The solubility of silver chromate in `0.01 M K_(2) CrO_(4)" is " 2 xx 10^(-8)` mol/1. The solubility product of silver chromate will be

Options

A. `8 xx 10^(-24)`
B. `16 xx 10^(-24)`
C. `1.6 xx 10^(-18)`
D. `16 xx 10^(-18)`

Answer

To find the solubility product (Ksp) of silver chromate (Ag2CrO4) given its solubility in a potassium chromate (K2CrO4) solution, we can follow these steps:

### Step 1: Write the dissociation equation for silver chromate
Silver chromate dissociates in water as follows:
\[ \text{Ag}_2\text{CrO}_4 (s) \rightleftharpoons 2 \text{Ag}^+ (aq) + \text{CrO}_4^{2-} (aq) \]

### Step 2: Define the solubility
Let the solubility of Ag2CrO4 in the solution be \( S \) (in mol/L). From the dissociation equation, we can see that:
- For every mole of Ag2CrO4 that dissolves, it produces 2 moles of Ag\(^+\) ions and 1 mole of CrO4\(^{2-}\) ions.
- Therefore, the concentration of Ag\(^+\) will be \( 2S \) and the concentration of CrO4\(^{2-}\) will be \( S \).

### Step 3: Consider the presence of K2CrO4
The solution also contains \( 0.01 \, M \) K2CrO4, which dissociates as follows:
\[ \text{K}_2\text{CrO}_4 (s) \rightleftharpoons 2 \text{K}^+ (aq) + \text{CrO}_4^{2-} (aq) \]
This means the concentration of CrO4\(^{2-}\) from K2CrO4 is \( 0.01 \, M \).

### Step 4: Calculate the total concentration of CrO4\(^{2-}\)
Since the solubility \( S \) of Ag2CrO4 is very small compared to \( 0.01 \, M \), we can approximate the total concentration of CrO4\(^{2-}\) as:
\[ \text{Total } [\text{CrO}_4^{2-}] = S + 0.01 \approx 0.01 \]

### Step 5: Write the expression for Ksp
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Ag}^+]^2 [\text{CrO}_4^{2-}] \]
Substituting the concentrations:
\[ K_{sp} = (2S)^2 \cdot (0.01) \]
\[ K_{sp} = 4S^2 \cdot 0.01 \]

### Step 6: Substitute the value of S
We are given that the solubility \( S \) of Ag2CrO4 in \( 0.01 \, M \) K2CrO4 is \( 2 \times 10^{-8} \, M \):
\[ K_{sp} = 4(2 \times 10^{-8})^2 \cdot 0.01 \]

### Step 7: Calculate Ksp
Calculating \( 4(2 \times 10^{-8})^2 \cdot 0.01 \):
1. Calculate \( (2 \times 10^{-8})^2 = 4 \times 10^{-16} \)
2. Then, \( 4 \times 4 \times 10^{-16} = 16 \times 10^{-16} \)
3. Finally, multiply by \( 0.01 \):
\[ K_{sp} = 16 \times 10^{-16} \times 0.01 = 16 \times 10^{-18} \]

Thus, the solubility product \( K_{sp} \) of silver chromate is:
\[ K_{sp} = 1.6 \times 10^{-17} \]

### Final Answer
The solubility product of silver chromate (Ag2CrO4) is \( 1.6 \times 10^{-17} \).