Determine the solubility of silver chromate at 298 K given its `K_(sp)` value is `1.1xx10^(-12)`?

Options

A. `6.5xx10^(-5)`
B. `2.4xx10^(-2)`
C. `3.6xx10^(-3)`
D. `8.9xx10^(-4)`

Answer

To determine the solubility of silver chromate (Ag2CrO4) at 298 K given its \( K_{sp} \) value of \( 1.1 \times 10^{-12} \), we can follow these steps:

### Step 1: Write the dissociation equation
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 silver chromate be \( S \) mol/L. This means:
- The concentration of \( \text{CrO}_4^{2-} \) ions will be \( S \).
- The concentration of \( \text{Ag}^+ \) ions will be \( 2S \) (since two silver ions are produced for each formula unit of silver chromate that dissolves).

### Step 3: Write the expression for \( K_{sp} \)
The solubility product constant (\( K_{sp} \)) is given by:
\[ K_{sp} = [\text{Ag}^+]^2 [\text{CrO}_4^{2-}] \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (2S)^2 (S) = 4S^2 \cdot S = 4S^3 \]

### Step 4: Substitute the \( K_{sp} \) value
Given \( K_{sp} = 1.1 \times 10^{-12} \):
\[ 4S^3 = 1.1 \times 10^{-12} \]

### Step 5: Solve for \( S^3 \)
To find \( S^3 \), we rearrange the equation:
\[ S^3 = \frac{1.1 \times 10^{-12}}{4} \]
Calculating this gives:
\[ S^3 = 2.75 \times 10^{-13} \]

### Step 6: Solve for \( S \)
Now, take the cube root to find \( S \):
\[ S = \sqrt[3]{2.75 \times 10^{-13}} \]
Calculating the cube root:
\[ S \approx 6.53 \times 10^{-5} \, \text{mol/L} \]

### Final Result
Thus, the solubility of silver chromate at 298 K is:
\[ \boxed{6.53 \times 10^{-5} \, \text{mol/L}} \]