Concentration of the `Ag^(+) `ions in a saturated solution of `Ag_2C_2O_4` is ` 2.2 xx 10 ^(-4) " mol " L^(-1) ` , Solubility product of `Ag_2C_2O_4` is
(a)` 2.42 xx 10^(-8)`
(b)` 2.66 xx 10^(-12)`
(c)` 4.5 xx 10^(-11)`
(d)` 5.3 xx 10 ^(-12)`

Options

A. ` 2.42 xx 10^(-8)`
B. ` 2.66 xx 10^(-12)`
C. ` 4.5 xx 10^(-11)`
D. ` 5.3 xx 10 ^(-12)`

Answer

To find the solubility product (Ksp) of silver oxalate (Ag2C2O4), we start by analyzing the dissociation of the compound in water.

### Step 1: Write the dissociation equation
The dissociation of silver oxalate can be represented as:
\[ \text{Ag}_2\text{C}_2\text{O}_4 (s) \rightleftharpoons 2 \text{Ag}^+ (aq) + \text{C}_2\text{O}_4^{2-} (aq) \]

### Step 2: Define the concentrations
Let the solubility of Ag2C2O4 be \( S \) mol/L. From the dissociation equation:
- The concentration of \( \text{Ag}^+ \) ions will be \( 2S \).
- The concentration of \( \text{C}_2\text{O}_4^{2-} \) ions will be \( S \).

### Step 3: Substitute the given concentration
We are given that the concentration of \( \text{Ag}^+ \) ions in a saturated solution is:
\[ [\text{Ag}^+] = 2.2 \times 10^{-4} \, \text{mol/L} \]

From our earlier definition, we have:
\[ 2S = 2.2 \times 10^{-4} \]
Thus, we can find \( S \):
\[ S = \frac{2.2 \times 10^{-4}}{2} = 1.1 \times 10^{-4} \, \text{mol/L} \]

### Step 4: Calculate the solubility product (Ksp)
The solubility product \( K_{sp} \) is given by the expression:
\[ K_{sp} = [\text{Ag}^+]^2 \times [\text{C}_2\text{O}_4^{2-}] \]

Substituting the values we found:
- \( [\text{Ag}^+] = 2.2 \times 10^{-4} \)
- \( [\text{C}_2\text{O}_4^{2-}] = 1.1 \times 10^{-4} \)

Now plug these into the Ksp expression:
\[ K_{sp} = (2.2 \times 10^{-4})^2 \times (1.1 \times 10^{-4}) \]

### Step 5: Perform the calculations
Calculating \( (2.2 \times 10^{-4})^2 \):
\[ (2.2 \times 10^{-4})^2 = 4.84 \times 10^{-8} \]

Now multiply this by \( 1.1 \times 10^{-4} \):
\[ K_{sp} = 4.84 \times 10^{-8} \times 1.1 \times 10^{-4} \]
\[ K_{sp} = 5.324 \times 10^{-12} \]

### Step 6: Round the answer
Rounding this value gives us:
\[ K_{sp} \approx 5.3 \times 10^{-12} \]

### Conclusion
The solubility product of \( \text{Ag}_2\text{C}_2\text{O}_4 \) is:
**(d) \( 5.3 \times 10^{-12} \)**

---