In basic medium `CrO_4^(2-)` oxidises `S_2O_3^(2-)` to form `SO_4^(2-)` and itself changes into `Cr(OH)_4^(-)`. The volume of 0.154 M `CrO_4^(2-)` required to react with 40 mL of 0.25 M `S_2O_3^(2-)` is `"___"` mL.

Answer

To solve the problem, we need to determine the volume of \( \text{CrO}_4^{2-} \) required to react with a given amount of \( \text{S}_2\text{O}_3^{2-} \). We will follow these steps:

### Step 1: Write the balanced redox reaction
The reaction can be represented as:
\[ \text{CrO}_4^{2-} + \text{S}_2\text{O}_3^{2-} \rightarrow \text{Cr(OH)}_4^{-} + \text{SO}_4^{2-} \]

### Step 2: Determine the oxidation states
- For \( \text{CrO}_4^{2-} \): Chromium is in the +6 oxidation state.
- For \( \text{Cr(OH)}_4^{-} \): Chromium is in the +3 oxidation state.
- For \( \text{S}_2\text{O}_3^{2-} \): Each sulfur is in the +2 oxidation state.
- For \( \text{SO}_4^{2-} \): Sulfur is in the +6 oxidation state.

### Step 3: Calculate the change in oxidation states
- Chromium changes from +6 to +3, which is a change of 3.
- Each sulfur changes from +2 to +6, which is a change of 4. Since there are 2 sulfur atoms, the total change for sulfur is \( 2 \times 4 = 8 \).

### Step 4: Determine the valency factors
- The valency factor for \( \text{CrO}_4^{2-} \) is 3 (since it gains 3 electrons).
- The valency factor for \( \text{S}_2\text{O}_3^{2-} \) is 8 (since it loses 8 electrons).

### Step 5: Calculate the milliequivalents
Using the formula for milliequivalents:
\[ \text{Milliequivalents} = \text{Molarity} \times \text{Volume} \times \text{Valency Factor} \]
For \( \text{CrO}_4^{2-} \):
\[ \text{Milliequivalents of } \text{CrO}_4^{2-} = 0.154 \, \text{M} \times V \, \text{mL} \times 3 \]

For \( \text{S}_2\text{O}_3^{2-} \):
\[ \text{Milliequivalents of } \text{S}_2\text{O}_3^{2-} = 0.25 \, \text{M} \times 40 \, \text{mL} \times 8 \]

### Step 6: Set the milliequivalents equal to each other
Equating the milliequivalents:
\[ 0.154 \times V \times 3 = 0.25 \times 40 \times 8 \]

### Step 7: Solve for \( V \)
Calculating the right side:
\[ 0.25 \times 40 \times 8 = 80 \, \text{meq} \]
Now, substituting back:
\[ 0.154 \times V \times 3 = 80 \]
\[ V = \frac{80}{0.154 \times 3} \]
\[ V = \frac{80}{0.462} \approx 173.16 \, \text{mL} \]

### Step 8: Round to the nearest integer
The final answer, rounded to the nearest integer, is:
\[ V \approx 173 \, \text{mL} \]