The solubility product of `Ag_2CrO_4` is `1.9xx10^(-12)` at 300 K. What is its solubility ?

Explanation

To find the solubility of \( \text{Ag}_2\text{CrO}_4 \) given its solubility product (\( K_{sp} \)), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of \( \text{Ag}_2\text{CrO}_4 \) 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 the solubility
Let the solubility of \( \text{Ag}_2\text{CrO}_4 \) be \( S \) moles per liter. Upon dissociation:
- The concentration of \( \text{Ag}^+ \) ions will be \( 2S \) (since there are 2 moles of \( \text{Ag}^+ \) for every mole of \( \text{Ag}_2\text{CrO}_4 \)).
- The concentration of \( \text{CrO}_4^{2-} \) ions will be \( S \).

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

### Step 4: Substitute the value of \( K_{sp} \)
We know that \( K_{sp} = 1.9 \times 10^{-12} \). Thus, we can set up the equation:
\[ 4S^3 = 1.9 \times 10^{-12} \]

### Step 5: Solve for \( S \)
To find \( S \), we rearrange the equation:
\[ S^3 = \frac{1.9 \times 10^{-12}}{4} \]
Calculating the right side:
\[ S^3 = 4.75 \times 10^{-13} \]
Now, take the cube root of both sides to find \( S \):
\[ S = \sqrt[3]{4.75 \times 10^{-13}} \]

### Step 6: Calculate \( S \)
Using a calculator:
\[ S \approx 7.8 \times 10^{-5} \text{ moles per liter} \]

### Final Answer
The solubility of \( \text{Ag}_2\text{CrO}_4 \) at 300 K is approximately:
\[ \boxed{7.8 \times 10^{-5} \text{ mol/L}} \]