The number of moles of NH$$_3$$, that must be added to 2 L of 0.80 M AgNO$$_3$$ in order to reduce the concentration of Ag$$^+$$ ions to $$5.0 \times 10^{-8}$$ M ($$K_{formation}$$ for $$[Ag(NH_3)_2]^+ = 1.0 \times 10^8$$) is _________. (Nearest integer)
[Assume no volume change on adding NH$$_3$$]

We are given a 2 L solution that is $$0.80\;{\rm M}$$ in $$AgNO_3$$. Since every mole of $$AgNO_3$$ furnishes one mole of $$Ag^+$$, the initial moles and concentration of silver-ion are

$$n_{Ag^+}^{\,\text{initial}}=0.80\;{\rm mol\,L^{-1}}\times 2\;{\rm L}=1.6\;{\rm mol},\qquad[Ag^+]_0=0.80\;{\rm M}.$$

Ammonia forms the complex ion $$[Ag(NH_3)_2]^+$$ according to

$$Ag^+ + 2\,NH_3 \rightleftharpoons [Ag(NH_3)_2]^+.$$

The formation (stability) constant is stated to be

$$K_f = \frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}=1.0\times10^{8}.$$

We must add enough $$NH_3$$ so that the free silver-ion concentration is reduced to

$$[Ag^+]_{\text{eq}} = 5.0\times10^{-8}\;{\rm M}.$$

Because this value is many orders of magnitude smaller than the initial $$[Ag^+]$$, essentially every silver ion becomes complexed. Hence at equilibrium

$$[Ag(NH_3)_2^+]_{\text{eq}} \approx [Ag^+]_0-[Ag^+]_{\text{eq}} =0.80-5.0\times10^{-8}\approx0.80\;{\rm M}.$$

Let $$[NH_3]_{\text{free}}$$ be the concentration of ammonia that remains unbound. Each complex consumes two molecules of ammonia, so the concentration of bound ammonia is

$$[NH_3]_{\text{bound}} = 2\,[Ag(NH_3)_2^+] = 2(0.80)=1.6\;{\rm M}.$$

Now apply the formation constant:

$$K_f = \frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]_{\text{free}}^{\,2}} =\frac{0.80}{(5.0\times10^{-8})[NH_3]_{\text{free}}^{\,2}} =1.0\times10^{8}.$$

Rearranging for $$[NH_3]_{\text{free}}^{\,2}$$ gives

$$[NH_3]_{\text{free}}^{\,2} =\frac{0.80}{(1.0\times10^{8})(5.0\times10^{-8})} =\frac{0.80}{5}=0.16.$$

Taking the square root:

$$[NH_3]_{\text{free}}=\sqrt{0.16}=0.40\;{\rm M}.$$

The total ammonia concentration that must be present is the sum of free and bound portions:

$$[NH_3]_{\text{total}} =[NH_3]_{\text{free}}+[NH_3]_{\text{bound}} =0.40+1.60=2.0\;{\rm M}.$$

Because the solution volume is fixed at 2 L (the problem tells us to neglect volume change), the total moles of ammonia required are

$$n_{NH_3}=([NH_3]_{\text{total}})\times V =(2.0\;{\rm mol\,L^{-1}})\times(2\;{\rm L}) =4.0\;{\rm mol}.$$

The nearest integer is 4.

Hence, the correct answer is Option 4.