The solubility product of `BaCl_(2)` is `3.2xx10^(-9)`. What will be solubility in mol `L^(-1)`

Options

A. `4xx10^(-3)`
B. `3.2xx10^(-9)`
C. `1xx10^(-3)`
D. `1xx10^(-9)`

Explanation

To find the solubility of `BaCl2` in moles per liter given its solubility product (Ksp), we can follow these steps:

### Step-by-Step Solution:

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

2. **Define the Solubility**:
Let the solubility of `BaCl2` be \( S \) mol/L. When `BaCl2` dissolves, it produces:
- \( [Ba^{2+}] = S \)
- \( [Cl^{-}] = 2S \)

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

4. **Substitute the Given Ksp Value**:
We know that \( Ksp = 3.2 \times 10^{-9} \). Therefore, we can set up the equation:
\[ 4S^3 = 3.2 \times 10^{-9} \]

5. **Solve for S**:
Rearranging the equation to find \( S \):
\[ S^3 = \frac{3.2 \times 10^{-9}}{4} \]
\[ S^3 = 0.8 \times 10^{-9} \]
Now, take the cube root of both sides:
\[ S = \sqrt[3]{0.8 \times 10^{-9}} \]

6. **Calculate the Value**:
Using a calculator:
\[ S \approx 0.8^{1/3} \times (10^{-9})^{1/3} \]
\[ S \approx 0.928 \times 10^{-3} \approx 1 \times 10^{-3} \, \text{mol/L} \]

### Final Answer:
The solubility of `BaCl2` is approximately \( 1 \times 10^{-3} \, \text{mol/L} \).

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