Bomai places 7 mol of ammonia in a 14 L At equilibrium 0.2M of ammonium was found in the flask.what is the equilibrium constant kc for this reaction?. what are the conjugate acid and base for this reaction?
The reaction for the dissociation of ammonia is:
NH3 (g) ⇌ NH4+ (aq) + OH- (aq)
Initially, there were 7 mol of ammonia in the flask. At equilibrium, 0.2 M of ammonium was found. This means that 0.2 mol of ammonia has dissociated to form 0.2 mol of ammonium.
Therefore, the equilibrium concentrations are:
[NH3] = 7 - 0.2 = 6.8 mol / 14 L = 0.4857 M
[NH4+] = 0.2 M
The equilibrium constant (Kc) for this reaction can be calculated as follows:
Kc = [NH4+][OH-] / [NH3]
Kc = (0.2) / (0.4857)
Kc ≈ 0.41
The conjugate acid of ammonia (NH3) in this reaction is ammonium (NH4+), and the conjugate base is hydroxide (OH-).