The solubility of solid silver chromate, `Ag_(2)Cro_(4)`, is determined in three solvents `K_(sp)` of `Ag_(2)CrO_(4) = 9 xx 10^(-12)`
I. pure water II. `0.1M gNO_(3)`
III. `0.1M Na_(2)CrO_(4)`
Predict the relative solubility of `Ag_(2)CrO_(4)` in the three solvents.

Options

A. `I =II =III`
B. `I lt II lt III`
C. `II = III lt I`
D. `II lt III lt I`

Explanation

To determine the relative solubility of silver chromate (`Ag2CrO4`) in three different solvents, we will analyze each case step by step based on the common ion effect and the solubility product constant (`Ksp`).

### Given:
- `Ksp` of `Ag2CrO4 = 9 x 10^(-12)`

### Step 1: Solubility in Pure Water
In pure water, there are no common ions present. The dissociation of `Ag2CrO4` can be represented as:
\[ Ag_2CrO_4 (s) \rightleftharpoons 2Ag^+ (aq) + CrO_4^{2-} (aq) \]

Let the solubility of `Ag2CrO4` in pure water be `s`. Therefore, the concentrations of the ions at equilibrium will be:
- \([Ag^+] = 2s\)
- \([CrO_4^{2-}] = s\)

Using the `Ksp` expression:
\[ Ksp = [Ag^+]^2 [CrO_4^{2-}] \]
Substituting the concentrations:
\[ 9 \times 10^{-12} = (2s)^2 (s) = 4s^3 \]

Now, solving for `s`:
\[ s^3 = \frac{9 \times 10^{-12}}{4} \]
\[ s^3 = 2.25 \times 10^{-12} \]
\[ s = \sqrt[3]{2.25 \times 10^{-12}} \approx 1.3 \times 10^{-4} \text{ M} \]

### Step 2: Solubility in `0.1 M AgNO3`
In `0.1 M AgNO3`, the `Ag^+` concentration is increased due to the presence of the common ion `Ag^+`. The dissociation of `Ag2CrO4` remains the same:
\[ Ag_2CrO_4 (s) \rightleftharpoons 2Ag^+ (aq) + CrO_4^{2-} (aq) \]

Let the solubility in this solution be `s`. The total concentration of `Ag^+` will be:
\[ [Ag^+] = 0.1 + 2s \approx 0.1 \text{ (since } s \text{ is small)} \]
And the concentration of `CrO_4^{2-}` will still be `s`.

Using the `Ksp` expression:
\[ Ksp = [Ag^+]^2 [CrO_4^{2-}] \]
Substituting the concentrations:
\[ 9 \times 10^{-12} = (0.1)^2 (s) \]
\[ 9 \times 10^{-12} = 0.01s \]
\[ s = \frac{9 \times 10^{-12}}{0.01} = 9 \times 10^{-10} \text{ M} \]

### Step 3: Solubility in `0.1 M Na2CrO4`
In `0.1 M Na2CrO4`, the `CrO4^{2-}` concentration is increased. The dissociation is:
\[ Ag_2CrO_4 (s) \rightleftharpoons 2Ag^+ (aq) + CrO_4^{2-} (aq) \]

Let the solubility be `s`. The total concentration of `CrO4^{2-}` will be:
\[ [CrO_4^{2-}] = 0.1 + s \approx 0.1 \text{ (since } s \text{ is small)} \]
And the concentration of `Ag^+` will be `2s`.

Using the `Ksp` expression:
\[ Ksp = [Ag^+]^2 [CrO_4^{2-}] \]
Substituting the concentrations:
\[ 9 \times 10^{-12} = (2s)^2 (0.1) \]
\[ 9 \times 10^{-12} = 0.1 \cdot 4s^2 \]
\[ s^2 = \frac{9 \times 10^{-12}}{0.4} = 2.25 \times 10^{-11} \]
\[ s = \sqrt{2.25 \times 10^{-11}} \approx 4.74 \times 10^{-6} \text{ M} \]

### Summary of Solubility in Each Solvent
1. **Pure Water**: \(s \approx 1.3 \times 10^{-4} \text{ M}\)
2. **0.1 M AgNO3**: \(s \approx 9 \times 10^{-10} \text{ M}\)
3. **0.1 M Na2CrO4**: \(s \approx 4.74 \times 10^{-6} \text{ M}\)

### Relative Solubility Order
- **Most soluble**: Pure Water > Na2CrO4 > AgNO3
- **Least soluble**: AgNO3

### Final Answer
The order of solubility is:
1. Pure Water (most soluble)
2. 0.1 M Na2CrO4
3. 0.1 M AgNO3 (least soluble)