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 mol `L^(-1)`, we will use the solubility product constant (Ksp) provided in the question. Here’s a step-by-step solution:

### Step 1: Write the dissociation equation for `BaCl₂`
When `BaCl₂` dissolves in water, it dissociates into its ions:
\[ BaCl_2 (s) \rightleftharpoons Ba^{2+} (aq) + 2Cl^{-} (aq) \]

### Step 2: Define the solubility product (Ksp)
The solubility product constant (Ksp) for the dissociation of `BaCl₂` can be expressed as:
\[ Ksp = [Ba^{2+}][Cl^{-}]^2 \]

### Step 3: Assign variables for ion concentrations
Let the solubility of `BaCl₂` be \( x \) mol/L. Then, the concentrations of the ions at equilibrium will be:
- For \( Ba^{2+} \): \( [Ba^{2+}] = x \)
- For \( Cl^{-} \): Since there are 2 moles of \( Cl^{-} \) for every mole of \( BaCl₂ \), \( [Cl^{-}] = 2x \)

### Step 4: Substitute the ion concentrations into the Ksp expression
Substituting these values into the Ksp expression gives:
\[ Ksp = [Ba^{2+}][Cl^{-}]^2 = (x)(2x)^2 \]
\[ Ksp = x(4x^2) = 4x^3 \]

### Step 5: Set Ksp equal to the given value
We know that the Ksp of `BaCl₂` is \( 3.2 \times 10^{-9} \):
\[ 4x^3 = 3.2 \times 10^{-9} \]

### Step 6: Solve for \( x \)
To find \( x \), we rearrange the equation:
\[ x^3 = \frac{3.2 \times 10^{-9}}{4} \]
\[ x^3 = 0.8 \times 10^{-9} \]
Taking the cube root of both sides:
\[ x = \sqrt[3]{0.8 \times 10^{-9}} \]

Calculating this gives:
\[ x \approx 1 \times 10^{-3} \text{ mol/L} \]

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

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