The solubility product of iron (III) hydroxide is `1.6xx10^(-19)`. If X is the solubility of iron (III) hydroxide, which one of the following expression can be used to calculate X ?

Options

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

Explanation

To solve the problem of finding the solubility (X) of iron (III) hydroxide (Fe(OH)₃) using the given solubility product (Ksp), we can follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of iron (III) hydroxide in water can be represented as:
\[ \text{Fe(OH)}_3 (s) \rightleftharpoons \text{Fe}^{3+} (aq) + 3 \text{OH}^- (aq) \]

### Step 2: Define the Solubility
Let \( X \) be the solubility of Fe(OH)₃ in moles per liter (M). At equilibrium:
- The concentration of \(\text{Fe}^{3+}\) ions will be \( X \).
- The concentration of \(\text{OH}^-\) ions will be \( 3X \) (since 1 mole of Fe(OH)₃ produces 3 moles of OH⁻).

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

### Step 4: Simplify the Expression
Now, simplify the expression:
\[ K_{sp} = X \cdot (27X^3) = 27X^4 \]

### Step 5: Substitute the Value of Ksp
Given that \( K_{sp} = 1.6 \times 10^{-19} \):
\[ 27X^4 = 1.6 \times 10^{-19} \]

### Step 6: Solve for X
Now, solve for \( X \):
\[ X^4 = \frac{1.6 \times 10^{-19}}{27} \]
\[ X^4 = 5.9259 \times 10^{-21} \]
Taking the fourth root:
\[ X = (5.9259 \times 10^{-21})^{1/4} \]

### Step 7: Calculate X
Now, calculate the value of \( X \):
\[ X \approx 1.34 \times 10^{-5} \, \text{M} \]

### Conclusion
The expression that can be used to calculate \( X \) is:
\[ X = \left( \frac{1.6 \times 10^{-19}}{27} \right)^{1/4} \]