Question

Silver chloride, AgCl (Ksp = 1.8 x 10‒10), can be dissolved in solutions containing ammonia due to the formation of the soluble complex ion Ag(NH3)2 + (Kf = 1.0 x 108 ). What is the minimum amount of NH3 that would need to be added to dissolve 0.010 mol AgCl in 1.00 L of solution?

Answer 1

Explanation:

The given reaction will be as follows.

           $AgCl(s) \rightarrow Ag^{+}(aq) + Cl^{-}(aq)$ ............. (1)

     $K_{sp}$ = $[Ag^{+}][Cl^{-}] = 1.8 \times 10^{-10}$

Reaction for the complex formation is as follows.

          $Ag^{+}(aq) + 2NH_{3}(aq) \rightleftharpoons [Ag(NH_{3})_{2}]^{+}(aq)$ ........... (2)

          $K_{f}$ = $\frac{[Ag(NH_{3})_{2}]}{[Ag^{+}][NH_{3}]^{2}} = 1.0 \times 10^{8}$

When we add both equations (1) and (2) then the resultant equation is as follows.

             $AgCl(s) + 2NH_{3}(aq) \rightarrow [Ag(NH_{3})_{2}]^{+}(aq) + Cl^{-}(aq)$ ............. (3)

Therefore, equilibrium constant will be as follows.

                       K = $K_{f} \times K_{sp}$

                          = $1.0 \times 10^{8} \times 1.8 \times 10^{-10}$

                          = $1.8 \times 10^{-2}$

Since, we need 0.010 mol of AgCl to be soluble in 1 liter of solution after after addition of $NH_{3}$ for complexation. This means we have to set

               $[Ag^{+}]$ = $[Cl^{-}]$

                          = $\frac{0.010 mol}{1 L}$

                          = 0.010 M

For the net reaction, $AgCl(s) + 2NH_{3}(aq) \rightarrow [Ag(NH_{3})_{2}]^{+}(aq) + Cl^{-}(aq)$

Initial :                             0.010         x                     0                           0

Change :                    -0.010         -0.020             +0.010                +0.010

Equilibrium :                   0            x - 0.020           0.010                 0.010

Hence, the equilibrium constant expression for this is as follows.

              K = $\frac{[Ag(NH_{3})^{+}_{2}][Cl^{-}]}{[NH_{3}]^{2}}$

     $1.8 \times 10^{-2}$ = $\frac{0.010 \times 0.010}{(x - 0.020)^{2}}$

             x = 0.0945 mol      

or,          x = 0.095 mol (approx)

Thus, we can conclude that the number of moles of $NH_{3}$ needed to be added is 0.095 mol.