What is the molarity of `Fe(CN)_(6)^(4-)` in a saturated solution of `Ag_(4)[Fe(CN)_(6)]`?
`(K_(sp)=1.6xx10^(-41))`

Options

A. `1.6xx10^(-8)`
B. `5.2xx10^(-8)`
C. `2.0xx10^(-8)`
D. `2.3xx10^(-9)`

Answer

To find the molarity of \( \text{Fe(CN)}_6^{4-} \) in a saturated solution of \( \text{Ag}_4[\text{Fe(CN)}_6] \), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of \( \text{Ag}_4[\text{Fe(CN)}_6] \) in water can be expressed as:
\[ \text{Ag}_4[\text{Fe(CN)}_6] (s) \rightleftharpoons 4 \text{Ag}^+ (aq) + \text{Fe(CN)}_6^{4-} (aq) \]

### Step 2: Define solubility
Let the solubility of \( \text{Ag}_4[\text{Fe(CN)}_6] \) in mol/L be \( s \). According to the dissociation equation, for every 1 mole of \( \text{Ag}_4[\text{Fe(CN)}_6] \) that dissolves, it produces 4 moles of \( \text{Ag}^+ \) and 1 mole of \( \text{Fe(CN)}_6^{4-} \). Therefore:
- The concentration of \( \text{Ag}^+ \) will be \( 4s \)
- The concentration of \( \text{Fe(CN)}_6^{4-} \) will be \( s \)

### Step 3: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for the dissociation can be expressed as:
\[ K_{sp} = [\text{Ag}^+]^4 [\text{Fe(CN)}_6^{4-}] \]
Substituting the concentrations in terms of \( s \):
\[ K_{sp} = (4s)^4 \cdot s = 256s^5 \]

### Step 4: Set up the equation with the given \( K_{sp} \)
We know that \( K_{sp} = 1.6 \times 10^{-41} \). Therefore, we can set up the equation:
\[ 256s^5 = 1.6 \times 10^{-41} \]

### Step 5: Solve for \( s \)
Now, we can solve for \( s \):
\[ s^5 = \frac{1.6 \times 10^{-41}}{256} \]
Calculating the right-hand side:
\[ s^5 = \frac{1.6 \times 10^{-41}}{256} = 6.25 \times 10^{-44} \]
Now, take the fifth root to find \( s \):
\[ s = (6.25 \times 10^{-44})^{1/5} \]

### Step 6: Calculate \( s \)
Calculating \( s \):
\[ s \approx 3.98 \times 10^{-9} \, \text{mol/L} \]

### Step 7: Conclusion
Since \( s \) represents the molarity of \( \text{Fe(CN)}_6^{4-} \), we conclude that:
\[ \text{Molarity of } \text{Fe(CN)}_6^{4-} \approx 3.98 \times 10^{-9} \, \text{mol/L} \]