How can I mask the Chromium (III) and Chromium (VI) ions in my solution?
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. $$\ce{2 Cr^{3+} + 3 Br2 + 16 OH^- -&gt; 2 CrO4^{2-} + 6 Br^- + 8 H2O}$$ 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 $$\ce{Ba^{2+} + 2 CrO4^{2-} -&gt; BaCrO4(s)}$$ 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.