1. Balanced chemical reaction: NH3 (g) ⇌ NH4+ (aq) + OH- (aq)
2. To calculate the equilibrium constant (Kc), we need to use the concentrations of the reactants and products at equilibrium. Since the initial concentration of NH3 was 7mol in a 14L flask, the concentration of NH3 at equilibrium is 0.5 M (7 mol / 14 L = 0.5 M). The concentration of NH4+ at equilibrium is given as 0.2 M. The concentration of OH- at equilibrium can be calculated using the fact that NH3 reacts with water to form NH4+ and OH-. Since every molecule of NH3 forms one NH4+ and one OH-, the concentration of OH- at equilibrium is also 0.5 M.
The equilibrium constant (Kc) is calculated as [NH4+][OH-] / [NH3]. Therefore, Kc = (0.2)(0.5) / 0.5 = 0.2
3. The conjugate acid of NH3 is NH4+ and the conjugate base of NH4+ is NH3.
Bomai places 7mol 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 a balanced chemical reaction for the above reaction?
2. What is the equilibrium constant (Kc) for this reaction?
3. What are conjugate acids and base for the reaction in 1?
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