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 Ag₂CrO₄ given its solubility product constant (Ksp = 4 x 10^(-12)), we can follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of Ag₂CrO₄ 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 the molar solubility of Ag₂CrO₄ be \( S \) mol/L. When Ag₂CrO₄ dissolves:
- The concentration of Ag⁺ ions will be \( 2S \) (since 2 moles of Ag⁺ are produced for every mole of Ag₂CrO₄ that dissolves).
- The concentration of CrO₄²⁻ ions will be \( S \).

### Step 3: Write the Ksp Expression
The solubility product constant (Ksp) expression for Ag₂CrO₄ is given by:
\[ K_{sp} = [\text{Ag}^+]^2 [\text{CrO}_4^{2-}] \]

Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (2S)^2 (S) \]
\[ K_{sp} = 4S^2 \cdot S = 4S^3 \]

### Step 4: Substitute the Ksp Value
Now, substitute the given Ksp value into the equation:
\[ 4S^3 = 4 \times 10^{-12} \]

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

Now take the cube root of both sides:
\[ S = (10^{-12})^{1/3} \]
\[ S = 10^{-4} \text{ mol/L} \]

### Conclusion
The molar solubility of Ag₂CrO₄ is:
\[ S = 1.0 \times 10^{-4} \text{ mol/L} \]