The specific conductance of a saturated solution of silver bromide is `kScm^(-1)` . The limiting ionic conductivity of `Ag^+` and `Br^-` ions are x and y respectively. The solubility of silver bromide in `gL^(-1)` is : (molar mass of AgBr=188)

Options

A. `(kxx1000)/(x-y)`
B. `k/(x+y)xx188`
C. `(kxx1000xx188)/(x+y)`
D. `(x+y)/(k)xx(1000)/(188)`

Answer

To solve the question regarding the solubility of silver bromide (AgBr) in g/L based on its specific conductance and limiting ionic conductivities, we can follow these steps:

### Step 1: Understand the relationship between conductivity, solubility, and ionic conductivities.

The specific conductance (k) of a solution is related to the concentration of ions and their respective ionic conductivities (λ) as follows:

\[ k = c \cdot (λ_{Ag^+} + λ_{Br^-}) \]

where:
- \( k \) = specific conductance of the solution (in S/cm)
- \( c \) = concentration of the solution (in mol/L)
- \( λ_{Ag^+} \) = limiting ionic conductivity of \( Ag^+ \) (in S cm²/mol)
- \( λ_{Br^-} \) = limiting ionic conductivity of \( Br^- \) (in S cm²/mol)

### Step 2: Calculate the concentration of the saturated solution.

Given that the molar mass of AgBr is 188 g/mol, we can express the concentration in terms of solubility (s) in g/L:

\[ c = \frac{s}{M} \]

where:
- \( s \) = solubility of AgBr in g/L
- \( M \) = molar mass of AgBr (188 g/mol)

Substituting this into the equation for specific conductance gives:

\[ k = \frac{s}{188} \cdot (λ_{Ag^+} + λ_{Br^-}) \]

### Step 3: Rearrange the equation to find solubility.

Rearranging the equation to solve for solubility (s):

\[ s = k \cdot 188 \cdot \frac{1}{(λ_{Ag^+} + λ_{Br^-})} \]

### Step 4: Substitute the known values.

If we know the specific conductance \( k \) and the values of \( λ_{Ag^+} \) and \( λ_{Br^-} \), we can substitute them into the equation.

Assuming:
- \( k \) = specific conductance (in S/cm)
- \( λ_{Ag^+} = x \) (in S cm²/mol)
- \( λ_{Br^-} = y \) (in S cm²/mol)

Then, we can express the solubility as:

\[ s = k \cdot 188 \cdot \frac{1}{(x + y)} \]

### Step 5: Final expression for solubility.

Thus, the final expression for the solubility of silver bromide in g/L is:

\[ s = \frac{188k}{(x + y)} \]