A compound `BA_2 " has " K_(sp)=4xx10^(-12)` solubility of this comp. will be :

Options

A. `10^-(3)`
B. `10^(-4)`
C. `10^(-5)`
D. `10^(-6)`

Answer

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

### Step 1: Write the Dissociation Equation
When the compound \( BA_2 \) dissolves in water, it dissociates into its ions:
\[ BA_2 (s) \rightleftharpoons B^{2+} (aq) + 2 A^{-} (aq) \]

### Step 2: Define the Solubility
Let the solubility of \( BA_2 \) be \( x \) mol/L. This means:
- The concentration of \( B^{2+} \) ions will be \( x \) mol/L.
- The concentration of \( A^{-} \) ions will be \( 2x \) mol/L (since there are two \( A^{-} \) ions for each formula unit of \( BA_2 \)).

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

### Step 4: Substitute the Given \( K_{sp} \) Value
We know that \( K_{sp} = 4 \times 10^{-12} \), so we can set up the equation:
\[ 4x^3 = 4 \times 10^{-12} \]

### Step 5: Solve for \( x \)
Dividing both sides by 4:
\[ x^3 = 10^{-12} \]
Now, taking the cube root of both sides:
\[ x = (10^{-12})^{1/3} = 10^{-4} \]

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