Bomai places 7M of ammonia (NH3) in 14L flask.At equilibrium 0.2M ammonium (NH4+) was found in the flask. What is the equilibrium constant Kc for this reaction.
1.write the balance chemical reaction for the above reaction
2.what is the equilibrium constant Kc of the reaction
3.what are the conjugate acid and base for the reaction.
1. The balanced chemical reaction for the dissociation of ammonia in water is:
NH3 + H2O ↔ NH4+ + OH-
2. The equilibrium constant Kc for this reaction can be calculated using the formula:
Kc = [NH4+][OH-] / [NH3][H2O]
Given that the concentration of NH4+ is 0.2M, NH3 is 7M, and the volume of the flask is 14L, we can calculate the initial concentration of OH- as 0M (as it is produced during the reaction):
Kc = (0.2)(0) / (7)(14)
Kc = 0
Therefore, the equilibrium constant Kc for this reaction is 0.
3. The conjugate acid of NH3 is NH4+ and the conjugate base of NH4+ is NH3. In the reverse reaction, NH4+ functions as an acid and donates a proton to form NH3, while NH3 functions as a base and accepts a proton to form NH4+.