Solubility product of `BaCl_(2)` is `4xx10^(-9)`. Its solubility in moles//litre would be

Options

A. `1 xx 10^(-3)`
B. `1 xx 10^(-9)`
C. `4xx10^(-27)`
D. `1 xx 10^(-27)`

Answer

To find the solubility of `BaCl2` given its solubility product (Ksp) of `4 x 10^(-9)`, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Dissociation Equation**:
The dissociation of barium chloride in water can be represented as:
\[ BaCl_2 (s) \rightleftharpoons Ba^{2+} (aq) + 2Cl^{-} (aq) \]

2. **Define Solubility**:
Let the solubility of `BaCl2` be \( S \) moles per liter. When `BaCl2` dissolves:
- The concentration of `Ba^{2+}` ions will be \( S \).
- The concentration of `Cl^{-}` ions will be \( 2S \) (since there are two chloride ions for every formula unit of `BaCl2`).

3. **Write the Expression for Ksp**:
The solubility product (Ksp) expression for `BaCl2` is given by:
\[ Ksp = [Ba^{2+}][Cl^{-}]^2 \]
Substituting the concentrations:
\[ Ksp = (S)(2S)^2 \]
This simplifies to:
\[ Ksp = S \cdot 4S^2 = 4S^3 \]

4. **Substitute the Given Ksp Value**:
We know that \( Ksp = 4 \times 10^{-9} \). Therefore:
\[ 4S^3 = 4 \times 10^{-9} \]

5. **Solve for S**:
Divide both sides by 4:
\[ S^3 = 10^{-9} \]
Now, take the cube root of both sides:
\[ S = (10^{-9})^{1/3} = 10^{-3} \]

6. **Final Result**:
The solubility of `BaCl2` in moles per liter is:
\[ S = 1 \times 10^{-3} \text{ moles/liter} \]

### Conclusion:
The solubility of `BaCl2` is \( 1 \times 10^{-3} \) moles/liter.