Zn solution I utilize in my back-titration. [I have confirmed my dilemma because the solution I am using was analyzed through Atomic Absorption Spectroscopy (AAS).]

How can I mask the chromium ions in my solution so I can proceed with the complexometric titration?

If I summarize, you want to make an EDTA titration of $$\ce{Al^{3+}}$$ in a solution containing also $$\ce{Cr^{3+}}$$ and $$\ce{Cr^{6+}}$$. The $$\ce{Cr^{6+}}$$ ions are usually in the form $$\ce{CrO4^{2-}}$$ and they don't interfere with $$\ce{Al^{3+}}$$ in EDTA titrations. But they can be eliminated by precipitation. The best thing to do is to oxidize $$\ce{Cr^{3+}}$$ into $$\ce{CrO4^{2-}}$$. In the old Treadwell's book, they propose to take $$5$$ mL $$\ce{Cr^{3+}}$$ solution, add $$0.5$$ mL of a saturated solution of $$\ce{Br_2}$$ + $$0.5$$ mL $$\ce{KOH}$$ $$4$$ mol/L. The reaction is finished at room temperature in $$5$$ minutes, and $$\ce{Al^{3+}}$$ ions do not interfere. $$\ce{2 Cr^{3+} + 3 Br2 + 16 OH^- -> 2 CrO4^{2-} + 6 Br^- + 8 H2O}$$ At the end of the reaction, a solution of barium chloride $$\ce{BaCl2}$$ or baryum hydroxide $$\ce{Ba(OH)2}$$ is slowly added to get rid of the chromate ions, by the reaction $$\ce{Ba^{2+} + 2 CrO4^{2-} -> BaCrO4(s)}$$ After filtration of the insoluble $$\ce{BaCrO4}$$, the obtained solution does not contain any more chromium, it can be acidified and the $$\ce{Al^{3+}}$$ ions can be titrated by EDTA.