Can someone help me with this question Explain based on Le Chatlier’s Principle
ID: 1082776 • Letter: C
Question
Can someone help me with this questionExplain based on Le Chatlier’s Principle for the following solutions:
1.) 3 drops of 1M K2CrO4 and 3 drops of 3 M H2SO4 turned the solution a bright orange
2.) 3 drops of 1 M K2CrO4, 3 drops of 3M H2SO4, 3 drops 6M NaOH resulted in the solution turning into a dark yellow color
Thank you for your help! Can someone help me with this question
Explain based on Le Chatlier’s Principle for the following solutions:
1.) 3 drops of 1M K2CrO4 and 3 drops of 3 M H2SO4 turned the solution a bright orange
2.) 3 drops of 1 M K2CrO4, 3 drops of 3M H2SO4, 3 drops 6M NaOH resulted in the solution turning into a dark yellow color
Thank you for your help!
Explain based on Le Chatlier’s Principle for the following solutions:
1.) 3 drops of 1M K2CrO4 and 3 drops of 3 M H2SO4 turned the solution a bright orange
2.) 3 drops of 1 M K2CrO4, 3 drops of 3M H2SO4, 3 drops 6M NaOH resulted in the solution turning into a dark yellow color
Thank you for your help!
Explanation / Answer
For K2CrO4 , following rection is seen :
K2CrO4 (yellow)+ H2SO4 ----------------> K2Cr2O7 (orange) + K2SO44+ H2O
Now , according to Le Chatlier’s , if there is an excess of a reactant on one side, the equilibrium will shift to the opposite side to counter the effects and bring balance to the system.
In first part , concentration of H+ is much more than K2CrO4 even though the quantity is same, this implies that the reaction will shift to the right hand side where the orange colured K2Cr2O7 is formed.
In the second part, again concentration of acid i.e H+ is more than K2Cr2O4 but the cincentration of base is even more than acid . So , what happens is that the base not only neutralises the acid but also makes the medium basic due to being in excess. This result in a lower concentration of H+ on left side , so to equiliberate , the sustem moves towards the formation of K2Cr2O4 which is yellow in colour.