What color will the solution be if we add H2SO4 to the K2CrO4 solution?
H2SO4 (aq) ----> H+ (aq) + HSO4- (aq)
HSO4- (aq) ---> H+(aq) + SO42- (aq)
How does increasing the concentration of H+ affect the chromate/dichromate equilibrium? What color will the solution be?
What if we add NaOH to the solution? How does it affect the equillibrium? What does OH- react with in the solution? What color will the solution be?
[CrO4]2- + 2H+ <---------> [Cr2O7]2- + H2O
[CrO4]2- => Chromate Ion (Yellow)
[Cr2O7]2- => DiChromate Ion (Orange)
The reason why this demonstrates Le Chatlier's principle is:
If you add more Hydrogen Ions (Sulphuric Acid, H2SO4) to Chromate (Yellow) you will produce more of the Di-Chromate Ion (Orange), If you add Sodium Hydroxide (OH-), you'll produce more of the Chromate Ion (Yellow).
The Sulphuric Acid adds more H+ ions, therefore more Di-Chromate Ions will be produced in order to mop up the excess H+ ions and equalise the system (we can observe this via the colour change, Yellow - Orange). If Sodium Hydroxide (OH-) is added, these will mop up the H+ ions and therefore more Chromate Ions (Yellow) will be produced since the system is equalising, (we can observe this via the colour change: Orange to Yellow).
I've also found this link for all your Chromium Needs and Le Chatlier's Principle.
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