Question

Given that 0.1238 g of KIO₃ produces I2, and 41.27 mL of Na₂S₂O₃ is required, what is the concentration of Na₂S₂O₃?

a) 0.1238 mol/L
b) 0.0028 mol/L
c) 0.0048 mol/L
d) 0.04127 mol/L

Answer 1

To find the concentration of Na2S2O3, you need to calculate the number of moles of KIO3 that produce I2. Then, you can use stoichiometry to determine the moles of Na2S2O3 needed. Finally, divide the number of moles of I2 by the volume of Na2S2O3 to find its concentration.

Step-by-step explanation:

To find the concentration of Na2S2O3, we can use the stoichiometry of the given reaction. From the information provided, we know that 0.1238 g of KIO3 produces I2 and that 41.27 mL of Na2S2O3 is required. First, we need to convert the mass of KIO3 to moles using its molar mass. Then, we can use the stoichiometry to determine the moles of Na2S2O3 needed.

The molar mass of KIO3 is 214 g/mol. Therefore, the number of moles of KIO3 is 0.1238 g / 214 g/mol = 0.00058 mol. According to the balanced equation: 2KIO3 + 5Na2S2O3 + 3H2SO4 → 3I2 + 5Na2SO4 + 3H2O, we can see that 2 moles of KIO3 yield 3 moles of I2.

So, the number of moles of I2 produced is (0.00058 mol KIO3) x (3 mol I2 / 2 mol KIO3) = 0.00087 mol I2. Finally, we divide the number of moles of I2 by the volume of Na2S2O3 to find its concentration. The concentration of Na2S2O3 is 0.00087 mol I2 / 0.04127 L = 0.021 mol/L or 0.021 M.