The conductivity of a saturated solution of `Ag_(3)PO_(4)` is `9 xx 10^(-6) S m^(-1)` and its equivalent conductivity is `1.50 xx 10^(-4) Sm^(2) "equivalent"^(-1)`. Th `K_(sp)` of `Ag_(3)PO_(4)` is:-

Options

A. `4.32 xx 10^(-18)`
B. `1.8 xx 10^(-9)`
C. `8.64 xx 10^(-13)`
D. None of these

Explanation

To find the solubility product constant (Ksp) of Ag₃PO₄, we can follow these steps:

### Step 1: Understand the relationship between conductivity, equivalent conductivity, and normality
The equivalent conductivity (Λ) is related to the conductivity (κ) and normality (N) by the formula:
\[ Λ = \frac{κ \times 1000}{N} \]
where:
- κ is the conductivity in S/m,
- N is the normality in equivalents/L.

### Step 2: Rearrange the formula to find normality
From the formula, we can rearrange it to find normality (N):
\[ N = \frac{κ \times 1000}{Λ} \]

### Step 3: Substitute the given values
Given:
- κ = \(9 \times 10^{-6} \, \text{S/m}\)
- Λ = \(1.50 \times 10^{-4} \, \text{S m}^2 \text{equivalent}^{-1}\)

Substituting these values into the equation:
\[ N = \frac{(9 \times 10^{-6}) \times 1000}{1.50 \times 10^{-4}} = \frac{9 \times 10^{-3}}{1.50 \times 10^{-4}} = 6 \times 10^{-5} \, \text{equivalents/L} \]

### Step 4: Convert normality to molarity
For Ag₃PO₄, the number of ions produced is 3 (3 Ag⁺ ions). Therefore, the molarity (S) can be calculated as:
\[ \text{Molarity (S)} = \frac{N}{n} = \frac{6 \times 10^{-5}}{3} = 2 \times 10^{-5} \, \text{mol/L} \]

### Step 5: Write the dissociation equation
The dissociation of Ag₃PO₄ in water can be represented as:
\[ \text{Ag}_3\text{PO}_4 (s) \rightleftharpoons 3 \text{Ag}^+ (aq) + \text{PO}_4^{3-} (aq) \]
Let the solubility of Ag₃PO₄ be S. Then:
- Concentration of Ag⁺ = 3S
- Concentration of PO₄³⁻ = S

### Step 6: Write the expression for Ksp
The solubility product (Ksp) can be expressed as:
\[ K_{sp} = [\text{Ag}^+]^3 [\text{PO}_4^{3-}] \]
Substituting the concentrations:
\[ K_{sp} = (3S)^3 \times S = 27S^4 \]

### Step 7: Substitute the value of S
Now substituting S = \(2 \times 10^{-5}\):
\[ K_{sp} = 27 \times (2 \times 10^{-5})^4 \]
Calculating \( (2 \times 10^{-5})^4 \):
\[ (2 \times 10^{-5})^4 = 16 \times 10^{-20} = 1.6 \times 10^{-19} \]
Thus:
\[ K_{sp} = 27 \times 1.6 \times 10^{-19} = 4.32 \times 10^{-18} \]

### Final Answer
The solubility product constant \(K_{sp}\) of Ag₃PO₄ is:
\[ K_{sp} = 4.32 \times 10^{-18} \]

---