The `K_(sp)` for a sparingly soluble `Ag_2CrO_4` is
`4xx10^(-12)` . The molar solubility of the salt is

Options

A. `2.0 xx 10^(-6) mol L^(-1)`
B. `1.0 xx 10^(-4) mol L^(-1)`
C. `2.0 xx 10^(-12) mol L^(-1)`
D. `1.0 xx 10^(-15) mol L^(-1)`

Answer

To find the molar solubility of the sparingly soluble salt \( \text{Ag}_2\text{CrO}_4 \) given its solubility product constant \( K_{sp} = 4 \times 10^{-12} \), 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 Molar Solubility
Let \( S \) be the molar solubility of \( \text{Ag}_2\text{CrO}_4 \). When \( S \) moles of \( \text{Ag}_2\text{CrO}_4 \) dissolve, it produces:
- \( 2S \) moles of \( \text{Ag}^+ \)
- \( S \) moles of \( \text{CrO}_4^{2-} \)

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

### Step 4: Substitute the Known Value of \( K_{sp} \)
Now substitute the value of \( K_{sp} \):
\[ 4S^3 = 4 \times 10^{-12} \]

### Step 5: Solve for \( S^3 \)
Dividing both sides by 4:
\[ S^3 = 10^{-12} \]

### Step 6: Calculate \( S \)
Taking the cube root of both sides:
\[ S = (10^{-12})^{1/3} = 10^{-4} \]

### Conclusion
Thus, the molar solubility of \( \text{Ag}_2\text{CrO}_4 \) is:
\[ S = 1 \times 10^{-4} \, \text{mol/L} \]