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:

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

1. **Identify the equilibrium expression**:
The equilibrium constant \( K \) for this reaction can be expressed as:
\[ K = \frac{[\text{Fe}^{3+}][\text{OH}^-]^3}{[\text{Fe(OH)}_3]} \]
Since \(\text{Fe(OH)}_3\) is a solid, its concentration does not appear in the equilibrium expression.

2. **Initial conditions**:
Let the initial concentration of \(\text{OH}^-\) ions be \([OH^-]_0\) and the initial concentration of \(\text{Fe}^{3+}\) ions be \([Fe^{3+}]_0\).

3. **Change in concentration of OH⁻**:
According to the problem, the concentration of \(\text{OH}^-\) ions is decreased by \( \frac{1}{3} \). Therefore, the new concentration of \(\text{OH}^-\) ions becomes:
\[ [OH^-] = \frac{[OH^-]_0}{3} \]

4. **Substituting into the equilibrium expression**:
The new equilibrium expression with the decreased concentration of \(\text{OH}^-\) becomes:
\[ K' = \frac{[\text{Fe}^{3+}] [\frac{[OH^-]_0}{3}]^3}{[\text{Fe(OH)}_3]} \]
Simplifying this gives:
\[ K' = \frac{[\text{Fe}^{3+}] \cdot \frac{[OH^-]_0^3}{27}}{[\text{Fe(OH)}_3]} \]

5. **Relating the new equilibrium constant to the original**:
Since \( K = \frac{[\text{Fe}^{3+}]_0 [OH^-]_0^3}{[\text{Fe(OH)}_3]} \), we can relate \( K' \) to \( K \):
\[ K' = \frac{[\text{Fe}^{3+}] \cdot \frac{[OH^-]_0^3}{27}}{[\text{Fe(OH)}_3]} = \frac{1}{27} K \]

6. **Finding the increase in concentration of Fe³⁺**:
From the relationship \( K' = \frac{1}{27} K \), we can see that if the concentration of \(\text{OH}^-\) decreases, the concentration of \(\text{Fe}^{3+}\) must increase to maintain the equilibrium constant. The increase in concentration of \(\text{Fe}^{3+}\) can be calculated as:
\[ \text{Increase factor} = \frac{K}{K'} = 27 \]

Thus, the equilibrium concentration of \(\text{Fe}^{3+}\) will increase by **27 times**.