If the solubility of a sparingly soluble saltof the type `BA_(2)` (Giving three ions on dissociation of a molecule ) is 'x' moles per litre , then its solubility product is given by

Options

A. `x^(2)`
B. `2x^(3)`
C. `4x^(2)`
D. `4x^(3)`

Answer

To find the solubility product (Ksp) of the sparingly soluble salt of the type \( BA_2 \) (which dissociates into three ions), follow these steps:

### Step-by-Step Solution

1. **Identify the Dissociation of the Salt**:
The salt \( BA_2 \) dissociates in water as follows:
\[ BA_2 (s) \rightleftharpoons B^{2+} (aq) + 2A^{-} (aq) \]
From this equation, we can see that one formula unit of \( BA_2 \) produces one \( B^{2+} \) ion and two \( A^{-} \) ions.

2. **Define the Solubility**:
Let the solubility of \( BA_2 \) be \( x \) moles per liter. This means that at equilibrium:
- The concentration of \( B^{2+} \) will be \( x \) moles per liter.
- The concentration of \( A^{-} \) will be \( 2x \) moles per liter (since there are two \( A^{-} \) ions produced for each \( BA_2 \) that dissolves).

3. **Write the Expression for Ksp**:
The solubility product \( K_{sp} \) is given by the product of the concentrations of the ions, each raised to the power of their coefficients in the balanced equation:
\[ K_{sp} = [B^{2+}][A^{-}]^2 \]

4. **Substitute the Concentrations**:
Substitute the concentrations we found in step 2 into the Ksp expression:
\[ K_{sp} = [B^{2+}][A^{-}]^2 = (x)(2x)^2 \]

5. **Simplify the Expression**:
Simplifying the expression:
\[ K_{sp} = x \cdot (2x)^2 = x \cdot 4x^2 = 4x^3 \]

6. **Final Result**:
Therefore, the solubility product \( K_{sp} \) for the salt \( BA_2 \) is:
\[ K_{sp} = 4x^3 \]