For the sparingly soluble salts, an equilibrium is established between the undissolved solid unit and ions of the dissolved salt. For a solution of the salt like `A_x B_y`.
`A_x B_y =xA^(y+)+yB^(x-)`
`K_(sp)=[A^(y+)]^x [B^(x-)]^y`
For precipitation to occur , the ionic product must be greater than `K_(sp)`.
`K_(sp)` for `A_2B_3` salt is `1.1xx10^(-23)` . Its solubility is

Options

A. `1xx10^(-8)`
B. `1xx10^(-5)`
C. `1xx10^(-6)`
D. `1xx10^(-7)`

Explanation

To find the solubility of the sparingly soluble salt \( A_2B_3 \) given its \( K_{sp} \) value, we will follow these steps:

### Step 1: Write the dissociation equation
The dissociation of the salt \( A_2B_3 \) in water can be represented as:
\[ A_2B_3 (s) \rightleftharpoons 2A^{3+} (aq) + 3B^{2-} (aq) \]

### Step 2: Define the solubility
Let the solubility of \( A_2B_3 \) be \( s \). This means that when \( A_2B_3 \) dissolves:
- The concentration of \( A^{3+} \) ions will be \( 2s \)
- The concentration of \( B^{2-} \) ions will be \( 3s \)

### Step 3: Write the expression for \( K_{sp} \)
The solubility product \( K_{sp} \) for the salt can be expressed as:
\[ K_{sp} = [A^{3+}]^2 [B^{2-}]^3 \]
Substituting the concentrations in terms of \( s \):
\[ K_{sp} = (2s)^2 (3s)^3 \]

### Step 4: Simplify the expression
Calculating the powers:
\[ K_{sp} = 4s^2 \cdot 27s^3 = 108s^5 \]

### Step 5: Set \( K_{sp} \) equal to the given value
We know from the problem statement that:
\[ K_{sp} = 1.1 \times 10^{-23} \]
Thus, we can set up the equation:
\[ 108s^5 = 1.1 \times 10^{-23} \]

### Step 6: Solve for \( s^5 \)
Rearranging gives:
\[ s^5 = \frac{1.1 \times 10^{-23}}{108} \]

### Step 7: Calculate \( s^5 \)
Calculating the right-hand side:
\[ s^5 = 1.0185 \times 10^{-25} \]

### Step 8: Solve for \( s \)
Taking the fifth root:
\[ s = (1.0185 \times 10^{-25})^{\frac{1}{5}} \]

### Step 9: Calculate \( s \)
Calculating \( s \):
\[ s \approx 1.0 \times 10^{-5} \]

### Final Answer
The solubility of the salt \( A_2B_3 \) is approximately:
\[ s \approx 1.0 \times 10^{-5} \, \text{mol/L} \]

---