Chile saltpetre, a source of NaNO₃, also contains NaI₃. The NaI₃ is a source of I₂ produced as shown in the following equations:

• Step (I) : IO₃⁻ + 3HSO₃⁻ + I⁻ + 3H⁻ + 3SO₄²⁻
• Step (II) : 5I⁻ + IO₃⁻ + 6H⁻ → 3I₂(S) + 3H₂O

One litre sample of chile saltpetre solution containing 6.6 g NaI₃ is treated with NaHSO₃. Now an additional amount of the same solution is added to the reaction mixture to bring about the second titration. Calculate the weight of NaHSO₃ required in step (I) and what additional volume of chile saltpetre must be added in step (II) to bring out complete conversion of I⁻ to I₂.

• A. 100 mL
• B. 400 mL
• C. 200 mL
• D. None of these

To solve the problem, we need to calculate the weight of NaHSO₃ required in Step (I) and the additional volume of chile saltpetre needed in Step (II) for complete conversion of I⁻ to I₂.

Step 1: Calculate Moles of NaI₃

First, we find the number of moles of NaI₃ in the solution:

• Molar mass of NaI₃ = 22.99 (Na) + 3 × 126.90 (I) = 402.69 g/mol
• Moles of NaI₃ = 6.6 g / 402.69 g/mol ≈ 0.0164 moles

Step 2: Determine Moles of I⁻

Since each mole of NaI₃ produces 3 moles of I⁻:

• Moles of I⁻ = 0.0164 moles × 3 = 0.0492 moles

Step 3: Calculate NaHSO₃ Required in Step (I)

From the reaction in Step (I), we see that 1 mole of IO₃⁻ reacts with 3 moles of HSO₃⁻:

• Moles of NaHSO₃ needed = 3 × moles of I⁻ = 3 × 0.0492 = 0.1476 moles

The molar mass of NaHSO₃ is approximately 104.06 g/mol. Thus, the weight of NaHSO₃ required is:

• Weight of NaHSO₃ = 0.1476 moles × 104.06 g/mol ≈ 15.34 g

Step 4: Additional Volume of Chile Saltpetre for Step (II)

In Step (II), we need to convert I⁻ to I₂. The reaction shows that 5 moles of I⁻ react with 1 mole of IO₃⁻:

• For 0.0492 moles of I⁻, moles of IO₃⁻ required = 0.0492 / 5 = 0.00984 moles

Now, we need to find out how much NaI₃ is needed to provide this amount of IO₃⁻:

• Moles of NaI₃ needed = 0.00984 moles (since 1 mole of NaI₃ gives 1 mole of IO₃⁻)
• Weight of NaI₃ = 0.00984 moles × 402.69 g/mol ≈ 3.96 g

Now, to find the volume of the original solution needed to obtain 3.96 g of NaI₃:

• Volume = (3.96 g / 6.6 g) × 1000 mL ≈ 600 mL

Final Calculation

Since we already have 1 L (1000 mL) of the solution, we need to add:

• Additional volume = 600 mL - 1000 mL = -400 mL

This indicates that we need to add 400 mL of the original solution to achieve complete conversion of I⁻ to I₂.

Answer

The weight of NaHSO₃ required is approximately 15.34 g, and the additional volume of chile saltpetre needed is 400 mL.