The solubility product of `BaCrO_(4)` is `2.4xx10^(-10)M^(2)`. The maximum concentration of `Ba(N0_(3))_(2)` possible without precipitation in a `6xx10^(-4)` M `K_(2)CrO_(4)` solution is :

Options

A. `4xx10^(-7)M`
B. `1.2xx10^(10)M`
C. `6xx10^(-4)M`
D. `3xx10^(-4)M`

Answer

To solve the problem, we need to determine the maximum concentration of \( Ba(NO_3)_2 \) that can be added to a \( 6 \times 10^{-4} \, M \) \( K_2CrO_4 \) solution without causing precipitation of \( BaCrO_4 \).

### Step-by-Step Solution:

1. **Write the dissociation equations**:
- The dissociation of \( K_2CrO_4 \) in solution:
\[ K_2CrO_4 \rightarrow 2K^+ + CrO_4^{2-} \]
- The dissociation of \( BaCrO_4 \):
\[ BaCrO_4 \rightarrow Ba^{2+} + CrO_4^{2-} \]

2. **Identify the solubility product (Ksp)**:
- Given \( K_{sp} \) for \( BaCrO_4 \) is \( 2.4 \times 10^{-10} \, M^2 \).

3. **Set up the expression for Ksp**:
- The solubility product expression is:
\[ K_{sp} = [Ba^{2+}][CrO_4^{2-}] \]
- From the question, we know that the concentration of \( CrO_4^{2-} \) from \( K_2CrO_4 \) is \( 6 \times 10^{-4} \, M \).

4. **Substitute known values into the Ksp expression**:
- Let \( [Ba^{2+}] = x \). Then,
\[ K_{sp} = x \cdot (6 \times 10^{-4}) = 2.4 \times 10^{-10} \]

5. **Solve for x**:
\[ x \cdot (6 \times 10^{-4}) = 2.4 \times 10^{-10} \]
\[ x = \frac{2.4 \times 10^{-10}}{6 \times 10^{-4}} = 4 \times 10^{-7} \, M \]
- Thus, the concentration of \( Ba^{2+} \) ions that can exist in solution without precipitating \( BaCrO_4 \) is \( 4 \times 10^{-7} \, M \).

6. **Relate \( Ba^{2+} \) concentration to \( Ba(NO_3)_2 \)**:
- The dissociation of \( Ba(NO_3)_2 \) is:
\[ Ba(NO_3)_2 \rightarrow Ba^{2+} + 2NO_3^{-} \]
- If the concentration of \( Ba(NO_3)_2 \) is \( C \), then \( [Ba^{2+}] = C \).

7. **Conclusion**:
- The maximum concentration of \( Ba(NO_3)_2 \) that can be added without causing precipitation is:
\[ C = 4 \times 10^{-7} \, M \]

### Final Answer:
The maximum concentration of \( Ba(NO_3)_2 \) possible without precipitation in a \( 6 \times 10^{-4} \, M \) \( K_2CrO_4 \) solution is \( 4 \times 10^{-7} \, M \).