How many moles of `NH_(3)` must be added to `1.0L` of `0.75M AgNO_(3)` in order to reduce the `[Ag^(o+)]` to `5.0 xx 10^(-8)M. K_(f) Ag (NH_(3))_(2)^(o+) = 1 xx 10^(8)`.

Explanation

To solve the question, we need to determine how many moles of \( NH_3 \) must be added to a solution of \( 0.75M \) \( AgNO_3 \) to reduce the concentration of \( [Ag^+] \) to \( 5.0 \times 10^{-8} M \). The formation constant \( K_f \) for the complex \( Ag(NH_3)_2^+ \) is given as \( 1 \times 10^8 \).

### Step-by-Step Solution:

**Step 1: Write the equilibrium expression for the formation of the complex.**
The reaction for the formation of the complex is:
\[ Ag^+ + 2NH_3 \rightleftharpoons Ag(NH_3)_2^+ \]
The formation constant \( K_f \) is given by:
\[ K_f = \frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2} \]

**Step 2: Define the concentrations.**
Let:
- The initial concentration of \( Ag^+ \) from \( AgNO_3 \) be \( 0.75 M \).
- The concentration of \( Ag^+ \) after adding \( NH_3 \) is \( 5.0 \times 10^{-8} M \).
- Let the concentration of \( NH_3 \) added be \( x \).

**Step 3: Substitute the known values into the \( K_f \) expression.**
Substituting the known values into the \( K_f \) expression:
\[ 1 \times 10^8 = \frac{0.75}{5.0 \times 10^{-8} \cdot x^2} \]

**Step 4: Rearrange the equation to solve for \( x^2 \).**
Rearranging gives:
\[ x^2 = \frac{0.75}{1 \times 10^8 \cdot 5.0 \times 10^{-8}} \]
Calculating the right side:
\[ x^2 = \frac{0.75}{5.0 \times 10^0} = 0.15 \]

**Step 5: Solve for \( x \).**
Taking the square root:
\[ x = \sqrt{0.15} \approx 0.387 M \]

**Step 6: Calculate the total moles of \( NH_3 \) required.**
Since \( 2 \) moles of \( NH_3 \) are required for \( 1 \) mole of \( Ag(NH_3)_2^+ \):
- For \( 0.75 \) moles of \( Ag(NH_3)_2^+ \), \( 2 \times 0.75 = 1.5 \) moles of \( NH_3 \) are needed.
- Therefore, the total moles of \( NH_3 \) required is:
\[ 1.5 + 0.387 \approx 1.887 \text{ moles} \]

### Final Answer:
The total moles of \( NH_3 \) that must be added is approximately \( 1.89 \) moles.