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. `0.1A 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 solid silver chromate, \( \text{Ag}_2\text{CrO}_4 \), in three different solvents, we will analyze the solubility product constant \( K_{sp} \) and the effect of common ions in each solvent.

### Step 1: Write the dissociation equation and expression for \( K_{sp} \)
The dissociation of silver chromate in water can be represented as follows:
\[ \text{Ag}_2\text{CrO}_4 (s) \rightleftharpoons 2 \text{Ag}^+ (aq) + \text{CrO}_4^{2-} (aq) \]

The expression for the solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Ag}^+]^2 [\text{CrO}_4^{2-}] \]

### Step 2: Calculate solubility in pure water
Let \( S \) be the solubility of \( \text{Ag}_2\text{CrO}_4 \) in pure water. At equilibrium, we have:
- \( [\text{Ag}^+] = 2S \)
- \( [\text{CrO}_4^{2-}] = S \)

Substituting these into the \( K_{sp} \) expression:
\[ K_{sp} = (2S)^2 (S) = 4S^3 \]

Given \( K_{sp} = 9 \times 10^{-12} \):
\[ 9 \times 10^{-12} = 4S^3 \]
\[ S^3 = \frac{9 \times 10^{-12}}{4} \]
\[ S^3 = 2.25 \times 10^{-12} \]
\[ S = \sqrt[3]{2.25 \times 10^{-12}} \approx 1.31 \times 10^{-4} \, \text{mol/L} \]

### Step 3: Calculate solubility in \( 0.1 \, M \, \text{AgNO}_3 \)
In \( 0.1 \, M \, \text{AgNO}_3 \), the concentration of \( \text{Ag}^+ \) ions is \( 0.1 \, M \). Using the common ion effect:
\[ K_{sp} = [\text{Ag}^+]^2 [\text{CrO}_4^{2-}] \]
\[ 9 \times 10^{-12} = (0.1)^2 [\text{CrO}_4^{2-}] \]
\[ 9 \times 10^{-12} = 0.01 [\text{CrO}_4^{2-}] \]
\[ [\text{CrO}_4^{2-}] = \frac{9 \times 10^{-12}}{0.01} = 9 \times 10^{-10} \, \text{mol/L} \]

### Step 4: Calculate solubility in \( 0.1 \, M \, \text{Na}_2\text{CrO}_4 \)
In \( 0.1 \, M \, \text{Na}_2\text{CrO}_4 \), the concentration of \( \text{CrO}_4^{2-} \) ions is \( 0.1 \, M \):
\[ K_{sp} = [\text{Ag}^+]^2 [\text{CrO}_4^{2-}] \]
\[ 9 \times 10^{-12} = [\text{Ag}^+]^2 (0.1) \]
\[ [\text{Ag}^+]^2 = \frac{9 \times 10^{-12}}{0.1} = 9 \times 10^{-11} \]
\[ [\text{Ag}^+] = \sqrt{9 \times 10^{-11}} = 3 \times 10^{-6} \, \text{mol/L} \]

### Step 5: Compare solubilities
1. **In pure water**: \( S \approx 1.31 \times 10^{-4} \, \text{mol/L} \)
2. **In \( 0.1 \, M \, \text{AgNO}_3 \)**: \( S \approx 9 \times 10^{-10} \, \text{mol/L} \)
3. **In \( 0.1 \, M \, \text{Na}_2\text{CrO}_4 \)**: \( S \approx 3 \times 10^{-6} \, \text{mol/L} \)

### Conclusion
The relative solubility of \( \text{Ag}_2\text{CrO}_4 \) in the three solvents is:
1. **Pure Water**: Highest solubility
2. **\( 0.1 \, M \, \text{Na}_2\text{CrO}_4 \)**: Moderate solubility
3. **\( 0.1 \, M \, \text{AgNO}_3 \)**: Lowest solubility