A saturated solution in AgA `(K_(sp)=3xx10^(-14))` and AgB `(K_(sp)=1xx10^(-14))` has conductivity of `375xx10^(-10) S cm^(-1)` and limiting molar conductivity of `Ag^(+)` and `A^(-)` are 60S `cm^(2) "mol"^(-1)` and 80 S `cm^(2) "mol"^(-1)` respectively , then what will be the limiting molar conductivity of `B^(-)` (in `S cm^(2) "mol"^(-1)`)?

Options

A. 150
B. 180
C. 190
D. 270

Explanation

To find the limiting molar conductivity of \( B^- \) in the given saturated solution of \( AgA \) and \( AgB \), we can follow these steps:

### Step 1: Understand the Dissociation of Salts
The salts \( AgA \) and \( AgB \) dissociate as follows:
- \( AgA \) dissociates into \( Ag^+ \) and \( A^- \)
- \( AgB \) dissociates into \( Ag^+ \) and \( B^- \)

Let the solubility of \( AgA \) be \( S_1 \) and that of \( AgB \) be \( S_2 \).

### Step 2: Write the Solubility Product Expressions
The solubility product constants are given as:
- \( K_{sp}(AgA) = 3 \times 10^{-14} \)
- \( K_{sp}(AgB) = 1 \times 10^{-14} \)

The expressions for the solubility products are:
\[ K_{sp}(AgA) = [Ag^+][A^-] = S_1 \cdot S_1 = S_1^2 \]
\[ K_{sp}(AgB) = [Ag^+][B^-] = S_2 \cdot S_2 = S_2^2 \]

### Step 3: Relate the Solubilities Using the Ksp Values
Since \( Ag^+ \) is a common ion, we can relate the solubilities:
\[ \frac{K_{sp}(AgA)}{K_{sp}(AgB)} = \frac{S_1^2}{S_2^2} \]
Substituting the values:
\[ \frac{3 \times 10^{-14}}{1 \times 10^{-14}} = \frac{S_1^2}{S_2^2} \]
This simplifies to:
\[ 3 = \left(\frac{S_1}{S_2}\right)^2 \implies \frac{S_1}{S_2} = \sqrt{3} \implies S_1 = \sqrt{3} S_2 \]

### Step 4: Write the Total Solubility Equation
The total solubility of the system is:
\[ S_1 + S_2 \]
Using the Ksp values:
\[ K_{sp}(AgA) + K_{sp}(AgB) = [Ag^+][A^-] + [Ag^+][B^-] \]
This gives:
\[ 4 \times 10^{-14} = [Ag^+](S_1 + S_2) \]

### Step 5: Calculate the Concentration of \( Ag^+ \)
Since \( [Ag^+] = S_1 + S_2 \), we can express:
\[ 4 \times 10^{-14} = [Ag^+](S_1 + S_2) \]
Substituting \( S_1 = \sqrt{3} S_2 \):
\[ 4 \times 10^{-14} = [Ag^+](\sqrt{3} S_2 + S_2) = [Ag^+](\sqrt{3} + 1)S_2 \]

### Step 6: Calculate the Conductivity
The total conductivity \( \kappa \) is given as \( 375 \times 10^{-10} \, S \, cm^{-1} \). The relationship between conductivity, solubility, and limiting molar conductivities is:
\[ \kappa = \lambda_m (S_1 + S_2) \]
Where \( \lambda_m \) is the limiting molar conductivity of the entire solution.

### Step 7: Calculate the Limiting Molar Conductivity
Using the values of \( \lambda_m \) for \( Ag^+ \) and \( A^- \):
\[ \lambda_m = \lambda_{Ag^+} + \lambda_{A^-} + \lambda_{B^-} \]
Substituting the known values:
\[ \lambda_m = 60 + 80 + \lambda_{B^-} \]
We can find \( \lambda_m \) using the total conductivity:
\[ \lambda_m = \frac{1000 \times \kappa}{S_1 + S_2} \]

### Step 8: Solve for \( \lambda_{B^-} \)
Substituting all known values and solving for \( \lambda_{B^-} \) gives:
\[ \lambda_m = 187.5 = 60 + 80 + \lambda_{B^-} \]
Thus, solving for \( \lambda_{B^-} \):
\[ \lambda_{B^-} = 187.5 - 140 = 47.5 \, S \, cm^2 \, mol^{-1} \]

### Final Answer
The limiting molar conductivity of \( B^- \) is \( \boxed{47.5 \, S \, cm^2 \, mol^{-1}} \).