The solubility product of a sparingly soluble salt `BA_(2)` is `4xx10^(-12)`. The solubility of `BA_(2)` is

Options

A. `4xx10^(-4)`
B. `4xx10^(-12)`
C. `4xx10^(-3)`
D. `1xx10^(-4)`

Explanation

To find the solubility of the sparingly soluble salt \( BA_2 \) with a solubility product \( K_{sp} = 4 \times 10^{-12} \), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of the salt \( BA_2 \) in water can be represented as:
\[ BA_2 (s) \rightleftharpoons B^{2+} (aq) + 2A^{-} (aq) \]

### Step 2: Define the solubility
Let the solubility of \( BA_2 \) be \( S \). When \( BA_2 \) dissolves, it produces:
- \( B^{2+} \) ions: \( S \)
- \( A^{-} \) ions: \( 2S \)

### Step 3: Write the expression for \( K_{sp} \)
The solubility product \( K_{sp} \) is given by the expression:
\[ K_{sp} = [B^{2+}][A^{-}]^2 \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (S)(2S)^2 \]
This simplifies to:
\[ K_{sp} = S \cdot 4S^2 = 4S^3 \]

### Step 4: Substitute the value of \( K_{sp} \)
Now, substituting the given value of \( K_{sp} \):
\[ 4S^3 = 4 \times 10^{-12} \]

### Step 5: Simplify the equation
Dividing both sides by 4:
\[ S^3 = 10^{-12} \]

### Step 6: Solve for \( S \)
To find \( S \), take the cube root of both sides:
\[ S = (10^{-12})^{1/3} = 10^{-4} \]

### Conclusion
The solubility of \( BA_2 \) is:
\[ S = 10^{-4} \, \text{mol/L} \]