If the concentration of `OH^(-)` ions in the reaction
`Fe(OH)_(3)(s)hArrFe^(3+)(aq.)+3OH^(-)(aq.)`
is decreased by `1//4` times, then the equilibrium concentration of `Fe^(3+)` will increase by

Options

A. (a) 8 times
B. (b) 16 times
C. (c) 64 times
D. (d) 4 times

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the equilibrium expression for the reaction.
The given reaction is:
\[ \text{Fe(OH)}_3(s) \rightleftharpoons \text{Fe}^{3+}(aq) + 3 \text{OH}^-(aq) \]

The equilibrium constant \( K_c \) for this reaction can be expressed as:
\[ K_c = \frac{[\text{Fe}^{3+}]}{[\text{OH}^-]^3} \]

### Step 2: Define the initial concentrations.
Let:
- The initial concentration of \(\text{OH}^-\) be \( [\text{OH}^-] \).
- The initial concentration of \(\text{Fe}^{3+}\) be \( [\text{Fe}^{3+}] \).

### Step 3: Analyze the change in concentration of \(\text{OH}^-\).
According to the problem, the concentration of \(\text{OH}^-\) is decreased by \( \frac{1}{4} \) times. Therefore, the new concentration of \(\text{OH}^-\) becomes:
\[ [\text{OH}^-]_{\text{new}} = \frac{[\text{OH}^-]}{4} \]

### Step 4: Set up the new equilibrium expression.
Let \( x \) be the factor by which the concentration of \(\text{Fe}^{3+}\) increases. Thus, the new concentration of \(\text{Fe}^{3+}\) will be:
\[ [\text{Fe}^{3+}]_{\text{new}} = x \cdot [\text{Fe}^{3+}] \]

Now, substituting these new concentrations into the equilibrium expression:
\[ K_c = \frac{x \cdot [\text{Fe}^{3+}]}{\left(\frac{[\text{OH}^-]}{4}\right)^3} \]

### Step 5: Simplify the new equilibrium expression.
Substituting the new concentration of \(\text{OH}^-\):
\[ K_c = \frac{x \cdot [\text{Fe}^{3+}]}{\frac{[\text{OH}^-]^3}{64}} \]
This simplifies to:
\[ K_c = \frac{64x \cdot [\text{Fe}^{3+}]}{[\text{OH}^-]^3} \]

### Step 6: Set the two expressions for \( K_c \) equal to each other.
Since \( K_c \) remains constant, we can set the original and new expressions equal:
\[ \frac{[\text{Fe}^{3+}]}{[\text{OH}^-]^3} = \frac{64x \cdot [\text{Fe}^{3+}]}{[\text{OH}^-]^3} \]

### Step 7: Cancel out common terms.
We can cancel \( [\text{Fe}^{3+}] \) and \( [\text{OH}^-]^3 \) from both sides:
\[ 1 = 64x \]

### Step 8: Solve for \( x \).
Now, solving for \( x \):
\[ x = \frac{1}{64} \]

### Conclusion
Thus, the equilibrium concentration of \(\text{Fe}^{3+}\) will increase by a factor of 64.

### Final Answer
The equilibrium concentration of \(\text{Fe}^{3+}\) will increase by **64 times**.

---