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 `BaaCl_(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 find the solubility product (Ksp) of \( Ba_3(PO_4)_2 \), we will follow these steps:

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

The limiting equivalent conductivity (\( \Lambda^0 \)) of a compound can be calculated using the contributions from its ions. The formula is:

\[ \Lambda^0_{Ba_3(PO_4)_2} = 3 \Lambda^0_{BaCl_2} + 2 \Lambda^0_{K_3PO_4} - 6 \Lambda^0_{KCl} \]

Where:
- \( \Lambda^0_{BaCl_2} = 160 \, \Omega^{-1} cm^2 eq^{-1} \)
- \( \Lambda^0_{K_3PO_4} = 140 \, \Omega^{-1} cm^2 eq^{-1} \)
- \( \Lambda^0_{KCl} = 100 \, \Omega^{-1} cm^2 eq^{-1} \)

Calculating each part:
- Contribution from \( BaCl_2 \): \( 3 \times 160 = 480 \)
- Contribution from \( K_3PO_4 \): \( 2 \times 140 = 280 \)
- Contribution from \( KCl \): \( 6 \times 100 = 600 \)

Now, substituting these values into the equation:

\[ \Lambda^0_{Ba_3(PO_4)_2} = 480 + 280 - 600 = 160 \, \Omega^{-1} cm^2 eq^{-1} \]

### Step 2: Relate Conductivity to Solubility

The conductivity (\( K \)) of the saturated solution is given as \( 1.2 \times 10^{-5} \, \Omega^{-1} cm^{-1} \). The relationship between conductivity, limiting equivalent conductivity, and solubility (\( s \)) is given by:

\[ K = \Lambda^0_{Ba_3(PO_4)_2} \times s \]

Rearranging for \( s \):

\[ s = \frac{K}{\Lambda^0_{Ba_3(PO_4)_2}} \]

Substituting the known values:

\[ s = \frac{1.2 \times 10^{-5}}{160} \]

Calculating \( s \):

\[ s = 7.5 \times 10^{-8} \, eq/L \]

### Step 3: Calculate the Solubility Product (Ksp)

The solubility product \( Ksp \) for \( Ba_3(PO_4)_2 \) can be expressed as:

\[ Ksp = [Ba^{2+}]^3 [PO_4^{3-}]^2 \]

From the dissociation of \( Ba_3(PO_4)_2 \):

- \( [Ba^{2+}] = 3s \)
- \( [PO_4^{3-}] = 2s \)

Substituting these into the Ksp expression:

\[ Ksp = (3s)^3 (2s)^2 = 27s^3 \times 4s^2 = 108s^5 \]

Now substituting the value of \( s \):

\[ Ksp = 108 \times (7.5 \times 10^{-8})^5 \]

Calculating \( Ksp \):

\[ Ksp = 108 \times 7.59375 \times 10^{-40} \approx 8.2 \times 10^{-38} \]

### Final Answer

Thus, the solubility product \( Ksp \) of \( Ba_3(PO_4)_2 \) is approximately:

\[ Ksp \approx 1.08 \times 10^{-25} \]