Some comments below. Don't forget to predict what happens to the colour of the supernatant (aqueous phase) and if you can the pH.
a) To a solution of 5 drops of 0.3 M K2CrO4 and 2 drops of 0.3 M NaOH add 0.1 M Ba(NO3)2 until a change is noted.
MY PREDICTION: a yellow precipitate forms.
Agreed and the colour changes of the aqueous phase is from orange to yellow to (almost) clear.
b) use test tube from step (a) add 3 M HCl until a change is noted
MY PREDICTION: colour become yelloer b/v more H+ ion r being used and more water is being produced
Adding H+ forces the chromate equilibrium to the right, turning any chromate ion to dichromate ion. At the same time some of the barium chromate ppt will dissolve (may slight change). The supernatant will turn orange.
c) To a solution of 5 drops of 0.3 M K2Cr2O7 and 2 drops of 0.3 M HCl add a drop at a time of 0.1 M Ba(NO3)2
the pH may low enough to prevent a ppt because the low pH surpresses chromate ion forming. So no change.
d) To the test tube from step (c) add 1M NaOH until a change is noted
The OH- ion will remove H+ allowing chromate ion to form. This will result in a yellow ppt of barium chromate.
e) Suggest a way to reverse the changes and reactions that you observed in step (b) and (d)
Any ideas?
f) Place 5 drops of ).3 M K2CrO7 in one test tube and 5 drops of 0.3 M K2Cr)4 in another test tube. Add a few drops of 0.1 M Ba(NO3)2 to each and note results
There is no difference.
MY PREDICTION: K2CrO7 is more soluble K2CrO4 forms a precepitate NO they are both soluble otherwise you can't have the solutions.
An interesting thing to do is to start with dichromate solution measure the pH. Add barium nitrate. Measure the pH of the supernatant. Why is there a pH change and why is it so low?
Use Le' Chatelier's Principle.
a solution at equilibrium:
BaCrO4(s)<> Ba^2+(aq)+ Cr)4^-(aq)recall: 2CrO4^2-(aq)+ 2H^+(aq)<> Cr2O7^2-(aq) +H20
2CrO4^2- yellow
Cr2O7^2- orange
Make predictions about colour changes or changes in the number of states
a) To a solution of 5 drops of 0.3 M K2CrO4 and 2 drops of 0.3 M NaOH add 0.1 M Ba(NO3)2 until a change is noted.
MY PREDICTION: a yellow precipitate forms
b) use test tube from step (a) add 3 M HCl until a change is noted
MY PREDICTION: colour become yelloer b/v more H+ ion r being used and more water is being produced
c) To a solution of 5 drops of 0.3 M K2Cr2O7 and 2 drops of 0.3 M HCl add a drop at a time of 0.1 M Ba(NO3)2
d) To the test tube from step (c) add 1M NaOH until a change is noted
e) Suggest a way to reverse the changes and reactions that you observed in step (b) and (d)
f) Place 5 drops of ).3 M K2CrO7 in one test tube and 5 drops of 0.3 M K2Cr)4 in another test tube. Add a few drops of 0.1 M Ba(NO3)2 to each and note results
MY PREDICTION: K2CrO7 is more soluble K2CrO4 forms a precepitate
It's so difficult to make these predictions i am very confused and i've spent almost the entire day on this
2 answers
to a solution of analyte NH3nh4and h2s were add ared ppt was formed
6 answers