The solublity product of iron (III) hydroxide is `1.6 xx 10^(-39)`. If X is the solublity of iron (III) hydroxide, then which one of the following expressions can be used to calculate X

Options

A. `K_(sp) = X^(4)`
B. `K_(sp) = 9 X^(4)`
C. `K_(sp) = 27 X^(3)`
D. `K_(sp) = 27 X^(4)`

Explanation

To solve the problem, we need to determine the expression that can be used to calculate the solubility (X) of iron (III) hydroxide, given its solubility product (Ksp) of \(1.6 \times 10^{-39}\).

### Step-by-Step Solution:

1. **Write the Dissociation Equation**:
Iron (III) hydroxide, \( \text{Fe(OH)}_3 \), dissociates in water according to the following equation:
\[ \text{Fe(OH)}_3 (s) \rightleftharpoons \text{Fe}^{3+} (aq) + 3 \text{OH}^- (aq) \]

2. **Define Solubility (X)**:
Let the solubility of \( \text{Fe(OH)}_3 \) be \( X \) mol/L. When \( \text{Fe(OH)}_3 \) dissolves, it produces:
- \( \text{Fe}^{3+} \) ions: \( X \) mol/L
- \( \text{OH}^- \) ions: \( 3X \) mol/L (since 3 hydroxide ions are produced for each formula unit of iron (III) hydroxide)

3. **Write the Expression for Ksp**:
The solubility product \( K_{sp} \) is given by the expression:
\[ K_{sp} = [\text{Fe}^{3+}][\text{OH}^-]^3 \]
Substituting the concentrations in terms of \( X \):
\[ K_{sp} = (X)(3X)^3 \]

4. **Simplify the Expression**:
Simplifying the expression:
\[ K_{sp} = X \cdot 27X^3 = 27X^4 \]

5. **Set Ksp Equal to the Given Value**:
We know that \( K_{sp} = 1.6 \times 10^{-39} \), so we can set up the equation:
\[ 27X^4 = 1.6 \times 10^{-39} \]

6. **Solve for X**:
To find \( X \), we rearrange the equation:
\[ X^4 = \frac{1.6 \times 10^{-39}}{27} \]
Then, take the fourth root:
\[ X = \left(\frac{1.6 \times 10^{-39}}{27}\right)^{1/4} \]

### Final Expression:
The expression that can be used to calculate the solubility \( X \) of iron (III) hydroxide is:
\[ X = \left(\frac{1.6 \times 10^{-39}}{27}\right)^{1/4} \]