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 find the equilibrium constant for the disproportionation of \( \text{HgCl}_2 \) into \( \text{HgCl}^+ \) and \( \text{HgCl}_3^- \), we will use the provided reactions and their equilibrium constants.

### Step-by-Step Solution:

1. **Write down the given reactions and their equilibrium constants:**
- Reaction 1:
\[ \text{HgCl}^+ + \text{Cl}^- \rightleftharpoons \text{HgCl}_2 \quad K_1 = 3 \times 10^6 \]
- Reaction 2:
\[ \text{HgCl}_2 + \text{Cl}^- \rightleftharpoons \text{HgCl}_3^- \quad K_2 = 9.0 \]

2. **Identify the target reaction:**
The target reaction for the disproportionation of \( \text{HgCl}_2 \) is:
\[ 2 \text{HgCl}_2 \rightleftharpoons \text{HgCl}^+ + \text{HgCl}_3^- \]

3. **Manipulate the given reactions:**
- From Reaction 1, we can rearrange it to express \( \text{HgCl}^+ \):
\[ \text{HgCl}_2 \rightleftharpoons \text{HgCl}^+ + \text{Cl}^- \quad (K = \frac{1}{K_1}) \]
- From Reaction 2, we can use it as is:
\[ \text{HgCl}_2 + \text{Cl}^- \rightleftharpoons \text{HgCl}_3^- \quad K_2 \]

4. **Subtract Reaction 1 from Reaction 2:**
- If we subtract Reaction 1 from Reaction 2, we get:
\[ (\text{HgCl}_2 + \text{Cl}^- \rightleftharpoons \text{HgCl}_3^-) - (\text{HgCl}^+ + \text{Cl}^- \rightleftharpoons \text{HgCl}_2) \]
- This simplifies to:
\[ 2 \text{HgCl}_2 \rightleftharpoons \text{HgCl}^+ + \text{HgCl}_3^- \]

5. **Determine the equilibrium constant for the target reaction:**
- When we subtract reactions, the equilibrium constants multiply (the reverse reaction's constant is taken as the reciprocal):
\[ K' = \frac{K_2}{K_1} \]
- Substitute the values:
\[ K' = \frac{9.0}{3 \times 10^6} \]

6. **Calculate \( K' \):**
\[ K' = 3 \times 10^{-6} \]

### Final Answer:
The equilibrium constant for the disproportionation of \( \text{HgCl}_2 \) into \( \text{HgCl}^+ \) and \( \text{HgCl}_3^- \) is:
\[ K' = 3 \times 10^{-6} \]