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) in a solution of potassium chromate (K2CrO4), 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 (S)
Let the solubility of silver chromate in the solution be \( S \) mol/L. From the dissociation equation, we can see that:
- For every 1 mole of Ag2CrO4 that dissolves, it produces 2 moles of Ag⁺ and 1 mole of CrO4²⁻.
- Therefore, if the solubility is \( S \), the concentrations at equilibrium will be:
- \([Ag^+] = 2S\)
- \([CrO_4^{2-}] = S\)

### Step 3: Consider the common ion effect
Since we have a \( 0.01 \, M \) solution of K2CrO4, which dissociates completely to give \( 0.01 \, M \) of \( CrO_4^{2-} \), we need to account for this in our calculations. Thus, the total concentration of \( CrO_4^{2-} \) becomes:
\[ [CrO_4^{2-}] = S + 0.01 \]

### Step 4: Substitute the known solubility value
From the problem, we know that the solubility \( S \) of silver chromate in \( 0.01 \, M \) K2CrO4 is given as \( 2 \times 10^{-8} \, mol/L \).

### Step 5: Write the expression for Ksp
The solubility product \( Ksp \) is given by:
\[ Ksp = [Ag^+]^2 \cdot [CrO_4^{2-}] \]
Substituting the values:
\[ Ksp = (2S)^2 \cdot (S + 0.01) \]

### Step 6: Substitute \( S \) into the Ksp expression
Substituting \( S = 2 \times 10^{-8} \):
\[ Ksp = (2 \times (2 \times 10^{-8}))^2 \cdot ((2 \times 10^{-8}) + 0.01) \]
\[ Ksp = (4 \times 10^{-8})^2 \cdot (0.01 + 2 \times 10^{-8}) \]

### Step 7: Calculate \( Ksp \)
Calculating \( (4 \times 10^{-8})^2 \):
\[ (4 \times 10^{-8})^2 = 16 \times 10^{-16} \]

Now, since \( 2 \times 10^{-8} \) is negligible compared to \( 0.01 \):
\[ Ksp \approx 16 \times 10^{-16} \cdot 0.01 \]
\[ Ksp \approx 16 \times 10^{-18} \]

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