The conductivity of a saturated solution of a sparingly soluble salt `MX_2` is found to be `4 xx 10^(-5) Omega^(-1) cm^(-1)." If "lambda_(m)^(oo) ( M^(2+))=50 Omega^(-1) cm^(2) mol^(-1) and lambda^(oo) (X^(-))= 50 Omega^(-1) cm^(2) mol^(-1)`, the solubility product of the salt is about

Options

A. `2 xx 10^(-10)`
B. `2 xx 10^(-12)`
C. `8 xx 10^(-12)`
D. `8 xx 10^(-11)`

Answer

To find the solubility product (Ksp) of the sparingly soluble salt \( MX_2 \), we can follow these steps:

### Step 1: Write the dissociation equation
The salt \( MX_2 \) dissociates in water as follows:
\[ MX_2 \rightleftharpoons M^{2+} + 2X^{-} \]

### Step 2: Determine the molar conductivity at infinite dilution
The molar conductivity at infinite dilution (\( \lambda_m^{\infty} \)) for the salt is given by:
\[ \lambda_m^{\infty} = \lambda_m^{\infty}(M^{2+}) + 2 \cdot \lambda_m^{\infty}(X^{-}) \]
Substituting the given values:
\[ \lambda_m^{\infty} = 50 \, \Omega^{-1} \, cm^2 \, mol^{-1} + 2 \cdot 50 \, \Omega^{-1} \, cm^2 \, mol^{-1} = 50 + 100 = 150 \, \Omega^{-1} \, cm^2 \, mol^{-1} \]

### Step 3: Relate conductivity to molar conductivity
The relationship between conductivity (\( K \)) and molar conductivity (\( \lambda_m \)) is given by:
\[ \lambda_m = \frac{K \cdot 1000}{s} \]
Where \( s \) is the solubility in mol/L.

### Step 4: Rearrange the equation to find solubility
Rearranging the equation for solubility:
\[ s = \frac{K \cdot 1000}{\lambda_m^{\infty}} \]
Substituting the given conductivity \( K = 4 \times 10^{-5} \, \Omega^{-1} \, cm^{-1} \) and \( \lambda_m^{\infty} = 150 \, \Omega^{-1} \, cm^2 \, mol^{-1} \):
\[ s = \frac{4 \times 10^{-5} \cdot 1000}{150} = \frac{4 \times 10^{-2}}{150} = 2.67 \times 10^{-4} \, mol/L \]

### Step 5: Write the expression for the solubility product
The solubility product \( K_{sp} \) for the dissociation can be expressed as:
\[ K_{sp} = [M^{2+}][X^{-}]^2 \]
From the dissociation, the concentration of \( M^{2+} \) is \( s \) and the concentration of \( X^{-} \) is \( 2s \):
\[ K_{sp} = s \cdot (2s)^2 = s \cdot 4s^2 = 4s^3 \]

### Step 6: Substitute the value of solubility into the Ksp expression
Substituting \( s = 2.67 \times 10^{-4} \):
\[ K_{sp} = 4 \cdot (2.67 \times 10^{-4})^3 \]
Calculating \( (2.67 \times 10^{-4})^3 \):
\[ (2.67 \times 10^{-4})^3 = 1.90 \times 10^{-11} \]
Thus,
\[ K_{sp} = 4 \cdot 1.90 \times 10^{-11} = 7.6 \times 10^{-11} \]

### Step 7: Round to appropriate significant figures
Rounding gives:
\[ K_{sp} \approx 8 \times 10^{-11} \]

### Final Answer
The solubility product \( K_{sp} \) of the salt \( MX_2 \) is approximately \( 8 \times 10^{-11} \).

---