Evaluate equivalent weight of `As_(2)O_(3)`:
`As_(2)O_(3) + 5H_(2)O to 2AsO_(4)^(3-) + 10 H^(+) + 4e^(-)`

Options

A. `E_(As_2O_3) = (M_(As_2O_3))/3`
B. `E_(As_2O_3) = (M_(As_2O_3))/4`
C. `E_(As_2O_3) = (M_(As_2O_3))/5`
D. `E_(As_2O_3) = (M_(As_2O_3))/2`

Answer

To evaluate the equivalent weight of \( \text{As}_2\text{O}_3 \), we will follow these steps:

### Step 1: Write the balanced redox reaction
The given reaction is:
\[ \text{As}_2\text{O}_3 + 5\text{H}_2\text{O} \rightarrow 2\text{AsO}_4^{3-} + 10\text{H}^+ + 4e^- \]

### Step 2: Determine the oxidation states
- In \( \text{As}_2\text{O}_3 \):
Let the oxidation state of As be \( x \). The equation can be set up as:
\[ 2x + 3(-2) = 0 \implies 2x - 6 = 0 \implies 2x = 6 \implies x = +3 \]
Thus, the oxidation state of As in \( \text{As}_2\text{O}_3 \) is +3.

- In \( \text{AsO}_4^{3-} \):
Let the oxidation state of As be \( x \). The equation can be set up as:
\[ x + 4(-2) = -3 \implies x - 8 = -3 \implies x = +5 \]
Thus, the oxidation state of As in \( \text{AsO}_4^{3-} \) is +5.

### Step 3: Calculate the change in oxidation state
The change in oxidation state for arsenic (As) is:
\[ \text{Change} = +5 - (+3) = 2 \]

### Step 4: Determine the number of electrons transferred
From the balanced reaction, we see that 4 electrons are transferred for 2 moles of \( \text{As}_2\text{O}_3 \). Therefore, the n-factor (number of electrons transferred per mole of \( \text{As}_2\text{O}_3 \)) is:
\[ \text{n-factor} = \frac{4 \text{ electrons}}{2 \text{ moles of } \text{As}_2\text{O}_3} = 2 \]

### Step 5: Calculate the molar mass of \( \text{As}_2\text{O}_3 \)
The molar mass of \( \text{As}_2\text{O}_3 \) can be calculated as follows:
- Atomic mass of As = 74.92 g/mol
- Atomic mass of O = 16.00 g/mol

Thus,
\[ \text{Molar mass of } \text{As}_2\text{O}_3 = 2(74.92) + 3(16.00) = 149.84 + 48.00 = 197.84 \text{ g/mol} \]

### Step 6: Calculate the equivalent weight
The equivalent weight of \( \text{As}_2\text{O}_3 \) is given by the formula:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{197.84 \text{ g/mol}}{2} = 98.92 \text{ g/equiv} \]

### Final Answer
The equivalent weight of \( \text{As}_2\text{O}_3 \) is \( 98.92 \text{ g/equiv} \).

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