The equilibrium constant for the disproportionation of `HgCl_2` into `HgCl^+` and `HgCl_3^-` of 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^(-6)`
C. `3.2xx10^(-7)`
D. `3xx10^(-7)`

Answer

To find the equilibrium constant for the disproportionation of \( \text{HgCl}_2 \) into \( \text{HgCl}^+ \) and \( \text{HgCl}_3^- \), we start by analyzing the given reactions and their equilibrium constants.

### Step 1: Write the reactions and their equilibrium constants

1. The first reaction is:
\[ \text{HgCl}^+ + \text{Cl}^- \rightleftharpoons \text{HgCl}_2 \quad K_1 = 3 \times 10^6 \]

2. The second reaction is:
\[ \text{HgCl}_2 + \text{Cl}^- \rightleftharpoons \text{HgCl}_3^- \quad K_2 = 9.0 \]

### Step 2: Reverse the first reaction

To find the equilibrium constant for the reaction we are interested in, we need to reverse the first reaction:
\[ \text{HgCl}_2 \rightleftharpoons \text{HgCl}^+ + \text{Cl}^- \]
The equilibrium constant for the reversed reaction, \( K' \), is given by:
\[ K' = \frac{1}{K_1} = \frac{1}{3 \times 10^6} \]

### Step 3: Calculate \( K' \)

Calculating \( K' \):
\[ K' = \frac{1}{3 \times 10^6} = 3.33 \times 10^{-7} \]

### Step 4: Combine the reactions

Now we can combine the reversed first reaction with the second reaction:
\[ \text{HgCl}_2 \rightleftharpoons \text{HgCl}^+ + \text{Cl}^- \quad (K' = 3.33 \times 10^{-7}) \]
\[ \text{HgCl}_2 + \text{Cl}^- \rightleftharpoons \text{HgCl}_3^- \quad (K_2 = 9.0) \]

Adding these two reactions gives:
\[ \text{HgCl}^+ + \text{Cl}^- + \text{HgCl}_2 \rightleftharpoons \text{HgCl}^+ + \text{HgCl}_3^- \]

### Step 5: Write the equilibrium constant for the overall reaction

The overall equilibrium constant \( K \) for the reaction:
\[ \text{HgCl}_2 \rightleftharpoons \text{HgCl}^+ + \text{HgCl}_3^- \]
is given by the product of the equilibrium constants of the two reactions:
\[ K = K' \times K_2 \]

### Step 6: Substitute the values

Substituting the values:
\[ K = (3.33 \times 10^{-7}) \times (9.0) \]

### Step 7: Calculate \( K \)

Calculating \( K \):
\[ K = 3.33 \times 9.0 \times 10^{-7} = 30.0 \times 10^{-7} = 3.0 \times 10^{-6} \]

### Final Answer

Thus, the equilibrium constant for the disproportionation of \( \text{HgCl}_2 \) into \( \text{HgCl}^+ \) and \( \text{HgCl}_3^- \) is:
\[ \boxed{3.0 \times 10^{-6}} \]