The expression for equilibrium constant, `K_(c)` for the following reaction is
`Fe_((aq))^(3+)+3OH_((aq))^(-)hArrFe(OH)_(3(s))`

Options

A. `K_(c)=([Fe(OH)_(3)])/([Fe^(3+)][OH^(-)]^(3))`
B. `K_(c)=([Fe(OH)_(3)])/([Fe^(3+)][OH^(-)])`
C. `K_(c)=(1)/([Fe^(3+)][OH^(-)]^(3))`
D. `K_(c)=[Fe(OH)_(3)]`

Explanation

To derive the expression for the equilibrium constant \( K_c \) for the reaction:

\[ \text{Fe}^{3+}_{(aq)} + 3 \text{OH}^-_{(aq)} \rightleftharpoons \text{Fe(OH)}_3_{(s)} \]

we will follow these steps:

### Step 1: Identify the components of the reaction
In the given reaction, we have:
- Reactants: \( \text{Fe}^{3+}_{(aq)} \) and \( 3 \text{OH}^-_{(aq)} \)
- Product: \( \text{Fe(OH)}_3_{(s)} \)

### Step 2: Understand the nature of the reaction
This reaction is a heterogeneous equilibrium because it involves both aqueous ions and a solid. In heterogeneous equilibria, the concentrations of solids and pure liquids are not included in the equilibrium expression.

### Step 3: Write the general form of the equilibrium constant expression
The general form of the equilibrium constant \( K_c \) for a reaction is given by:

\[ K_c = \frac{[\text{Products}]^{\text{stoichiometric coefficient}}}{[\text{Reactants}]^{\text{stoichiometric coefficient}}} \]

### Step 4: Apply the equilibrium constant expression to the reaction
For our specific reaction, the equilibrium constant expression will be:

\[ K_c = \frac{[\text{Fe(OH)}_3]^{1}}{[\text{Fe}^{3+}]^{1} [\text{OH}^-]^{3}} \]

### Step 5: Consider the state of the products and reactants
Since \( \text{Fe(OH)}_3 \) is a solid, its concentration is not included in the expression. Therefore, the expression simplifies to:

\[ K_c = \frac{1}{[\text{Fe}^{3+}] [\text{OH}^-]^3} \]

### Step 6: Finalize the expression for \( K_c \)
Thus, the final expression for the equilibrium constant \( K_c \) for the reaction is:

\[ K_c = \frac{1}{[\text{Fe}^{3+}] [\text{OH}^-]^3} \]