`Ag^(+) + NH_(3) ltimplies [Ag(NH_(3))]^(+), k_(1)=6.8 xx 10^(-5)`
`[Ag(NH_(3))]^(+) + NH_(3) ltimplies [Ag(NH_(3))_(2)]^(+)`,
`k_(2) = 1.6xx10^(-3)`
The formation constant of `[Ag(NH_(3))_(2)]^(+)` is :

Options

A. `6.8xx10^(-6)`
B. `1.08xx10^(-5)`
C. `1.08xx10^(-6)`
D. `6.8xx10^(-5)`

Explanation

To find the formation constant of the complex \([Ag(NH_3)_2]^+\), we will follow these steps:

### Step 1: Write the reactions and their equilibrium constants
We have two reactions:

1. \( Ag^+ + NH_3 \rightleftharpoons [Ag(NH_3)]^+ \) with \( K_1 = 6.8 \times 10^{-5} \)
2. \( [Ag(NH_3)]^+ + NH_3 \rightleftharpoons [Ag(NH_3)_2]^+ \) with \( K_2 = 1.6 \times 10^{-3} \)

### Step 2: Combine the reactions
When we add the two reactions, we can cancel out the intermediate complex \([Ag(NH_3)]^+\):

\[ Ag^+ + NH_3 + [Ag(NH_3)]^+ + NH_3 \rightleftharpoons [Ag(NH_3)_2]^+ \]

This simplifies to:

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

### Step 3: Calculate the overall equilibrium constant
The overall equilibrium constant \( K \) for the combined reaction is the product of the individual equilibrium constants:

\[ K = K_1 \times K_2 \]

Substituting the values:

\[ K = (6.8 \times 10^{-5}) \times (1.6 \times 10^{-3}) \]

### Step 4: Perform the multiplication
Calculating the product:

\[ K = 6.8 \times 1.6 \times 10^{-5} \times 10^{-3} = 10.88 \times 10^{-8} \]

### Step 5: Adjust the scientific notation
Now, we can express this in proper scientific notation:

\[ K = 1.088 \times 10^{-7} \]

### Final Answer
Thus, the formation constant of \([Ag(NH_3)_2]^+\) is:

\[ K = 1.088 \times 10^{-7} \]

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