If the concentration of `CrO_4^(2-)` ions in a saturated solution of silver chromate is `2xx10^(-4)` , solubility product of silver chromate will be:

Options

A. `32xx10^(-12)`
B. `16xx10^(-12)`
C. `8xx10^(-12)`
D. `4xx10^(-8)`

Explanation

To find the solubility product (Ksp) of silver chromate (Ag2CrO4) given the concentration of CrO4^(2-) ions in a saturated solution, we can follow these steps:

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

### Step 2: Define solubility (S)
Let the solubility of silver chromate be \( S \). This means that in a saturated solution:
- The concentration of \( \text{CrO}_4^{2-} \) ions will be \( S \).
- The concentration of \( \text{Ag}^+ \) ions will be \( 2S \) (since 2 moles of Ag+ are produced for every mole of Ag2CrO4 that dissolves).

### Step 3: Use the given concentration
According to the problem, the concentration of \( \text{CrO}_4^{2-} \) ions is given as:
\[ [\text{CrO}_4^{2-}] = 2 \times 10^{-4} \, \text{mol/L} \]
Thus, we can set:
\[ S = 2 \times 10^{-4} \]

### Step 4: Calculate the concentration of \( \text{Ag}^+ \) ions
From the dissociation equation, we know that:
\[ [\text{Ag}^+] = 2S = 2 \times (2 \times 10^{-4}) = 4 \times 10^{-4} \, \text{mol/L} \]

### Step 5: Write the expression for Ksp
The solubility product \( K_{sp} \) is given by the formula:
\[ K_{sp} = [\text{Ag}^+]^2 \times [\text{CrO}_4^{2-}] \]

### Step 6: Substitute the concentrations into the Ksp expression
Substituting the values we found:
\[ K_{sp} = (4 \times 10^{-4})^2 \times (2 \times 10^{-4}) \]

### Step 7: Calculate \( Ksp \)
Calculating the values:
1. Calculate \( (4 \times 10^{-4})^2 \):
\[ (4 \times 10^{-4})^2 = 16 \times 10^{-8} \]
2. Now multiply by \( (2 \times 10^{-4}) \):
\[ K_{sp} = 16 \times 10^{-8} \times 2 \times 10^{-4} = 32 \times 10^{-12} \]

### Final Answer
Thus, the solubility product \( K_{sp} \) of silver chromate is:
\[ K_{sp} = 32 \times 10^{-12} \]

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