Calculate the molar solubility of `Fe(OH)_(3)` in a buffer solution that is `0.1M` in `NH_(4)OH` and `0.1M` in `NH_(4)Cl` `(K_(b)` of `NH_(4)OH=1.8xx10^(-5)`,`K_(sp)` of `Fe(OH)_(3)=2.6xx10^(-39)`)

Options

A. `4.46xx10^(-22)M`
B. `8.92xx10^(-25)M`
C. `2.23xx10^(-24)M`
D. `4.46xx10^(-25)M`

Explanation

To calculate the molar solubility of `Fe(OH)₃` in a buffer solution that is `0.1M` in `NH₄OH` and `0.1M` in `NH₄Cl`, we will follow these steps:

### Step 1: Calculate pOH of the Buffer Solution
The pOH of the buffer solution can be calculated using the formula:
\[ \text{pOH} = \text{pK}_B + \log \left( \frac{[\text{Salt}]}{[\text{Base}]} \right) \]

Given:
- \( K_b \) for \( NH_4OH = 1.8 \times 10^{-5} \)

First, calculate \( \text{pK}_B \):
\[ \text{pK}_B = -\log(1.8 \times 10^{-5}) \]

Now, since the concentrations of salt (\( NH_4Cl \)) and base (\( NH_4OH \)) are equal (both are 0.1 M), the log term becomes:
\[ \log \left( \frac{0.1}{0.1} \right) = \log(1) = 0 \]

Thus, the pOH is:
\[ \text{pOH} = -\log(1.8 \times 10^{-5}) \]

### Step 2: Calculate OH⁻ Ion Concentration
From the pOH, we can find the concentration of OH⁻ ions:
\[ [OH^-] = 10^{-\text{pOH}} \]

Since we already calculated pOH, we can directly find:
\[ [OH^-] = 1.8 \times 10^{-5} \, \text{M} \]

### Step 3: Write the Ksp Expression for Fe(OH)₃
The dissociation of \( Fe(OH)_3 \) in water can be represented as:
\[ Fe(OH)_3 (s) \rightleftharpoons Fe^{3+} (aq) + 3OH^- (aq) \]

The solubility product constant \( K_{sp} \) is given by:
\[ K_{sp} = [Fe^{3+}][OH^-]^3 \]

### Step 4: Substitute Known Values into Ksp Expression
Let \( s \) be the molar solubility of \( Fe(OH)_3 \). Then:
\[ K_{sp} = s \cdot [OH^-]^3 \]

Given:
- \( K_{sp} = 2.6 \times 10^{-39} \)
- \( [OH^-] = 1.8 \times 10^{-5} \)

Substituting these values into the Ksp expression:
\[ 2.6 \times 10^{-39} = s \cdot (1.8 \times 10^{-5})^3 \]

### Step 5: Solve for s
Calculate \( (1.8 \times 10^{-5})^3 \):
\[ (1.8 \times 10^{-5})^3 = 5.832 \times 10^{-15} \]

Now substitute this back into the equation:
\[ 2.6 \times 10^{-39} = s \cdot 5.832 \times 10^{-15} \]

Solving for \( s \):
\[ s = \frac{2.6 \times 10^{-39}}{5.832 \times 10^{-15}} \]

Calculating this gives:
\[ s \approx 4.46 \times 10^{-25} \, \text{M} \]

### Conclusion
The molar solubility of \( Fe(OH)_3 \) in the given buffer solution is:
\[ s \approx 4.46 \times 10^{-25} \, \text{M} \]

### Final Answer
**Option D: 4.46 x 10⁻²⁵ M**

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