What is the molar solubility product for `V_3(PO_4)_5` in terms of `K_(SP)`?

Options

A. `S = (K_(SP)/84375)^(1/8)`
B. `S = (K_(SP)/155)^(1/8)`
C. `S = (K_(SP)/108)^(1/8)`
D. `S = (K_(SP))^(1/8)`

Answer

To find the molar solubility product for \( V_3(PO_4)_5 \) in terms of \( K_{sp} \), we can follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of \( V_3(PO_4)_5 \) in water can be represented as:
\[ V_3(PO_4)_5 (s) \rightleftharpoons 3 V^{5+} (aq) + 5 PO_4^{3-} (aq) \]

### Step 2: Define the Solubility
Let the molar solubility of \( V_3(PO_4)_5 \) be \( s \). This means:
- The concentration of \( V^{5+} \) ions will be \( 3s \) (since 3 moles of \( V^{5+} \) are produced).
- The concentration of \( PO_4^{3-} \) ions will be \( 5s \) (since 5 moles of \( PO_4^{3-} \) are produced).

### Step 3: Write the Expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) is given by the expression:
\[ K_{sp} = [V^{5+}]^3 \times [PO_4^{3-}]^5 \]
Substituting the concentrations in terms of \( s \):
\[ K_{sp} = (3s)^3 \times (5s)^5 \]

### Step 4: Simplify the Expression
Now, we can simplify the expression:
\[ K_{sp} = (27s^3) \times (3125s^5) \]
\[ K_{sp} = 27 \times 3125 \times s^{3+5} = 84375 \times s^8 \]

### Step 5: Solve for \( s \)
Now, we can express \( s \) in terms of \( K_{sp} \):
\[ s^8 = \frac{K_{sp}}{84375} \]
Taking the eighth root of both sides gives:
\[ s = \left( \frac{K_{sp}}{84375} \right)^{\frac{1}{8}} \]

### Final Answer
Thus, the molar solubility \( s \) of \( V_3(PO_4)_5 \) in terms of \( K_{sp} \) is:
\[ s = \left( \frac{K_{sp}}{84375} \right)^{\frac{1}{8}} \]

---