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 a sparingly soluble salt of the type \( BA_2 \), where \( B \) is a cation and \( A \) is an anion, we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of the salt \( BA_2 \) in water can be represented as:
\[ BA_2 (s) \rightleftharpoons B^{2+} (aq) + 2A^{-} (aq) \]
This shows that one formula unit of \( BA_2 \) produces one \( B^{2+} \) ion and two \( A^{-} \) ions.

### Step 2: Define the solubility
Let the solubility of \( BA_2 \) be \( x \) moles per liter. This means that in a saturated solution:
- The concentration of \( B^{2+} \) ions will be \( x \) moles per liter.
- The concentration of \( A^{-} \) ions will be \( 2x \) moles per liter (since there are two \( A^{-} \) ions for every one \( BA_2 \) that dissolves).

### Step 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:
\[ K_{sp} = [B^{2+}][A^{-}]^2 \]

### Step 4: Substitute the concentrations
Now, substituting the concentrations we found:
\[ K_{sp} = [B^{2+}][A^{-}]^2 = (x)(2x)^2 \]

### Step 5: Simplify the expression
Now, simplifying the expression:
\[ K_{sp} = x \cdot (4x^2) = 4x^3 \]

### Conclusion
Thus, the solubility product \( K_{sp} \) of the salt \( BA_2 \) is given by:
\[ K_{sp} = 4x^3 \]