The molar conductivity of a saturated solution of `BaCl_(2)` is `10^(-12) ohm^(-1) cm^(2) mol^(-1)`. If the value for specific conductivity of the solution is `10^(-14) ohm^(-1) cm^(-1)`, the value for `K_(sp)` for `BaCl_(2)` is

Options

A. `10^(3) xx 4`
B. `10^(2)`
C. 10
D. `10^(5)`

Explanation

To find the solubility product constant (Ksp) for BaCl₂ given the molar conductivity (Λm) and specific conductivity (κ) of a saturated solution, we can follow these steps:

### Step 1: Write the dissociation equation for BaCl₂
BaCl₂ dissociates in water as follows:
\[ \text{BaCl}_2 (s) \rightarrow \text{Ba}^{2+} (aq) + 2 \text{Cl}^- (aq) \]

### Step 2: Define solubility
Let the solubility of BaCl₂ be \( S \) mol/L. Therefore, at equilibrium:
- The concentration of \(\text{Ba}^{2+}\) will be \( S \) mol/L.
- The concentration of \(\text{Cl}^-\) will be \( 2S \) mol/L.

### Step 3: Write the expression for Ksp
The expression for Ksp can be written as:
\[ K_{sp} = [\text{Ba}^{2+}][\text{Cl}^-]^2 \]
Substituting the concentrations:
\[ K_{sp} = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \]

### Step 4: Relate molar conductivity and specific conductivity
The relationship between molar conductivity (Λm) and specific conductivity (κ) is given by:
\[ \Lambda_m = \frac{\kappa \times 1000}{S} \]
Where:
- \( \Lambda_m = 10^{-12} \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \)
- \( \kappa = 10^{-14} \, \Omega^{-1} \, \text{cm}^{-1} \)

### Step 5: Solve for S
Rearranging the equation to find \( S \):
\[ S = \frac{\kappa \times 1000}{\Lambda_m} \]
Substituting the known values:
\[ S = \frac{10^{-14} \times 1000}{10^{-12}} \]
\[ S = \frac{10^{-11}}{10^{-12}} = 10 \, \text{mol/L} \]

### Step 6: Substitute S into the Ksp expression
Now substitute \( S = 10 \) mol/L into the Ksp expression:
\[ K_{sp} = 4S^3 = 4 \times (10)^3 = 4 \times 1000 = 4000 \]

### Final Answer
Thus, the value of \( K_{sp} \) for BaCl₂ is:
\[ K_{sp} = 4000 \]

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