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 in terms of \( K_{sp} \) for the salt \( A_3B \), we will follow these steps:

### Step 1: Write the dissociation equation
The salt \( A_3B \) dissociates in water to form ions. The dissociation can be represented as follows:
\[ A_3B_{(aq)} \rightarrow 3A^{+} + B^{-3} \]

### Step 2: Define solubility
Let the solubility of \( A_3B \) in water be \( S \). This means that when \( A_3B \) dissolves, it produces \( 3 \) moles of \( A^{+} \) ions and \( 1 \) mole of \( B^{-3} \) ions.

### Step 3: Express concentrations in terms of solubility
From the dissociation, we can express the concentrations of the ions in terms of \( S \):
- The concentration of \( A^{+} \) ions will be \( 3S \) (since there are 3 moles of \( A^{+} \)).
- The concentration of \( B^{-3} \) ions will be \( S \).

### Step 4: Write the expression for \( K_{sp} \)
The solubility product \( K_{sp} \) is given by the product of the concentrations of the ions, each raised to the power of their coefficients in the balanced equation:
\[ K_{sp} = [A^{+}]^3 [B^{-3}] \]
Substituting the concentrations we found:
\[ K_{sp} = (3S)^3 \cdot (S) \]

### Step 5: Simplify the expression
Now, simplify the expression:
\[ K_{sp} = 27S^3 \cdot S = 27S^4 \]

### Step 6: Solve for \( S \) in terms of \( K_{sp} \)
To find the solubility \( S \) in terms of \( K_{sp} \), we rearrange the equation:
\[ S^4 = \frac{K_{sp}}{27} \]
Taking the fourth root of both sides gives:
\[ S = \left(\frac{K_{sp}}{27}\right)^{\frac{1}{4}} \]

### Final Answer
Thus, the solubility \( S \) in terms of \( K_{sp} \) for the salt \( A_3B \) is:
\[ S = \left(\frac{K_{sp}}{27}\right)^{\frac{1}{4}} \]

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