To solve the problem of calculating the strength of the sodium thiosulfate (Na2S2O3) solution, we need to break down the reaction step by step. The key here is to understand the stoichiometry involved in the reaction between potassium iodate (KIO3), potassium iodide (KI), and sodium thiosulfate.
Understanding the Reaction
When KIO3 is treated with KI in an acidic medium, iodine (I2) is liberated. The reaction can be summarized as follows:
- KIO3 + 5 KI + 6 HCl → 3 I2 + 3 KCl + 3 H2O
From this equation, we see that 1 mole of KIO3 reacts with 5 moles of KI to produce 3 moles of I2. This is crucial for determining how much iodine is produced from the 10 moles of KIO3.
Calculating Moles of Iodine Produced
Given that we have 10 moles of KIO3, we can calculate the moles of I2 produced:
- 10 moles KIO3 × (3 moles I2 / 1 mole KIO3) = 30 moles I2
Reaction with Sodium Thiosulfate
The liberated iodine (I2) then reacts with sodium thiosulfate according to the following reaction:
- I2 + 2 Na2S2O3 → 2 NaI + Na2S4O6
This indicates that 1 mole of I2 reacts with 2 moles of Na2S2O3. Therefore, for 30 moles of I2, the moles of Na2S2O3 required would be:
- 30 moles I2 × (2 moles Na2S2O3 / 1 mole I2) = 60 moles Na2S2O3
Volume of Sodium Thiosulfate Solution
Now, we know that this 60 moles of Na2S2O3 is consumed in 20 liters of the sodium thiosulfate solution. To find the concentration in moles per liter (Molarity), we can use the formula:
- Molarity (M) = Moles of solute / Volume of solution in liters
Substituting the values:
- M = 60 moles / 20 L = 3 M
Calculating Strength in Gram per Liter
Next, we need to convert this molarity into grams per liter. The molar mass of sodium thiosulfate (Na2S2O3) is approximately 158.11 g/mol. Thus, the strength in grams per liter can be calculated as follows:
- Strength (g/L) = Molarity (mol/L) × Molar mass (g/mol)
Substituting the values:
- Strength = 3 mol/L × 158.11 g/mol = 474.33 g/L
Final Result
Therefore, the strength of the sodium thiosulfate solution is approximately 474.33 grams per liter. This calculation illustrates the stoichiometric relationships in the reactions and how to convert between moles and grams effectively.
