In the system `Fe(OH)_(3(s))hArr Fe_((aq)^(3+) +3OH_((aq))^-` , decreasing the conc. Of `OH^-` ions `1/3` times cause the equilibrium conc. Of `Fe^(3+)` to increase ……. Times

Options

A. 3
B. 9
C. 18
D. 27

Answer

To solve the problem, we need to analyze the equilibrium reaction given:

\[ \text{Fe(OH)}_3(s) \rightleftharpoons \text{Fe}^{3+}(aq) + 3\text{OH}^-(aq) \]

We are tasked with determining how the equilibrium concentration of \(\text{Fe}^{3+}\) changes when the concentration of \(\text{OH}^-\) ions is decreased to one-third of its initial value.

### Step-by-Step Solution:

1. **Write the Expression for the Equilibrium Constant (K)**:
The equilibrium constant \( K \) for the reaction can be expressed as:
\[ K = \frac{[\text{Fe}^{3+}]}{[\text{OH}^-]^3} \]
Here, \([\text{Fe}^{3+}]\) is the concentration of iron ions, and \([\text{OH}^-]\) is the concentration of hydroxide ions.

2. **Define Initial Concentrations**:
Let the initial concentration of \(\text{OH}^-\) be \([\text{OH}^-]_{initial}\) and the initial concentration of \(\text{Fe}^{3+}\) be \([\text{Fe}^{3+}]_{initial}\).

3. **Change in Concentration of \(\text{OH}^-\)**:
According to the problem, the concentration of \(\text{OH}^-\) is decreased to one-third:
\[ [\text{OH}^-]_{final} = \frac{1}{3} [\text{OH}^-]_{initial} \]

4. **Substituting into the Equilibrium Constant Expression**:
Since \( K \) remains constant, we can set up the equation:
\[ K = \frac{[\text{Fe}^{3+}]_{initial}}{[\text{OH}^-]_{initial}^3} = \frac{[\text{Fe}^{3+}]_{final}}{[\text{OH}^-]_{final}^3} \]

5. **Substituting the Final Concentration of \(\text{OH}^-\)**:
Substitute \([\text{OH}^-]_{final}\) into the equation:
\[ K = \frac{[\text{Fe}^{3+}]_{final}}{\left(\frac{1}{3} [\text{OH}^-]_{initial}\right)^3} \]
This simplifies to:
\[ K = \frac{[\text{Fe}^{3+}]_{final}}{\frac{1}{27} [\text{OH}^-]_{initial}^3} \]

6. **Setting the Two Expressions for K Equal**:
Now we can set the two expressions for \( K \) equal to each other:
\[ \frac{[\text{Fe}^{3+}]_{initial}}{[\text{OH}^-]_{initial}^3} = \frac{[\text{Fe}^{3+}]_{final}}{\frac{1}{27} [\text{OH}^-]_{initial}^3} \]

7. **Cross-Multiplying to Solve for \([\text{Fe}^{3+}]_{final}\)**:
Cross-multiplying gives:
\[ [\text{Fe}^{3+}]_{initial} \cdot \frac{1}{27} = [\text{Fe}^{3+}]_{final} \]
Therefore:
\[ [\text{Fe}^{3+}]_{final} = 27 \cdot [\text{Fe}^{3+}]_{initial} \]

8. **Conclusion**:
The equilibrium concentration of \(\text{Fe}^{3+}\) increases by a factor of 27 when the concentration of \(\text{OH}^-\) ions is decreased to one-third of its initial value.

### Final Answer:
The equilibrium concentration of \(\text{Fe}^{3+}\) increases **27 times**.