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?

Answer 1

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-).