The value of K_f for the complex ion Ag(NH₃)₂⁺ is 1.7 × 10⁷. K_sp for AgCl is 1.6 × 10^-10. Calculate the molar solubility of AgCl in 1.0 M NH₃. A) 1.7 × 10^-10 B) 2.9 × 10^-3 C) 4.7 × 10^-2 D) 1.3 × 10^-5 E) 5.2 × 10^-2

Question

Quizplus · Accepted Answer

Step 1: Write the solubility product expression:

AgCl(s) ⇌ Ag+(aq) + Cl-(aq)

Ksp = [Ag+][Cl-]

Step 2: Write the formation constant expression for the complex ion:

Ag(NHS13-)(aq) + 12S2-(aq) ⇌ Ag(NHS13-)(S2)12(aq)

Kf = [Ag(NHS13-)(S2)12] / [Ag(NHS13-)][S2-]12

Step 3: Write the overall equilibrium expression for the dissolution of AgCl in the presence of Ag(NHS13-) and S2-:

AgCl(s) + Ag(NHS13-)(aq) + 12S2-(aq) ⇌ Ag(NHS13-)(S2)12(aq) + Cl-(aq) + Ag+(aq)

Koverall = [Ag(NHS13-)(S2)12][Ag+][Cl-] / [Ag(NHS13-)][S2-]12[AgCl(s)]

Step 4: Use the values given in the problem to solve for the molar solubility of AgCl:

Ksp = 1.6 × 10-10
Kf = 1.7 × 1017
[NHS13-] = 1.0 M
[S2-] = 1.0 M

Koverall = (1.7 × 1017)(1.0 M)(1.0 M)12 / (1.0 M)(1.0 M)12(1.6 × 10-10) = 4.7 × 10-12

Solving for [Ag+] and [Cl-], assuming they are equal because AgCl is a 1:1 electrolyte:

Ksp = [Ag+]2
[Ag+] = [Cl-] = (1.6 × 10-10)1/2 = 1.3 × 10-5

Substituting into the overall equilibrium expression:

4.7 × 10-12 = (1.7 × 1017)(1.3 × 10-5)2 / (1.0 M)(1.0 M)12[S]

[S] = 4.7 × 10-2 M = 4.7 × 10-3 mol/L

Therefore, the molar solubility of AgCl in 1.0 M NHS13- is 4.7 × 10-3 mol/L.

The best choice is C, as it is the only option with a value close to 4.7 × 10-3 mol/L.

The Value of Kf for the Complex Ion Ag(NH3)2+ Is

The Value of K_f for the Complex Ion Ag(NH₃)₂⁺ Is