The solubility product of ferric hydroxide is `1.1xx10^(-36)` at 298 K. Calculate the solubility of ferric hydroxide in gram per litre. (Atomic mass, F=56 , O=16)

Explanation

To solve the problem of calculating the solubility of ferric hydroxide (Fe(OH)₃) in grams per liter given its solubility product (Ksp), we can follow these steps:

### Step 1: Write the Dissociation Equation
Ferric hydroxide dissociates in water as follows:
\[ \text{Fe(OH)}_3 (s) \rightleftharpoons \text{Fe}^{3+} (aq) + 3 \text{OH}^- (aq) \]

### Step 2: Define the Solubility
Let the solubility of ferric hydroxide be \( x \) moles per liter. This means:
- The concentration of \(\text{Fe}^{3+}\) ions will be \( x \).
- The concentration of \(\text{OH}^-\) ions will be \( 3x \) (since there are 3 hydroxide ions for each formula unit of ferric hydroxide).

### Step 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 from Step 2:
\[ K_{sp} = (x)(3x)^3 \]
\[ K_{sp} = x(27x^3) = 27x^4 \]

### Step 4: Substitute the Value of Ksp
We know that \( K_{sp} = 1.1 \times 10^{-36} \). Therefore, we can set up the equation:
\[ 27x^4 = 1.1 \times 10^{-36} \]

### Step 5: Solve for x
Rearranging the equation gives:
\[ x^4 = \frac{1.1 \times 10^{-36}}{27} \]
Calculating the right side:
\[ x^4 = \frac{1.1 \times 10^{-36}}{27} \approx 4.07 \times 10^{-38} \]
Now take the fourth root to find \( x \):
\[ x = \left(4.07 \times 10^{-38}\right)^{1/4} \]
Calculating this gives:
\[ x \approx 0.449 \times 10^{-9} \text{ moles per liter} \]

### Step 6: Convert Moles to Grams
To convert the solubility from moles per liter to grams per liter, we need the molar mass of ferric hydroxide:
- Molar mass of Fe = 56 g/mol
- Molar mass of O = 16 g/mol
- Molar mass of H = 1 g/mol

The molar mass of Fe(OH)₃ is:
\[ 56 + 3(16 + 1) = 56 + 3(17) = 56 + 51 = 107 \text{ g/mol} \]

Now, we can calculate the solubility in grams per liter:
\[ \text{Solubility (g/L)} = x \times \text{Molar mass} \]
\[ \text{Solubility (g/L)} = (0.449 \times 10^{-9}) \times 107 \]
Calculating this gives:
\[ \text{Solubility (g/L)} \approx 4.8 \times 10^{-8} \text{ g/L} \]

### Final Answer
The solubility of ferric hydroxide in grams per liter is approximately:
\[ \boxed{4.8 \times 10^{-8} \text{ g/L}} \]