If solubility product of `A_3 B_2` is 'K' then the solubility of `A_3 B_2` will be

Options

A. `K^1`/`5`
B. `2K^1`/`5`
C. `(108 K)^1`/`5`
D. `(K/108)^1`/`5`

Answer

To find the solubility of the salt \( A_3B_2 \) given that its solubility product \( K_{sp} \) is \( K \), we can follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of the salt \( A_3B_2 \) in water can be represented as:
\[ A_3B_2 (s) \rightleftharpoons 3A^{3+} (aq) + 2B^{2-} (aq) \]

### Step 2: Define the Solubility
Let the solubility of \( A_3B_2 \) be \( S \) moles per liter. When \( A_3B_2 \) dissolves, it produces:
- \( 3S \) moles of \( A^{3+} \)
- \( 2S \) moles of \( B^{2-} \)

### Step 3: Write the Expression for \( K_{sp} \)
The solubility product \( K_{sp} \) is given by the formula:
\[ K_{sp} = [A^{3+}]^3 \times [B^{2-}]^2 \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (3S)^3 \times (2S)^2 \]

### Step 4: Simplify the Expression
Now, simplify the expression:
\[ K_{sp} = 27S^3 \times 4S^2 = 108S^5 \]

### Step 5: Set \( K_{sp} \) Equal to \( K \)
Since we know that \( K_{sp} = K \), we can write:
\[ K = 108S^5 \]

### Step 6: Solve for \( S \)
To find the solubility \( S \), rearrange the equation:
\[ S^5 = \frac{K}{108} \]
Now take the fifth root of both sides:
\[ S = \left(\frac{K}{108}\right)^{\frac{1}{5}} \]

### Final Answer
The solubility of \( A_3B_2 \) is:
\[ S = \left(\frac{K}{108}\right)^{\frac{1}{5}} \]