Explain the observation based on Le Chatlier’s principle for the following solut

Question

Explain the observation based on Le Chatlier’s principle for the following solutions:
1.) 5mL of 0.1 M BaCl2 and 1 mL of 0.1 M (NH4)2C2O4 with 5 drops of 1 M HCl formed a precipitate on top that slowly disappeared and became slightly clear.
2 .) 5mL of 0.1 M BaCl2 and 1 mL of 0.1 M (NH4)2C2O4 with 5 drops of 1 M NH3 forms a precipitate layer on top. Slowly the cloudy part sinks to the bottom. Explain the observation based on Le Chatlier’s principle for the following solutions:
1.) 5mL of 0.1 M BaCl2 and 1 mL of 0.1 M (NH4)2C2O4 with 5 drops of 1 M HCl formed a precipitate on top that slowly disappeared and became slightly clear.
2 .) 5mL of 0.1 M BaCl2 and 1 mL of 0.1 M (NH4)2C2O4 with 5 drops of 1 M NH3 forms a precipitate layer on top. Slowly the cloudy part sinks to the bottom.
1.) 5mL of 0.1 M BaCl2 and 1 mL of 0.1 M (NH4)2C2O4 with 5 drops of 1 M HCl formed a precipitate on top that slowly disappeared and became slightly clear.
2 .) 5mL of 0.1 M BaCl2 and 1 mL of 0.1 M (NH4)2C2O4 with 5 drops of 1 M NH3 forms a precipitate layer on top. Slowly the cloudy part sinks to the bottom.

Explanation / Answer

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.

The following rections are seen :

a)
BaCl2 + (NH4)2C2O4 ------> BaC2O4 (precipitate) + 2NH4Cl

In this case, the precipitate is formed due to reaction of BaCl2 and NH4)2C2O4 , it stays for some time due volume of BaCl2 is higher , but it goes back slowly due to the presence of HCl , which decreases it's solubility in the solution and moves the equilibrium to the left .

In the second part, Due to addition of ammonia , it acts as a proton sponge , taking up H+ from the product side and shifting the reaction back to left side where the precipitate is not dissolved

NH3 + H(+) ----> NH4(+)

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