Question

What is the equilibrium constant of the reaction between Ag+ and NH3 to form Ag(NH3)2+ if a solution has an initial concentration of 0.1 M Ag+ and 0.2 M NH3, and at equilibrium, the concentration of Ag(NH3)2+ is 0.02 M?

Answer 1

The reaction between Ag+ and NH3 to form Ag NH3 2+ can be represented as:Ag+ + 2NH3  Ag NH3 2+To find the equilibrium constant  K  for this reaction, we can use the equilibrium concentrations of the reactants and products. First, we need to determine the change in concentration for each species during the reaction:Initial concentrations:[Ag+] = 0.1 M[NH3] = 0.2 M[Ag NH3 2+] = 0 MEquilibrium concentrations:[Ag NH3 2+] = 0.02 MSince 1 mole of Ag+ reacts with 2 moles of NH3 to form 1 mole of Ag NH3 2+, the change in concentration for each species is:[Ag+] = -0.02 M[NH3] = -2 * 0.02 M = -0.04 M[Ag NH3 2+] = +0.02 MNow we can find the equilibrium concentrations for Ag+ and NH3:[Ag+]_eq = 0.1 M - 0.02 M = 0.08 M[NH3]_eq = 0.2 M - 0.04 M = 0.16 MNow we can calculate the equilibrium constant  K  using the expression:K = [Ag NH3 2+] /  [Ag+] * [NH3]^2 K =  0.02 M  /   0.08 M  *  0.16 M ^2 K  97.66The equilibrium constant for the reaction between Ag+ and NH3 to form Ag NH3 2+ is approximately 97.66.