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Step-by-Step Solution
Step 1: Identify the given quantities
• Volume of AgNO3 solution, V = 2 L
• Molarity of AgNO3 solution, M(AgNO3) = 0.80 M
• Target [Ag+] after adding NH3 = 5.0 × 10−8 M
• Formation constant, Kf for [Ag(NH3)2]+ = 1.0 × 108
Step 2: Calculate initial moles of Ag+
Initial moles of Ag+ (from AgNO3) = molarity × volume = 0.80 × 2 = 1.60 moles.
Step 3: Express final concentrations in terms of equilibrium variables
Let the total moles of NH3 added be a. When the complex [Ag(NH3)2]+ forms, each Ag+ ion requires 2 NH3 molecules. We want the final [Ag+] to be 5.0 × 10−8 M in a total volume of 2 L.
Hence, the final free Ag+ moles = (5.0 × 10−8) × 2 = 1.0 × 10−7 moles.
Because we started with 1.60 moles of Ag+, nearly all of it is converted into the complex, and only 1.0 × 10−7 moles stay uncomplexed. The rest (approximately 1.60 moles) form [Ag(NH3)2]+.
Step 4: Write the expression for the formation constant
The formation (stability) constant is given by:
K_f = \frac{[\text{Ag(NH}_3)_2^+]}{[\text{Ag}^+][\text{NH}_3]^2} = 1.0 \times 10^8.
• [Ag(NH3)2]+ concentration = (moles of complex) / (volume in L) ≈ 1.60 / 2 = 0.80 M (since almost 1.60 moles of Ag+ are complexed).
• [Ag+] = 5.0 × 10−8 M (given).
• If a moles of NH3 were added, then total NH3 used in complexes is 3.2 moles (because 1.60 moles of Ag+ each require 2 NH3, i.e. 1.60 × 2 = 3.2). Thus, leftover free NH3 moles = a − 3.2.
• Hence, [NH3] (free) = (a − 3.2) / 2 (because the total volume is 2 L).
Step 5: Plug equilibrium values into the Kf expression
1.0 \times 10^8 \;=\; \frac{0.80}{\bigl(5.0 \times 10^{-8}\bigr)\,\bigl(\frac{a - 3.2}{2}\bigr)^2}.
Step 6: Solve for a
Rearrange and solve for (a − 3.2) / 2:
\bigl(5.0 \times 10^{-8}\bigr)\,\bigl(\frac{a - 3.2}{2}\bigr)^2 \;=\; \frac{0.80}{1.0 \times 10^8} \;=\; 8.0 \times 10^{-9}.
\bigl(\frac{a - 3.2}{2}\bigr)^2 = \frac{8.0 \times 10^{-9}}{5.0 \times 10^{-8}} = 0.16.
\frac{a - 3.2}{2} = \sqrt{0.16} = 0.4.
a - 3.2 = 0.8 \quad \Longrightarrow \quad a = 4.0.
Step 7: State the final answer
The number of moles of NH3 that must be added is 4 (nearest integer).
