What is the molar solubility of AgCl in 0.10 M NH₃? K_sp for AgCl is 1.8 × 10^-10 and the K_f for Ag(NH₃)₂⁺ is 1.7 × 10⁷. A)1.3 × 10^-5 M B)5.0 × 10^-3 M C)5.5 × 10^-3 M D)5.5 × 10^-2 M

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Quizplus · Accepted Answer

Using the information given, we can set up the following equilibrium expression: AgCl(s) ⇌ Ag+(aq) + Cl-(aq) Ksp = [Ag+][Cl-] We know that the concentration of Cl- comes from the dissociation of NH₃: [NH₃] = [Cl-] We also know that Ag+ reacts with NH₃ to form a complex ion: Ag+(aq) + NH₃(aq) ⇌ Ag(NH₃)(aq) Kf = [Ag(NH₃)]/[Ag+][NH₃] Now we can use the equilibrium constant expressions to set up an expression for the molar solubility of AgCl: Ksp = [Ag+][Cl-] = [Ag+][NH₃] Kf = [Ag(NH₃)]/[Ag+][NH₃] Substituting [Cl-] = [NH₃] into the Ksp expression and solving for [Ag+], we get: [Ag+] = sqrt(Ksp/[Cl-]) = sqrt(1.8 × 10^-10/0.10) = 1.34 × 10^-5 M Substituting [Ag+] = 1.34 × 10^-5 M into the Kf expression and solving for [Ag(NH₃)], we get: [Ag(NH₃)] = Kf[Ag+][NH₃] = (1.7 × 10⁷)(1.34 × 10^-5)(0.10) = 2.41 × 10^-7 M Finally, we can use the stoichiometry of the reaction to find the molar solubility of AgCl: AgCl(s) ⇌ Ag+(aq) + Cl-(aq) -1 +1 +1 If [Ag+] = 1.34 × 10^-5 M, then [Cl-] = [NH₃] = 1.34 × 10^-5 M And if [Ag(NH₃)] = 2.41 × 10^-7 M, then [AgCl] = [Ag+] = 1.34 × 10^-5 M Therefore, the molar solubility of AgCl is 1.34 × 10^-5 M, which corresponds to answer choice B.

What Is the Molar Solubility of AgCl in 0