Respuesta :
Explanation:
The given reaction will be as follows.
      [tex]AgCl(s) \rightarrow Ag^{+}(aq) + Cl^{-}(aq)[/tex] ............. (1)
   [tex]K_{sp}[/tex] = [tex][Ag^{+}][Cl^{-}] = 1.8 \times 10^{-10}[/tex]
Reaction for the complex formation is as follows.
     [tex]Ag^{+}(aq) + 2NH_{3}(aq) \rightleftharpoons [Ag(NH_{3})_{2}]^{+}(aq)[/tex] ........... (2)
     [tex]K_{f}[/tex] = [tex]\frac{[Ag(NH_{3})_{2}]}{[Ag^{+}][NH_{3}]^{2}} = 1.0 \times 10^{8}[/tex]
When we add both equations (1) and (2) then the resultant equation is as follows.
       [tex]AgCl(s) + 2NH_{3}(aq) \rightarrow [Ag(NH_{3})_{2}]^{+}(aq) + Cl^{-}(aq)[/tex] ............. (3)
Therefore, equilibrium constant will be as follows.
            K = [tex]K_{f} \times K_{sp}[/tex]
             = [tex]1.0 \times 10^{8} \times 1.8 \times 10^{-10}[/tex]
             = [tex]1.8 \times 10^{-2}[/tex]
Since, we need 0.010 mol of AgCl to be soluble in 1 liter of solution after after addition of [tex]NH_{3}[/tex] for complexation. This means we have to set
        [tex][Ag^{+}][/tex] = [tex][Cl^{-}][/tex]
             = [tex]\frac{0.010 mol}{1 L}[/tex]
             = 0.010 M
For the net reaction, [tex]AgCl(s) + 2NH_{3}(aq) \rightarrow [Ag(NH_{3})_{2}]^{+}(aq) + Cl^{-}(aq)[/tex]
Initial :               0.010     x           0              0
Change : Â Â Â Â Â Â Â Â Â Â -0.010 Â Â Â Â -0.020 Â Â Â Â Â Â +0.010 Â Â Â Â Â Â Â Â +0.010
Equilibrium : Â Â Â Â Â Â Â Â Â 0 Â Â Â Â Â Â x - 0.020 Â Â Â Â Â 0.010 Â Â Â Â Â Â Â Â 0.010
Hence, the equilibrium constant expression for this is as follows.
       K = [tex]\frac{[Ag(NH_{3})^{+}_{2}][Cl^{-}]}{[NH_{3}]^{2}}[/tex]
   [tex]1.8 \times 10^{-2}[/tex] = [tex]\frac{0.010 \times 0.010}{(x - 0.020)^{2}}[/tex]
       x = 0.0945 mol   Â
or, Â Â Â Â Â x = 0.095 mol (approx)
Thus, we can conclude that the number of moles of [tex]NH_{3}[/tex] needed to be added is 0.095 mol.
