If the solubility of `Ag_(2)SO_(4)` in `10^(-2)M Na_(2)SO_(4)` solution be `2xx10^(-8)M` then `K_(sp)` of `Ag_(2)SO_(4)` will be

Options

A. `32xx10^(-24)`
B. `16xx10^(-18)`
C. `32xx10^(-18)`
D. `16xx10^(-24)`

Explanation

To find the solubility product \( K_{sp} \) of \( Ag_2SO_4 \) in a \( 10^{-2} M \) \( Na_2SO_4 \) solution, we can follow these steps:

### Step 1: Understand the dissociation of \( Ag_2SO_4 \)
The dissociation of silver sulfate in water can be represented as:
\[ Ag_2SO_4 (s) \rightleftharpoons 2Ag^+ (aq) + SO_4^{2-} (aq) \]
From this equation, we see that 1 mole of \( Ag_2SO_4 \) produces 2 moles of \( Ag^+ \) ions and 1 mole of \( SO_4^{2-} \) ions.

### Step 2: Define solubility in the presence of \( Na_2SO_4 \)
Let the solubility of \( Ag_2SO_4 \) in \( 10^{-2} M \) \( Na_2SO_4 \) be \( S \). According to the problem, \( S = 2 \times 10^{-8} M \).

### Step 3: Calculate the concentration of ions
In the presence of \( Na_2SO_4 \), which dissociates completely to give \( Na^+ \) and \( SO_4^{2-} \):
\[ Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-} \]
The concentration of \( SO_4^{2-} \) from \( Na_2SO_4 \) is \( 10^{-2} M \).

When \( Ag_2SO_4 \) dissolves, it contributes additional \( SO_4^{2-} \) ions, but since \( S \) is very small compared to \( 10^{-2} M \), we can approximate the total concentration of \( SO_4^{2-} \) as:
\[ [SO_4^{2-}]_{total} = 10^{-2} M + S \approx 10^{-2} M \]

### Step 4: Write the expression for \( K_{sp} \)
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [Ag^+]^2 [SO_4^{2-}] \]
From the dissociation, we know:
- \( [Ag^+] = 2S \)
- \( [SO_4^{2-}] = 10^{-2} M \)

### Step 5: Substitute the values into the \( K_{sp} \) expression
Substituting \( S = 2 \times 10^{-8} M \):
\[ [Ag^+] = 2S = 2 \times (2 \times 10^{-8}) = 4 \times 10^{-8} M \]
Now substituting into the \( K_{sp} \) expression:
\[ K_{sp} = (4 \times 10^{-8})^2 \times (10^{-2}) \]
Calculating this gives:
\[ K_{sp} = 16 \times 10^{-16} \times 10^{-2} = 16 \times 10^{-18} \]

### Final Answer
Thus, the solubility product \( K_{sp} \) of \( Ag_2SO_4 \) is:
\[ K_{sp} = 1.6 \times 10^{-17} \]