Calculate the solubility of silver chloride in a solution that is 0.100 M in NH3. Using Ksp

Question

Calculate the solubility of silver chloride in a solution that is 0.100 M in NH3.

(Use standard Ksp AgCl(s) = 1.77 x 10-10 and

Kf Ag(NH3)2+ = 1.7 x 107 )

Answer 1

AgCl(s) <==> Ag⁺(aq) + Cl^-(aq) ... dissolution of AgCl ... Ksp = 1.77x10^-10

Ag⁺(aq) + 2NH₃(aq) <==> Ag(NH₃)₂⁺(aq) ... complex formation of Ag(NH₃)₂⁺ ... Kf = 1.7x10⁷

---------------------------------------------------

AgCl(s) + 2NH₃(aq) <==> Ag(NH₃)₂⁺ + Cl^- ... K = (1.77x10^-10)(1.7x10⁷) = 0.00301

Let x = molar solubility of AgCl

K = [Ag(NH₃)₂⁺][Cl^-] / [NH₃]²

0.00301 = (x)(x) / (0.100 - x)² ... and taking the square root of both sides, we have...

0.0549 = x / 0.100 - x

x = 0.00520 M = molar solubility of AgCl in 0.100 M NH₃

Answer 2

Step 1: Write the relevant equations

The dissolution of AgCl in water is represented by:

AgCl (s)⇌Ag+(aq)+Cl−(aq)\text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq)AgCl (s)⇌Ag+(aq)+Cl−(aq)with the Ksp expression:

Ksp=[Ag+][Cl−]=1.77×10−10K_{\text{sp}} = [\text{Ag}^+][\text{Cl}^-] = 1.77 \times 10^{-10}Ksp​=[Ag+][Cl−]=1.77×10−10In the presence of NH₃, the Ag⁺ ions can form a complex with NH₃:

Ag+(aq)+2NH3(aq)⇌[Ag(NH3)2]+(aq)\text{Ag}^+ (aq) + 2 \text{NH}_3 (aq) \rightleftharpoons [\text{Ag(NH}_3)_2]^+ (aq)Ag+(aq)+2NH3​(aq)⇌[Ag(NH3​)2​]+(aq)with the formation constant:

Kf=[Ag(NH3)2+]/[Ag+][NH3]2=1.7×107K_f = [\text{Ag(NH}_3)_2^+]/[\text{Ag}^+][\text{NH}_3]^2 = 1.7 \times 10^7Kf​=[Ag(NH3​)2+​]/[Ag+][NH3​]2=1.7×107Step 2: Set up the solubility equilibrium expressions

Let the solubility of AgCl in the NH₃ solution be sss mol/L. At equilibrium:

1. The concentration of Ag⁺ will be sss mol/L from the dissociation of AgCl.
2. The concentration of Cl⁻ will also be sss mol/L from the dissociation of AgCl.
3. The amount of Ag⁺ that reacts with NH₃ to form the complex [Ag(NH3)2]+[Ag(NH₃)_2]^+[Ag(NH3​)2​]+ is determined by the amount of NH₃, which is 0.100 M.

Now, let's define the change in concentration:

1. Initially, there are no [Ag(NH3)2]+[Ag(NH₃)_2]^+[Ag(NH3​)2​]+ ions, so the concentration of [Ag(NH3)2]+[Ag(NH₃)_2]^+[Ag(NH3​)2​]+ is 0.
2. At equilibrium, the concentration of [Ag(NH3)2]+[Ag(NH₃)_2]^+[Ag(NH3​)2​]+ will be xxx, where xxx is the amount of Ag⁺ that reacted with NH₃ to form the complex.
3. The concentration of Ag⁺ will then be s−xs - xs−x, and the concentration of NH₃ will decrease by 2x2x2x, so it will be 0.100−2x0.100 - 2x0.100−2x.

Step 3: Set up the equilibrium equations

1. For the dissociation of AgCl:
2. Ksp=[Ag+][Cl−]=(s−x)s=1.77×10−10K_{\text{sp}} = [\text{Ag}^+][\text{Cl}^-] = (s - x)s = 1.77 \times 10^{-10}Ksp​=[Ag+][Cl−]=(s−x)s=1.77×10−10For the formation of [Ag(NH3)2]+[Ag(NH₃)_2]^+[Ag(NH3​)2​]+:

Kf=[Ag(NH3)2+][Ag+][NH3]2=x(s−x)(0.100−2x)2=1.7×107K_f = \frac{[Ag(NH₃)_2^+]}{[\text{Ag}^+][\text{NH}_3]^2} = \frac{x}{(s - x)(0.100 - 2x)^2} = 1.7 \times 10^7Kf​=[Ag+][NH3​]2[Ag(NH3​)2+​]​=(s−x)(0.100−2x)2x​=1.7×107Step 4: Approximate the solution

Since KfK_fKf​ is large, we expect a significant amount of Ag⁺ to react with NH₃, making xxx relatively large compared to sss. Therefore, we can approximate s−x≈0s - x \approx 0s−x≈0 and 0.100−2x≈0.1000.100 - 2x \approx 0.1000.100−2x≈0.100.

Thus, the equilibrium equation for KfK_fKf​ simplifies to:

Kf=x(0)(0.100)2=1.7×107K_f = \frac{x}{(0)(0.100)^2} = 1.7 \times 10^7Kf​=(0)(0.100)2x​=1.7×107