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 compound \( A_3B_2 \) given its solubility product \( K \), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of \( A_3B_2 \) 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 \( S \). When \( A_3B_2 \) dissolves, it produces:
- 3 moles of \( A^{3+} \) ions, which gives a concentration of \( 3S \)
- 2 moles of \( B^{2-} \) ions, which gives a concentration of \( 2S \)

### Step 3: Write the expression for the solubility product \( K \)
The solubility product \( K \) can be expressed in terms of the concentrations of the ions:
\[ K = [A^{3+}]^3 \cdot [B^{2-}]^2 \]
Substituting the concentrations from the solubility:
\[ K = (3S)^3 \cdot (2S)^2 \]

### Step 4: Simplify the expression
Calculating the powers:
\[ K = 27S^3 \cdot 4S^2 = 108S^5 \]

### Step 5: Solve for \( S \)
Now, we can rearrange the equation to solve for \( S \):
\[ S^5 = \frac{K}{108} \]
Taking the fifth root of both sides gives:
\[ S = \left(\frac{K}{108}\right)^{\frac{1}{5}} \]

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

---