The `K_(sp)` of `Ag_(2)CrO_(4)` is `1.1xx10^(-12)` at `298K`. The solubility (in mol/L) of `Ag_(2)CrO_(4)` in a `0.1 M AgNO_(3)` solution is

Options

A. `1.1xx10^(-11)`
B. `1.1xx10^(-10)`
C. `1.1xx10^(-12)`
D. `1.1xx10^(-9)`

Answer

To solve the problem, we need to determine the solubility of `Ag2CrO4` in a `0.1 M AgNO3` solution, given that the `Ksp` of `Ag2CrO4` is `1.1 x 10^(-12)` at `298 K`. The presence of `AgNO3` introduces a common ion effect due to the `Ag^+` ions.

### Step-by-Step Solution:

1. **Write the Dissociation Reaction:**
The dissociation of `Ag2CrO4` in water can be represented as:
\[ Ag_2CrO_4 (s) \rightleftharpoons 2Ag^+ (aq) + CrO_4^{2-} (aq) \]

2. **Define the Solubility Product (Ksp):**
The solubility product expression for `Ag2CrO4` is given by:
\[ K_{sp} = [Ag^+]^2 [CrO_4^{2-}] \]
Given that `Ksp = 1.1 x 10^{-12}`.

3. **Set Up the Initial Concentrations:**
In a `0.1 M AgNO3` solution, the concentration of `Ag^+` ions is `0.1 M`. The initial concentration of `CrO_4^{2-}` ions is `0 M`.

4. **Define the Change in Concentration:**
Let the solubility of `Ag2CrO4` in the `0.1 M AgNO3` solution be `S`. At equilibrium:
- The concentration of `Ag^+` will be `0.1 + 2S` (since 2 moles of `Ag^+` are produced for every mole of `Ag2CrO4` that dissolves).
- The concentration of `CrO_4^{2-}` will be `S`.

5. **Substitute into the Ksp Expression:**
Substitute the equilibrium concentrations into the Ksp expression:
\[ K_{sp} = (0.1 + 2S)^2 (S) \]
Since `0.1` is much larger than `2S`, we can approximate:
\[ K_{sp} \approx (0.1)^2 (S) \]

6. **Set Up the Equation:**
Now, substituting the value of `Ksp`:
\[ 1.1 x 10^{-12} = (0.1)^2 (S) \]
\[ 1.1 x 10^{-12} = 0.01 S \]

7. **Solve for S:**
Rearranging the equation to solve for `S`:
\[ S = \frac{1.1 x 10^{-12}}{0.01} = 1.1 x 10^{-10} \text{ mol/L} \]

### Final Answer:
The solubility of `Ag2CrO4` in a `0.1 M AgNO3` solution is:
\[ S = 1.1 x 10^{-10} \text{ mol/L} \]