The equilibrium constant for the disproportionation of `HgCl_2` into `HgCl^+ and HgCl_3^(-)` is
Given `HgCl^(+)+Cl^(-)hArrHgCl_2,K_1=3xx10^6,HgCl_2+Cl^(-)hArrHgCl_3^(-),K_2=9.0`

Options

A. `27xx10^6`
B. `3.3xx10^(-7)`
C. `3.3xx10^(-6)`
D. `3xx10^(-6)`

Answer

To solve the problem, we need to find the equilibrium constant for the disproportionation of \( \text{HgCl}_2 \) into \( \text{HgCl}^+ \) and \( \text{HgCl}_3^- \). We have two reactions with their equilibrium constants:

1. \( \text{HgCl}^+ + \text{Cl}^- \rightleftharpoons \text{HgCl}_2 \) with \( K_1 = 3 \times 10^6 \)
2. \( \text{HgCl}_2 + \text{Cl}^- \rightleftharpoons \text{HgCl}_3^- \) with \( K_2 = 9.0 \)

We can derive the equilibrium constant for the overall reaction using the relationship between the constants of the individual reactions.

### Step 1: Write the overall reaction
The overall reaction for the disproportionation of \( \text{HgCl}_2 \) can be represented as:
\[ \text{HgCl}_2 \rightleftharpoons \text{HgCl}^+ + \text{HgCl}_3^- \]

### Step 2: Relate the equilibrium constants
The equilibrium constant \( K_3 \) for the overall reaction can be expressed in terms of \( K_1 \) and \( K_2 \):
\[ K_3 = \frac{K_2}{K_1} \]

### Step 3: Substitute the values of \( K_1 \) and \( K_2 \)
Now we substitute the values of \( K_1 \) and \( K_2 \):
\[ K_3 = \frac{9.0}{3 \times 10^6} \]

### Step 4: Calculate \( K_3 \)
Perform the division:
\[ K_3 = \frac{9.0}{3 \times 10^6} = 3 \times 10^{-6} \]

### Conclusion
Thus, the equilibrium constant for the disproportionation of \( \text{HgCl}_2 \) into \( \text{HgCl}^+ \) and \( \text{HgCl}_3^- \) is:
\[ K_3 = 3 \times 10^{-6} \]

### Final Answer
The correct option is \( 3 \times 10^{-6} \).

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