The concentration of `Ag^(o+)` ions in a saturated solution of `Ag_(2)C_(2)O_(4)` is `2.0 xx 10^(-4)M`. Calculate the solubility of `Ag_(2)C_(2)O_(4)` in a solution which is `0.01M` in `H_(2)C_(2)O_(4)`.

Answer

To solve the problem, we need to calculate the solubility of `Ag2C2O4` in a solution that contains `0.01 M` of `H2C2O4`. Let's go through the solution step by step.

### Step 1: Understand the Dissociation of `Ag2C2O4`
The dissociation of `Ag2C2O4` can be represented as:
\[ Ag_2C_2O_4 (s) \rightleftharpoons 2 Ag^+ (aq) + C_2O_4^{2-} (aq) \]
From this equation, we can see that for every mole of `Ag2C2O4` that dissolves, it produces 2 moles of `Ag^+` ions and 1 mole of `C2O4^{2-}` ions.

### Step 2: Define Solubility
Let the solubility of `Ag2C2O4` in the solution be \( S \). Therefore, the concentrations of the ions at equilibrium will be:
- \([Ag^+] = 2S\)
- \([C_2O_4^{2-}] = S\)

### Step 3: Use Given Information
We are given that the concentration of `Ag^+` ions in a saturated solution of `Ag2C2O4` is \( 2.0 \times 10^{-4} \, M \). This means:
\[ 2S = 2.0 \times 10^{-4} \]
From this, we can find \( S \):
\[ S = \frac{2.0 \times 10^{-4}}{2} = 1.0 \times 10^{-4} \, M \]

### Step 4: Calculate the Solubility Product (Ksp)
The solubility product \( K_{sp} \) for `Ag2C2O4` can be calculated using the concentrations at saturation:
\[ K_{sp} = [Ag^+]^2[C_2O_4^{2-}] \]
Substituting the values:
\[ K_{sp} = (2.0 \times 10^{-4})^2 \times (1.0 \times 10^{-4}) \]
Calculating this gives:
\[ K_{sp} = 4.0 \times 10^{-12} \]

### Step 5: Consider the Effect of `H2C2O4`
When we add `0.01 M` of `H2C2O4`, it partially dissociates to produce `C2O4^{2-}` ions. The concentration of `C2O4^{2-}` from `H2C2O4` will be approximately `0.01 M`. Therefore, the total concentration of `C2O4^{2-}` in the solution will be:
\[ [C_2O_4^{2-}] = S + 0.01 \approx 0.01 \, M \]
since \( S \) is very small compared to `0.01 M`.

### Step 6: Set Up the Ksp Expression in the New Solution
Now, we can set up the \( K_{sp} \) expression again:
\[ K_{sp} = [Ag^+]^2[C_2O_4^{2-}] \]
Let the new solubility of `Ag2C2O4` in the presence of `H2C2O4` be \( S' \):
- \([Ag^+] = 2S'\)
- \([C_2O_4^{2-}] \approx 0.01\)

Thus, we have:
\[ 4.0 \times 10^{-12} = (2S')^2 \times 0.01 \]

### Step 7: Solve for \( S' \)
Rearranging gives:
\[ 4.0 \times 10^{-12} = 4S'^2 \times 0.01 \]
\[ 4.0 \times 10^{-12} = 4S'^2 \times 0.01 \]
\[ S'^2 = \frac{4.0 \times 10^{-12}}{0.04} = 1.0 \times 10^{-10} \]
Taking the square root:
\[ S' = \sqrt{1.0 \times 10^{-10}} = 1.0 \times 10^{-5} \, M \]

### Final Answer
The solubility of `Ag2C2O4` in a solution that is `0.01 M` in `H2C2O4` is:
\[ S' = 1.0 \times 10^{-5} \, M \]