`Fe(OH)_(2)` is diacidic base has `K_(b1)=10^(-4)` and `K_(b2)=2.5xx10^(-6)` What is the concentration of `Fe(OH)_(2)` in 0.1 M `Fe(NO_(3))_(2)` solution?

Options

A. `4xx10^(-9)`
B. `2.5xx10^(-6)`
C. `10^(-10)`
D. `10^(-14)`

Answer

To find the concentration of `Fe(OH)₂` in a 0.1 M `Fe(NO₃)₂` solution, we will follow these steps:

### Step 1: Understand the Dissociation of `Fe(OH)₂`
`Fe(OH)₂` is a diacidic base, meaning it can dissociate in two steps:
1. The first dissociation:
\[ \text{Fe}^{2+} + \text{H}_2\text{O} \rightleftharpoons \text{Fe(OH)}^+ + \text{H}^+ \quad (K_{b1} = 10^{-4}) \]
2. The second dissociation:
\[ \text{Fe(OH)}^+ + \text{H}_2\text{O} \rightleftharpoons \text{Fe(OH)}_2 + \text{H}^+ \quad (K_{b2} = 2.5 \times 10^{-6}) \]

### Step 2: Set Up the Equilibrium Expressions
Let the initial concentration of `Fe(NO₃)₂` be 0.1 M. This means the concentration of `Fe²⁺` ions is also 0.1 M.

For the first dissociation:
- Initial: `[Fe²⁺] = 0.1`, `[Fe(OH)⁺] = 0`, `[H⁺] = 0`
- Change: `[Fe²⁺] = 0.1 - x`, `[Fe(OH)⁺] = x`, `[H⁺] = x`
- At equilibrium: `[Fe²⁺] = 0.1 - x`, `[Fe(OH)⁺] = x`, `[H⁺] = x`

For the second dissociation:
- Initial: `[Fe(OH)⁺] = x`, `[Fe(OH)₂] = 0`, `[H⁺] = x`
- Change: `[Fe(OH)⁺] = x - y`, `[Fe(OH)₂] = y`, `[H⁺] = x + y`
- At equilibrium: `[Fe(OH)⁺] = x - y`, `[Fe(OH)₂] = y`, `[H⁺] = x + y`

### Step 3: Apply the Common Ion Effect
Since `Fe²⁺` ions are present from `Fe(NO₃)₂`, the `H⁺` concentration will be affected by the common ion effect. We can neglect `y` in comparison to `x` because `K_{b1}` is much larger than `K_{b2}`.

### Step 4: Calculate `K_h`
Using the relationship:
\[ K_h = \frac{K_w}{K_{b1}} \]
where \( K_w = 10^{-14} \) at 25°C, we can substitute the values:
\[ K_h = \frac{10^{-14}}{10^{-4}} = 10^{-10} \]

### Step 5: Equate `K_h` to `y`
Since `K_h` is equal to the concentration of `Fe(OH)₂` at equilibrium, we have:
\[ [Fe(OH)₂] = K_h = 10^{-10} \text{ M} \]

### Conclusion
Thus, the concentration of `Fe(OH)₂` in a 0.1 M `Fe(NO₃)₂` solution is:
\[ \text{Concentration of } Fe(OH)₂ = 10^{-10} \text{ M} \]