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 \( \text{Ag}_3\text{PO}_4 \), we will follow these steps:

### Step 1: Understand the given data
- Conductivity of saturated solution, \( \kappa = 9 \times 10^{-6} \, \text{S m}^{-1} \)
- Equivalent conductivity, \( \Lambda_e = 1.50 \times 10^{-4} \, \text{S m}^2 \, \text{equivalent}^{-1} \)

### Step 2: Calculate the molar conductivity
The molar conductivity (\( \Lambda_m \)) can be calculated using the formula:
\[ \Lambda_m = \Lambda_e \times n \]
where \( n \) is the number of equivalents. For \( \text{Ag}_3\text{PO}_4 \), it dissociates into \( 3 \, \text{Ag}^+ \) and \( \text{PO}_4^{3-} \), giving a total of 4 ions. Thus, \( n = 4 \).

Calculating \( \Lambda_m \):
\[ \Lambda_m = 1.50 \times 10^{-4} \, \text{S m}^2 \, \text{equivalent}^{-1} \times 4 = 6.00 \times 10^{-4} \, \text{S m}^2 \, \text{mol}^{-1} \]

### Step 3: Relate molar conductivity to conductivity and solubility
The relationship between molar conductivity (\( \Lambda_m \)), conductivity (\( \kappa \)), and solubility (\( s \)) is given by:
\[ \Lambda_m = \frac{\kappa}{s} \]
Rearranging gives:
\[ s = \frac{\kappa}{\Lambda_m} \]

### Step 4: Substitute the values to find solubility
Substituting the known values:
\[ s = \frac{9 \times 10^{-6} \, \text{S m}^{-1}}{6.00 \times 10^{-4} \, \text{S m}^2 \, \text{mol}^{-1}} = 1.5 \times 10^{-2} \, \text{mol m}^{-3} \]

### Step 5: Write the expression for Ksp
The dissociation of \( \text{Ag}_3\text{PO}_4 \) can be represented as:
\[ \text{Ag}_3\text{PO}_4 (s) \rightleftharpoons 3 \, \text{Ag}^+ (aq) + \text{PO}_4^{3-} (aq) \]
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Ag}^+]^3 [\text{PO}_4^{3-}] \]
If \( s \) is the solubility, then:
\[ [\text{Ag}^+] = 3s \quad \text{and} \quad [\text{PO}_4^{3-}] = s \]
Thus,
\[ K_{sp} = (3s)^3 \cdot s = 27s^4 \]

### Step 6: Substitute the value of solubility into Ksp
Substituting \( s = 1.5 \times 10^{-2} \):
\[ K_{sp} = 27 \cdot (1.5 \times 10^{-2})^4 \]
Calculating \( (1.5 \times 10^{-2})^4 \):
\[ (1.5 \times 10^{-2})^4 = 5.0625 \times 10^{-8} \]
Thus,
\[ K_{sp} = 27 \cdot 5.0625 \times 10^{-8} = 1.365 \times 10^{-6} \]

### Final Answer
The \( K_{sp} \) of \( \text{Ag}_3\text{PO}_4 \) is approximately:
\[ K_{sp} = 1.365 \times 10^{-6} \]

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