The solubility in terms of `K_(sp) "for " A_3B_((aq))` is

Options

A. `(K_(sp)/3)^(1//4)`
B. `(K_(sp)/27)^(1//4)`
C. `(27K_(sp)) ^(1//4)`
D. `(3K_(sp)) ^(1//4)`

Answer

To find the solubility of the compound \( A_3B \) in terms of its solubility product \( K_{sp} \), we can follow these steps:

### Step 1: Write the dissociation equation
When \( A_3B \) dissolves in water, it dissociates into its ions:
\[ A_3B \rightarrow 3A^+ + B^{3-} \]

### Step 2: Define molar solubility
Let the molar solubility of \( A_3B \) be \( S \). This means that when \( A_3B \) dissolves, it produces \( 3S \) moles of \( A^+ \) ions and \( S \) moles of \( B^{3-} \) ions.

### Step 3: Write the expression for \( K_{sp} \)
The solubility product \( K_{sp} \) is defined as:
\[ K_{sp} = [A^+]^3 \cdot [B^{3-}] \]
Substituting the concentrations in terms of \( S \):
\[ [A^+] = 3S \quad \text{and} \quad [B^{3-}] = S \]
Thus, we can write:
\[ K_{sp} = (3S)^3 \cdot (S) = 27S^3 \cdot S = 27S^4 \]

### Step 4: Solve for solubility \( S \) in terms of \( K_{sp} \)
To express \( S \) in terms of \( K_{sp} \), we rearrange the equation:
\[ S^4 = \frac{K_{sp}}{27} \]
Taking the fourth root of both sides gives:
\[ S = \left(\frac{K_{sp}}{27}\right)^{1/4} \]

### Conclusion
The solubility \( S \) of \( A_3B \) in terms of \( K_{sp} \) is:
\[ S = \left(\frac{K_{sp}}{27}\right)^{1/4} \]