Question

1)Info:
a)Consider the equilibrium:
Ag (NH3)2^+ + Cl^- <------> AgCl + 2NH3

b) Based on the stoichiometry of the reaction, which of the conditions constitutes an excess of NH3 over Cl-? (3:1 or 30:1)
     Both ratios work
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Question:
c) Based on your answer to part (b), which ion (Cl-/NH3) forms a stronger bond to Ag+. State your reasoning.

d) Derive and calculate the Keq value for the equilibrium above (part (a)) using the following values of the equilibrium constants:
Kf[Ag (NH3)2+] = 1.7 x 10^7 and
Ksp(AgCl) = 1.7 x 10^-10

Answer 1

b) The balanced equation is

Ag (NH₃)₂⁺ + Cl^- <------> AgCl + 2NH₃

From the above balanced reaction, 1 mole of Ag (NH₃)₂⁺ reacts with 1 mole of Cl^- to produce 1 mole of AgCl & 2 moles of NH_3.

Ratio of Ag (NH₃)₂⁺ & Cl^- = 3:1

(3 mol Ag (NH₃)₂⁺ * 1 mol Cl^-) / 1 mol Ag (NH₃)₂⁺ = 3 mol Cl^-

3 mol of Cl^- required but there is only 1 mol of Cl^-. So, Cl^- is the limiting reagent.

Amount of NH₃ formed = (1 mol Cl^- * 2 mol NH₃) / 1 mol Cl^-= 2 mol NH₃

Amount of NH₃ formed = 2 mol NH₃ for Ratio of Ag (NH₃)₂⁺ & Cl^- = 3:1

Ratio of Ag (NH₃)₂⁺ & Cl^- = 30:1

(30 mol Ag (NH₃)₂⁺ * 1 mol Cl^-) / 1 mol Ag (NH₃)₂⁺ = 30 mol Cl^-

30 mol of Cl^- required but there is only 1 mol of Cl^-. So, Cl^- is the limiting reagent.

Amount of NH₃ formed = (1 mol Cl^- * 2 mol NH₃) / 1 mol Cl^-= 2 mol NH₃

Amount of NH₃ formed = 2 mol NH₃ for Ratio of Ag (NH₃)₂⁺ & Cl^- = 30:1

Both ratio 3:1 & 30:1 produce the same amount (2 mol) of NH₃

c)

• Like charges repel & opposite charges attract.
• Positively charged Ag⁺ forms stronger bond with negatively charged Cl^- resulting in the formation of ionic bond between Ag⁺ &Cl^-

So, Ag⁺ forms stronger bond with Cl^- compared to NH₃

d) dissolution

AgCl <------> Ag⁺ + Cl^-

K_sp = [Ag⁺] [Cl^-] = 1.7 x 10^-10

complex formation

Ag⁺ + (NH₃)²⁺ <------> Ag (NH₃)²⁺

K_f = [Ag (NH₃)²⁺] / [Ag⁺][NH₃)²⁺]

keq = K_sp * K_f