The overall K_f for the complex ion Ag(NH₃)₂⁺ is 1.7 * 10⁷. K_sp for AgI is 1.5 *10^-¹⁶. What is the molar solubility of AgI in a solution that is 2.0 M in NH₃?

Question

Quizplus · Accepted Answer

The equation for the solubility product of AgI is: AgI(s) ⇌ Ag+ (aq) + I− (aq) The equilibrium constant expression is: K_sp = [Ag+][I−] The equation for the dissociation of Ag(NH₃)₂⁺ is: Ag(NH₃)₂⁺ ⇌ Ag+ (aq) + NH₃(aq) The equilibrium constant expression is: K_f = [Ag+][NH₃] Using the given equilibrium constants and the stoichiometry of the reaction, we can write: K_f / K_sp = ([Ag+][NH₃])/([Ag+][I−]) Solving for [I−]: [I−] = K_sp([Ag+][NH₃])/K_f Substituting the given values: [I−] = (1.5 x 10^-10)(2.0 M)/(1.7 x 10^-10) = 1.765 x 10^-10 M Therefore, the molar solubility of AgI is: [AgI] = [Ag+] + [I−] = [I−] = 1.765 x 10^-10 M The answer is choice C, 1.0 x 10^-10 M, which is the closest option to the calculated value.

The Overall Kf for the Complex Ion Ag(NH3)2+ Is 1

The Overall K_f for the Complex Ion Ag(NH₃)₂⁺ Is 1