Question

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 (\(K_{\text{formation}}\) for \([Ag(NH_3)_2]^+ = 1.0 \times 10^8\)) is _________. (Nearest integer)
[ Assume no volume change on adding \(NH_3\) ]

Hint:

For very large \(K_f\), always assume complete conversion of the limiting reactant to the complex to simplify the equilibrium calculation.

Answer 1

Correct Answer: 4

Step 1: Understanding the Concept:
This is a complexation equilibrium problem. Since \(K_f\) is very high, we assume almost all \(Ag^+\) is converted to the complex.
Step 2: Detailed Explanation:
Reaction: \(Ag^+ + 2NH_3 \rightleftharpoons [Ag(NH_3)_2]^+\)
Initial \([Ag^+] = 0.80 \, M\).
Since \(K_f = 10^8\) is very large, \([Ag(NH_3)_2]^+ \approx 0.80 \, M\).
Let the total concentration of added \(NH_3\) be \(C\).
Free \([NH_3] = C - 2 \times 0.80 = C - 1.60\).
Given remaining \([Ag^+] = 5.0 \times 10^{-8} \, M\).
\[ K_f = \frac{[[Ag(NH_3)_2]^+]}{[Ag^+][NH_3]^2} \]
\[ 1.0 \times 10^8 = \frac{0.80}{(5.0 \times 10^{-8}) [NH_3]^2} \]
\[ [NH_3]^2 = \frac{0.80}{1.0 \times 10^8 \times 5.0 \times 10^{-8}} = \frac{0.80}{5} = 0.16 \]
\[ [NH_3]_{\text{free}} = \sqrt{0.16} = 0.40 \, M \]
Total concentration \(C = [NH_3]_{\text{free}} + 2[Complex] = 0.40 + 2(0.80) = 2.00 \, M\).
Total moles of \(NH_3 = \text{Molarity} \times \text{Volume} = 2.00 \, M \times 2 \, L = 4.0 \, \text{moles}\).
Step 3: Final Answer:
The number of moles of \(NH_3\) is 4.