The solubility product of `Ag CrO_(4)` is `32xx10^(-12).` What is the concentration of `CrO_(4)^(2-)` ions in that solution ?

Options

A. `2 xx 10^(-4) M`
B. `16 xx 10^(-4) M`
C. `8 xx 10^(-4) M`
D. `8 xx 10^(-8) M`

Answer

To find the concentration of \( \text{CrO}_4^{2-} \) ions in a solution 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 solubility and ion concentrations
Let the solubility of \( \text{Ag}_2\text{CrO}_4 \) be \( S \) mol/L. From the dissociation equation:
- The concentration of \( \text{Ag}^+ \) ions will be \( 2S \) (since 2 moles of \( \text{Ag}^+ \) are produced 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 the solubility product
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Ag}^+]^2 [\text{CrO}_4^{2-}] \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (2S)^2 (S) = 4S^2 \cdot S = 4S^3 \]

### Step 4: Substitute the given \( K_{sp} \) value
We know that \( K_{sp} = 32 \times 10^{-12} \):
\[ 4S^3 = 32 \times 10^{-12} \]

### Step 5: Solve for \( S^3 \)
Dividing both sides by 4:
\[ S^3 = \frac{32 \times 10^{-12}}{4} = 8 \times 10^{-12} \]

### Step 6: Calculate \( S \)
Now, take the cube root of both sides to find \( S \):
\[ S = (8 \times 10^{-12})^{1/3} \]
Calculating the cube root:
\[ S = 2 \times 10^{-4} \text{ mol/L} \]

### Step 7: Determine the concentration of \( \text{CrO}_4^{2-} \)
Since the concentration of \( \text{CrO}_4^{2-} \) ions is equal to \( S \):
\[ [\text{CrO}_4^{2-}] = S = 2 \times 10^{-4} \text{ mol/L} \]

### Final Answer
The concentration of \( \text{CrO}_4^{2-} \) ions in the solution is:
\[ \boxed{2 \times 10^{-4} \text{ mol/L}} \]