The number of moles of ${\mathrm{NH}}_3$, that must be added to $2 \mathrm{L}\text{ of }0.80\mathrm{M} {\mathrm{AgNO}}_3$ in order to reduce the concentration of Ag+ ions$5.0\times {10}^{-8}\mathrm{M} \left({\mathrm{K}}_{{\text{formation }}_{ }}\text{for }{\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]}^{+}=1.0\times {10}^8\right)$ is _____ .

We are tasked with determining the number of moles of NH₃ that must be added to a 2L solution of 0.80M AgNO₃ to reduce the concentration of Ag⁺ ions to 5.0 × 10^-8 M. The formation constant (K_f) for the complex ion (Ag(NH₃)₂)+ is given as 1.0 × 10⁸.

Step 1: Calculate Initial Moles of Ag⁺

The initial moles of Ag⁺ can be calculated using the formula:

Moles of Ag+ = Concentration × Volume = 0.80 M × 2 L = 1.6 moles

Step 2: Define the Number of Moles of NH3 Added

Let A represent the moles of NH₃ added. The reaction can be written as:

Ag+ + 2 NH3 ⇌ (Ag(NH3)2)+

From the stoichiometry, for every mole of Ag⁺, 2 moles of NH₃ are required.

Step 3: Determine the Change in Concentration

After the reaction reaches equilibrium, the concentration of Ag⁺ will be:

[Ag+] = 1.6 - A / 2

Where A is the number of moles of NH₃ added, and A / 2 is the amount of Ag⁺ that reacts with the NH₃.

Step 4: Set Up the Equilibrium Expression

The equilibrium expression for the formation constant K_f is:

Kf = ([Ag(NH3)2]+) / ([Ag+][NH3]^2)

At equilibrium, the concentration of Ag(NH₃)₂+ can be approximated as A / 2 (since 2 moles of NH₃ react with 1 mole of Ag⁺).

Step 5: Substitute Known Values into the Equilibrium Expression

We know that at equilibrium:

[Ag+] = 5.0 × 10-8 M

Substituting into the equilibrium expression:

1.0 × 108 = (A / 2) × (5.0 × 10-8) × (A / 2)2

Step 6: Simplify the Equation

Rearranging the equation gives:

1.0 × 108 = (A / 2) × (5.0 × 10-8) × A2 / 4

Continue simplifying:

1.0 × 108 = (2A) / (5.0 × 10-8) × A2

Multiplying both sides:

5A2 = 2A

Step 7: Solve the Equation

We factor the equation:

A(5A - 2) = 0

Thus, A can be either 0 or 0.4 moles.

Step 8: Final Answer

The number of moles of NH₃ that must be added is 0.4 moles (rounded to the nearest integer).