Hello, I'm currently working on a lab, and I'm unsure how to proceed on a certain question. The lab is detailed here:
http://www.carlton.srsd119.ca/chemical/equilibrium/dichromate/LCP_procedureP.htmMy question concerns this part of the lab:
9. Put approximately 1 mL (10 drops) of 0.1 M CrO42-(aq) solution into one clean 13 x 100 mm test tube, and the same amount of 0.1 M Cr2O72-(aq) into another. Add no acid or base to either test tube, but add about 5 drops of Ba+2 (aq) solution to each. Record your observations, and compare them to the above results when you added HCl or NaOH before adding the Ba+2 (aq) ions.
In steps 5 and 6 of the lab we did almost exactly the same thing, except we purified either chromate or the dichromate (using sodium hydroxide and hydrochloric acid, respectfully). In step 9, the exact same thing is done, without the purification.
This means there in the chromate, there is still the presence of some dichromate, and vice versa with the dichromate. How does factor into adding the Ba+ to form a non-soluble precipitate with the CrO4 and a soluble precipitate with the Cr2O7.
If the non-soluble precipitate that is formed from Ba and CrO4 is a non-reversible reaction, and the soluble precipitate formed with the Cr2O7 is reversible, then both test tube, regardless of the equilibrium constant, will both eventually contain exclusively CrO4.
So my question basically is, is BaCrO4 non-reversible, and is BaCr2O7 reversible? If they aren't, how does the equilibrium shape out as you're introducing Ba to a test tube containing mostly CrO4 and a little Cr2O7 and vice versa.
Thanks.