What is the solubility (in mol/L) of `Fe(OH)_3` in a solution of pH= 8.0 ? `[K_(sp)" for "Fe(OH)_3= 1.0 xx 10^(-36)]`

Answer

To find the solubility of `Fe(OH)₃` in a solution with a pH of 8.0, we can follow these steps:

### Step 1: Understand the dissociation of `Fe(OH)₃`
`Fe(OH)₃` dissociates in water as follows:
\[ \text{Fe(OH)}_3 (s) \rightleftharpoons \text{Fe}^{3+} (aq) + 3 \text{OH}^- (aq) \]

### Step 2: Define solubility
Let the solubility of `Fe(OH)₃` be \( S \) mol/L. According to the dissociation equation:
- The concentration of `Fe^{3+}` ions will be \( S \).
- The concentration of `OH^-` ions will be \( 3S \).

### Step 3: Calculate the concentration of `OH^-` from pH
Given that the pH of the solution is 8.0, we can find the concentration of hydrogen ions \([H^+]\):
\[ [H^+] = 10^{-pH} = 10^{-8} \, \text{mol/L} \]

Using the ion product of water, we can find the concentration of hydroxide ions \([OH^-]\):
\[ [OH^-] = \frac{10^{-14}}{[H^+]} = \frac{10^{-14}}{10^{-8}} = 10^{-6} \, \text{mol/L} \]

### Step 4: Set up the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for `Fe(OH)₃` is given as:
\[ K_{sp} = [Fe^{3+}][OH^-]^3 \]
Substituting the values we have:
\[ K_{sp} = S \cdot (3S)^3 = S \cdot 27S^3 = 27S^4 \]

### Step 5: Substitute \( K_{sp} \) value
We know:
\[ K_{sp} = 1.0 \times 10^{-36} \]
Thus, we can set up the equation:
\[ 27S^4 = 1.0 \times 10^{-36} \]

### Step 6: Solve for \( S \)
Rearranging the equation gives:
\[ S^4 = \frac{1.0 \times 10^{-36}}{27} \]
Calculating this:
\[ S^4 = 3.70 \times 10^{-38} \]
Now, taking the fourth root:
\[ S = \left(3.70 \times 10^{-38}\right)^{1/4} \]

Calculating \( S \):
\[ S \approx 1.0 \times 10^{-9.5} \]
\[ S \approx 1.0 \times 10^{-9} \, \text{mol/L} \]

### Final Answer
The solubility of `Fe(OH)₃` in a solution with a pH of 8.0 is approximately:
\[ S \approx 1.0 \times 10^{-9} \, \text{mol/L} \]

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