`CrO_4^(-2)` (yellow) changes to `Cr_2O_7^(2-)` (orange) in pH = x and vice - vera in pH = y Hence , x and y are

Options

A. 6 , 8
B. 6 , 5
C. 8 , 6
D. 7 , 7

Answer

To solve the problem, we need to understand the relationship between the two chromium species, `CrO_4^(-2)` (yellow) and `Cr_2O_7^(2-)` (orange), and how they change with pH.

### Step-by-step Solution:

1. **Identify the Species and Their Colors**:
- `CrO_4^(-2)` is yellow.
- `Cr_2O_7^(2-)` is orange.

2. **Understand the pH Relationship**:
- The conversion from `CrO_4^(-2)` to `Cr_2O_7^(2-)` occurs when we add an acid (H⁺ ions), which lowers the pH. This means that at this point (let's call it pH = x), the solution is acidic.
- Conversely, the conversion from `Cr_2O_7^(2-)` back to `CrO_4^(-2)` occurs when we add a base (OH⁻ ions), which raises the pH. This means that at this point (let's call it pH = y), the solution is basic.

3. **Determine the pH Values**:
- Since `CrO_4^(-2)` is converted to `Cr_2O_7^(2-)` in an acidic medium, we can conclude that x (the pH at which the yellow solution turns orange) must be less than 7.
- Conversely, since `Cr_2O_7^(2-)` is converted back to `CrO_4^(-2)` in a basic medium, we can conclude that y (the pH at which the orange solution turns yellow) must be greater than 7.

4. **Choosing Specific pH Values**:
- A common choice for acidic pH (x) is 6, which is less than 7.
- A common choice for basic pH (y) is 8, which is greater than 7.

5. **Final Values**:
- Therefore, we have:
- x = 6 (pH at which `CrO_4^(-2)` changes to `Cr_2O_7^(2-)`)
- y = 8 (pH at which `Cr_2O_7^(2-)` changes back to `CrO_4^(-2)`)

### Conclusion:
The values of x and y are:
- x = 6
- y = 8