The solubility of silver chromate `Ag_2CrO_4` is `8.0 xx 10^(-5) " mol L"^(-1)`. Calculate the solubility product.

Explanation

To calculate the solubility product (Ksp) of silver chromate (Ag2CrO4) given its solubility, we can follow these steps:

### Step 1: Write the dissociation equation
Silver chromate (Ag2CrO4) 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 solubility
Let the solubility of Ag2CrO4 be \( S = 8.0 \times 10^{-5} \, \text{mol/L} \).

### Step 3: Determine the concentrations of ions at equilibrium
From the dissociation equation:
- For every 1 mole of Ag2CrO4 that dissolves, it produces 2 moles of Ag⁺ and 1 mole of CrO₄²⁻.
- Therefore, at equilibrium:
- The concentration of Ag⁺ will be \( 2S = 2 \times 8.0 \times 10^{-5} = 1.6 \times 10^{-4} \, \text{mol/L} \)
- The concentration of CrO₄²⁻ will be \( S = 8.0 \times 10^{-5} \, \text{mol/L} \)

### Step 4: Write the expression for Ksp
The solubility product (Ksp) expression for the dissociation of Ag2CrO4 is given by:
\[ K_{sp} = [\text{Ag}^+]^2 [\text{CrO}_4^{2-}] \]

### Step 5: Substitute the equilibrium concentrations into the Ksp expression
Substituting the values we found:
\[ K_{sp} = (1.6 \times 10^{-4})^2 \times (8.0 \times 10^{-5}) \]

### Step 6: Calculate Ksp
1. Calculate \( (1.6 \times 10^{-4})^2 \):
\[ (1.6 \times 10^{-4})^2 = 2.56 \times 10^{-8} \]
2. Now multiply by \( 8.0 \times 10^{-5} \):
\[ K_{sp} = 2.56 \times 10^{-8} \times 8.0 \times 10^{-5} \]
\[ K_{sp} = 2.048 \times 10^{-12} \]

### Final Answer
Thus, the solubility product \( K_{sp} \) of silver chromate (Ag2CrO4) is:
\[ K_{sp} = 2.048 \times 10^{-12} \]

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