For an experiment I had to add 1 drop of 1.0M HCl into a test tube that had 2.5mL of 0.10M Na2CrO4 and record the color.
The original color of the Na2CrO4 was yellow. After adding the HCl the color became a very dark yellow. I know that the equilibrium shifted shifted to the left.I think that the color got darker because more of 2CrO4 was being formed which is why i think that it shifted left but I am not sure if this is correct?
The equation then give is
2CrO4^2-^(Aq) + 2H^+^(aq) == Cr2O7^2-^(aq) + h2O(l)
Could someone please explain and let me know if I am on the right track?
Thank you for your help!!!
If you add H^+ to the CrO4^2- it shifts to the right (to get rid of the extra acid you've added). The darker color is due to the dichromate that is forming as more acid is added. It should go from yellow to orange.
Ok so when the HCl is added it shifts to the right which is an exothermic process correct? It shifted to the right because the reactant is being used up??
Also the next question was which chromium ion is orange and which is yellow ans how do you know. So the dichromate is yellow and the Cr207^2-^ is yellow correct??
no and no. I don't know how exothermic that reaction is. It shifts to the right to use the added H^+ but that doesn't have anything to do with it being exothermic or not, at least I don't think so. CrO4^2- is chromate. It is yellow.
Cr2O7^2- is dichromate. It is orange.
oh ok I have to include in my answer if the process is exo or endothermic but I am not sure how I am suppose to know that.
We then had to add 1 drop of 1.0M NaOh to a new test tube with 2.5mL of Na2CrO4 in it. I guess the color change was suppose to be orange even though my color was still yellow. So does this mean that the shift would still be to the right which is towards the products because the NaOH was being used up? The NaOH would react with the dichromate which woudl lower the concentration of the other ions??