A sample containing `HAsO_(2)` (mol. Mass=108) and weighing 3.78 g is dissolved and diluted to 250 mL in a volumetric flask. A 50 mL sample (aliquot) is withdrawn with a pipet and titrated with 25 mL of 0.05 M solution of `I_(2)`. Calculate the percentage `HAsO_(2)` in the sample :

Options

A. 0.25
B. 0.2
C. 0.1
D. None of these

Answer

To solve the problem step by step, we will follow the stoichiometric calculations involved in determining the percentage of HAsO₂ in the sample.

### Step 1: Calculate the number of moles of I₂ used in the titration.

Given:
- Volume of I₂ solution = 25 mL = 0.025 L
- Molarity of I₂ solution = 0.05 M

Using the formula:
\[ \text{Moles of } I_2 = \text{Molarity} \times \text{Volume} = 0.05 \, \text{mol/L} \times 0.025 \, \text{L} = 0.00125 \, \text{mol} \]

### Step 2: Determine the number of equivalents of I₂.

The n-factor for I₂ in this reaction is 2 (since it can accept 2 electrons). Therefore, the number of equivalents of I₂ is:
\[ \text{Equivalents of } I_2 = \text{Moles of } I_2 \times \text{n-factor} = 0.00125 \, \text{mol} \times 2 = 0.0025 \, \text{equivalents} \]

### Step 3: Relate the equivalents of I₂ to the equivalents of HAsO₂.

From the stoichiometry of the reaction, 1 equivalent of HAsO₂ reacts with 1 equivalent of I₂. Thus, the equivalents of HAsO₂ will also be 0.0025.

### Step 4: Calculate the moles of HAsO₂ in the 50 mL aliquot.

Since the number of equivalents of HAsO₂ is equal to the number of equivalents of I₂, we have:
\[ \text{Equivalents of HAsO}_2 = 0.0025 \]
Using the n-factor for HAsO₂, which is 2 (as it can donate 2 electrons), we can find the moles of HAsO₂:
\[ \text{Moles of HAsO}_2 = \frac{\text{Equivalents}}{\text{n-factor}} = \frac{0.0025}{2} = 0.00125 \, \text{mol} \]

### Step 5: Calculate the mass of HAsO₂ in the 50 mL aliquot.

Using the molar mass of HAsO₂ (108 g/mol):
\[ \text{Mass of HAsO}_2 = \text{Moles} \times \text{Molar Mass} = 0.00125 \, \text{mol} \times 108 \, \text{g/mol} = 0.135 \, \text{g} \]

### Step 6: Calculate the mass of HAsO₂ in the original 250 mL solution.

Since the 50 mL aliquot is \(\frac{50}{250} = \frac{1}{5}\) of the total solution, the total mass of HAsO₂ in the 250 mL solution is:
\[ \text{Total mass of HAsO}_2 = 0.135 \, \text{g} \times 5 = 0.675 \, \text{g} \]

### Step 7: Calculate the percentage of HAsO₂ in the original sample.

Given the total mass of the sample is 3.78 g:
\[ \text{Percentage of HAsO}_2 = \left( \frac{\text{Mass of HAsO}_2}{\text{Total mass of sample}} \right) \times 100 = \left( \frac{0.675 \, \text{g}}{3.78 \, \text{g}} \right) \times 100 \approx 17.8\% \]

### Final Answer:
The percentage of HAsO₂ in the sample is approximately **17.8%**.

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