Post a New Question


posted by .

Given the following information:

Ag^+(aq) + Cl^-(aq)<-->AgCl(s) Keq=1.8x10^-1

Ag^+(aq) + 2NH3(aq)<->Ag(NH3)2^+(aq) Keq=8.2x10^3

Explain why the AgCl(s) dissolved when NH3 was added to the test tube.

  • Chemistry -

    eqn 1. Ag^+(aq) + Cl^-(aq)<-->AgCl(s) Keq=1.8x10^-1

    eqn 2. Ag^+(aq) + 2NH3(aq)<->Ag(NH3)2^+(aq) Keq=8.2x10^3

    I think you made a typo in Keq (= Ksp) = 1.8 x 10^-10 and not -1.
    Note the equation 2 has a large Keq which means the equilibrium is far to the right (more products and fewer reactants). Equation 1 is a small K so the reaction is far to the left (which is why AgCl is not very soluble and forms a ppt in water solvent. From Le Chatelier's Principle, equation 2 is far to the right which means (Ag^+) is decreased significantly. That makes Ag^+ small in equation 1 and the reaction shifts to the right (meaning the AgCl solid dissolves to increase the Ag^+ removed by equation 2. That continues until all of the AgCl solid is dissolved. Usually a relatively large excess of NH3 is used which shifts that equilibrium even farther to the right and that ends up dissolving all of the AgCl solid from equation 1.

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question