The conductivity of saturated solution of `Ba_(3) (PO_(4))_(2)` is `1.2 xx 10^(-5) Omega^(-1) cm^(-1)`. The limiting equivalent conductivities of `BaCl_(2), K_(3)PO_(4)` and KCl are `160, 140` and `100 Omega^(-1) cm^(2)eq^(-1)`, respectively. The solubility product of `Ba_(3)(PO_(4))_(2)` is

Options

A. `10^(-5)`
B. `1.08 xx 10^(-23)`
C. `1.08 xx 10^(-25)`
D. `1.08 xx 10^(-27)`

Explanation

To solve the problem, we need to find the solubility product (Ksp) of barium phosphate, \( Ba_3(PO_4)_2 \), using the given conductivity data and the limiting equivalent conductivities of the relevant salts.

### Step 1: Calculate the Limiting Equivalent Conductivity of \( Ba_3(PO_4)_2 \)

The limiting equivalent conductivity (\( \Lambda^0_m \)) of \( Ba_3(PO_4)_2 \) can be calculated using the limiting equivalent conductivities of its constituent ions from the salts \( BaCl_2 \), \( K_3PO_4 \), and \( KCl \).

1. **Calculate \( \Lambda^0_m \) for each salt:**
- For \( BaCl_2 \):
\[ \Lambda^0_{BaCl_2} = 160 \, \Omega^{-1} cm^2 eq^{-1} \times 2 = 320 \, \Omega^{-1} cm^2 eq^{-1} \]
- For \( K_3PO_4 \):
\[ \Lambda^0_{K_3PO_4} = 140 \, \Omega^{-1} cm^2 eq^{-1} \times 3 = 420 \, \Omega^{-1} cm^2 eq^{-1} \]
- For \( KCl \):
\[ \Lambda^0_{KCl} = 100 \, \Omega^{-1} cm^2 eq^{-1} \times 1 = 100 \, \Omega^{-1} cm^2 eq^{-1} \]

### Step 2: Combine the Conductivities

Using the contributions of the ions, we can express the limiting equivalent conductivity of \( Ba_3(PO_4)_2 \):
\[ \Lambda^0_{Ba_3(PO_4)_2} = 3 \times \Lambda^0_{BaCl_2} + 2 \times \Lambda^0_{K_3PO_4} - 6 \times \Lambda^0_{KCl} \]
Substituting the values:
\[ \Lambda^0_{Ba_3(PO_4)_2} = 3 \times 320 + 2 \times 420 - 6 \times 100 \]
Calculating:
\[ = 960 + 840 - 600 = 1200 \, \Omega^{-1} cm^2 eq^{-1} \]

### Step 3: Calculate the Solubility

Using the formula for the solubility (\( s \)):
\[ s = \frac{k \times 1000}{\Lambda^0_m} \]
Where \( k \) is the conductivity of the saturated solution:
\[ k = 1.2 \times 10^{-5} \, \Omega^{-1} cm^{-1} \]
Substituting the values:
\[ s = \frac{1.2 \times 10^{-5} \times 1000}{1200} \]
Calculating:
\[ s = \frac{1.2 \times 10^{-2}}{1200} = 1 \times 10^{-5} \, mol/L \]

### Step 4: Calculate the Solubility Product (Ksp)

The solubility product \( Ksp \) for \( Ba_3(PO_4)_2 \) can be expressed as:
\[ Ksp = [Ba^{2+}]^3 [PO_4^{3-}]^2 \]
Given that the dissociation of \( Ba_3(PO_4)_2 \) produces 3 moles of \( Ba^{2+} \) and 2 moles of \( PO_4^{3-} \):
\[ Ksp = (3s)^3 (2s)^2 \]
Substituting \( s = 1 \times 10^{-5} \):
\[ Ksp = (3 \times 1 \times 10^{-5})^3 (2 \times 1 \times 10^{-5})^2 \]
Calculating:
\[ = (9 \times 10^{-15}) \times (4 \times 10^{-10}) = 36 \times 10^{-25} = 3.6 \times 10^{-24} \]

### Final Answer

Thus, the solubility product \( Ksp \) of \( Ba_3(PO_4)_2 \) is:
\[ Ksp = 1.08 \times 10^{-23} \]